2.1 Number fields background

Recall that in this paper, $K=\mathbb Q(i)$ is the Gaussian field. We denote by $U_{K}$ the group of units in ${\mathcal O}_{K}$ , so that $U_{K}=\{ \pm 1, \pm i\}$ . As it is well known that K has class number one, we do not distinguish n and $(n)$ when this causes no confusion from the context. We therefore write $\mu _{[i]}(n)$ to mean the Möbius function $\mu _{[i]}((n))$ . We use $\varpi $ to denote a prime (or prime ideal) in K and write $\Lambda (n)$ for the von Mangoldt function on $\mathcal {O}_{K}$ so that

$$ \begin{align*} \Lambda(n)=\begin{cases} \log N(\varpi) & n=\varpi^{k}, \; \varpi\text{ prime}, \; k \geq 1, \\ 0 & \text{otherwise}. \end{cases} \end{align*} $$

Let $({\cdot }/{n})_{4}$ stand for the quartic residue symbol on $\mathcal {O}_{K}$ . For a prime $\varpi \in \mathcal {O}_{K}$ with $N(\varpi ) \neq 2$ , the quartic symbol is defined for $a \in \mathcal {O}_{K}$ , $(a, \varpi )=1$ by $({a}/{\varpi })_{4} \equiv a^{(N(\varpi )-1)/4} ({\operatorname {mod}}\,\,\varpi )$ , with $({a}/{\varpi })_{4} \in \{ \pm 1, \pm i \}$ . When $\varpi | a$ , we define $({a}/{\varpi })_{4} =0$ . Then the quartic symbol can be extended to any composite n with $(N(n), 2)=1$ multiplicatively. We further define $({\cdot }/{n})=({\cdot }/{n})^{2}_{4}$ to be the quadratic residue symbol for these n.

We say that an element $c \in \mathcal {O}_{K}$ (or the ideal $(c)$ ) is odd if $(c, 1+i)=1$ . Note that in $\mathcal {O}_{K}$ , every odd ideal has a unique generator congruent to $1$ modulo $(1+i)^{3}$ . Such a generator is called primary. For two primary integers $m, n \in \mathcal {O}_{K}$ , we note that the quadratic reciprocity law (see [Reference Gao and Zhao15, formula (2.1)]) gives

(2-1) $$ \begin{align} \bigg(\frac{m}{n}\bigg) = \bigg(\frac{n}{m}\bigg). \end{align} $$

Let $\chi $ denote a Hecke character of K. We say that $\chi $ is of trivial infinite type if its component at the infinite place of K is trivial. In particular, $\chi _{c}$ defined earlier is a Hecke character of trivial infinite type. It is further shown in [Reference Gao and Zhao15, Section 2.1] that when c is square-free and coprime to $1+i$ , $\chi _{i(1+i)^{5}c}$ defines a primitive Hecke character $\text {mod} ((1+i)^{5}c)$ of trivial infinite type.

For any primitive Hecke character $\chi ({\operatorname {mod}}\,\,m)$ of trivial infinite type, let

$$ \begin{align*} \Lambda(s, \chi) = (|D_{K}|N(m))^{s/2}(2\pi)^{-s}\Gamma(s)L(s, \chi), \end{align*} $$

where $L(s, \chi )$ is the L-function attached to $\chi $ and $D_{K}=-4$ is the discriminant of K. In particular, we use $\zeta _{K}(s)$ to denote the Dedekind zeta function of K.

It was shown by Hecke that $\Lambda (s, \chi )$ is an entire function and satisfies the functional equation [Reference Iwaniec and Kowalski22, Theorem 3.8]

$$ \begin{align*} \Lambda(s, \chi) = W(\chi)(N(m))^{-1/2}\Lambda(1-s, \overline{\chi}), \end{align*} $$

where $|W(\chi )|=(N(m))^{1/2}$ .

2.2 Poisson summation

For any $r,n \in \mathcal {O}_{K}$ with n odd, we define the Gauss sum $g(r,n)$ as

$$ \begin{align*} g(r,n) = \sum_{x \bmod{n}} \bigg(\frac{x}{n}\bigg)\, \widetilde{e}\,\bigg(\frac{rx}{n}\bigg) \end{align*} $$

where $ \widetilde {e}(z) =\exp ( 2\pi i (( {z}/{2i}) - ({\overline {z}}/{2i}) ))$ . It is shown in [Reference Gao and Zhao15, Lemma 2.2] that, for a primary prime $\varpi $ ,

$$ \begin{align*} g(r, \varpi)=\bigg(\frac{ir}{\varpi}\bigg)N(\varpi)^{1/2}. \end{align*} $$

We quote the following Poisson summation formula from [Reference Gao and Zhao15, Lemma 2.7].

Lemma 2.1. Let $n \in \mathcal {O}_{K}$ be primary and $\chi $ a quadratic character $ ({\operatorname {mod}}\,\,n)$ of trivial infinite type. For any Schwartz class function W, we have

(2-2) $$ \begin{align} \sum_{m \in \mathcal{O}_{K}}\chi(m)W\bigg(\frac {N(m)}{X}\bigg)=\frac {X}{N(n)}\sum_{k \in \mathcal{O}_{K}}g(k,n)\widetilde{W}\bigg(\sqrt{\frac {N(k)X}{N(n)}}\bigg) \end{align} $$

and

(2-3) $$ \begin{align} \sum_{m \in \mathcal{O}_{K}} W\bigg(\frac {N(m)}{X}\bigg)=X \sum_{k \in \mathcal{O}_{K}}\widetilde{W}(\sqrt{N(k)X}\,), \end{align} $$

where

(2-4) $$ \begin{align} \widetilde{W}(t) &=\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}W(N(x+yi))\, \widetilde{e}\,(- t(x+yi))\,{d} x \,{d} y, \quad t \geq 0. \end{align} $$

We include here our conventions for various transforms used in this paper. For any function W, we write $\widehat {W}$ for the Fourier transform of W and recall that its Mellin transform $ \mathcal M w$ is defined in (1-5). We note that, for any Schwartz class function W and any integer $E \geq 0$ , integration by parts $E+1$ times yields that for $\mathrm {Re}(s)>0$ ,

(2-5) $$ \begin{align} \mathcal M {W}(s) \ll \frac{1}{|s| (1+|s|)^{E}}. \end{align} $$

Furthermore, we set

When $W(t)$ is a real smooth function, one follows the arguments that lead to the bounds given in [Reference Gao and Zhao15, formula (2.12)] to get that both $\widetilde {W}$ and are real and

(2-6)

for all integers $\mu \geq 0$ , $j \geq 1$ and all real t.

2.3 Some consequences of the GRH

In this section we state a few results that are derived by using the GRH. The first one is for sums over primes.

Lemma 2.2. Suppose that the GRH is true. For any Hecke character $\chi ({\operatorname {mod}}\,\,m)$ of trivial infinite type, we have for $X \geq 1$ ,

(2-7) $$ \begin{align} \sum_{\substack {N(\varpi) \leq X\\ \varpi \equiv 1 \bmod {(1+i)^{3}}}} \chi (\varpi) \log N(\varpi) =\delta_{\chi} X+ O( X^{1/2} \log^{2} (2X) \log N(m) ), \end{align} $$

where $\delta _{\chi }=1$ if $\chi $ is principal and $\delta _{\chi }=0$ otherwise. Moreover, we have

(2-8) $$ \begin{align} \sum_{\substack{ N(\varpi) \leq X \\ \varpi \equiv 1 \bmod {(1+i)^{3}}}} \frac {\log N(\varpi)}{N(\varpi) }= \log X+O(1). \end{align} $$

Our next two lemmas provide estimations on certain weighted quadratic character sums. The following one is a generalization of [Reference Fiorilli, Parks and Södergren9, Lemma 2.10].

Lemma 2.3. Suppose that the GRH is true. For any even, nonzero and nonnegative Schwartz function w, we have for any primary $n \in \mathcal {O}_{K}$ and $\epsilon>0$ ,

$$ \begin{align*} \underset{(c, 1+i) =1 }{\sum \nolimits^{*}} w\bigg( \frac {N(c)}X \bigg) \bigg(\frac {i(1+i)^{5}c}{n} \bigg)&= \delta_{\chi_{n} } \frac{\pi X}{3\zeta_{K}(2)} \widehat w(0) \prod_{\varpi\mid n} \bigg(1+\frac 1{N(\varpi)} \bigg)^{-1} \\[3pt] &\quad +O(N(n)^{3(1-\delta_{\chi_{n} })/8 +\varepsilon}X^{1/4+\varepsilon}). \end{align*} $$

Here we recall that $\sum ^{*}$ denotes the sum over square-free elements in $\mathcal {O}_{K}$ .

Proof. Since each c coprime to $1+i$ can be uniquely written as the product of a unit and a primary element, it follows that

$$ \begin{align*} &\underset{(c, 1+i) =1 }{\sum \nolimits^{*}} w\bigg( \frac {N(c)}X \bigg) \bigg(\frac {(1+i)^{5}c}{n} \bigg) \\[1pt] &\quad= \bigg(1+ \bigg(\frac{-1}{n}\bigg)+\bigg(\frac{i}{n}\bigg)+\bigg(\frac{i^{3}}{n}\bigg) \bigg) \sideset{}{^{*}}\sum_{c \equiv 1 \ {\operatorname{mod}}\ (1+i)^{3} } w\bigg( \frac {N(c)}X \bigg) \bigg(\frac {(1+i)c}{n}\bigg) \nonumber \\&\quad= 2\bigg(1+ \bigg(\frac{i}{n}\bigg) \bigg)\bigg(\frac {1+i}{n} \bigg) \sideset{}{^{*}}\sum_{c \equiv 1 \ {\operatorname{mod}}\ (1+i)^{3} } w\bigg( \frac {N(c)}X \bigg) \bigg(\frac {c}{n} \bigg). \end{align*} $$

Note further that the quadratic reciprocity law (2-1) allows us to write $\chi _{n}(c)$ for $({c}/{n})$ , and $\chi _{n}$ is a Hecke character. We then apply the Mellin inversion formula and get

$$ \begin{align*} &\sideset{}{^{*}}\sum_{c \equiv 1 \ {\operatorname{mod}}\ (1+i)^{3} } w\bigg( \frac {N(c)}X \bigg)\,\chi_{n}(c) \\[3pt] &\qquad = \frac 1{2\pi i } \int_{(2)} \sum_{\substack{c \equiv 1 \ {\operatorname{mod}}\ (1+i)^{3}}} \frac{\mu_{[i]}^{2}(c) \chi_{n}(c) }{N(c)^{s}} X^{s} \mathcal M w(s)\, {d} s \\[3pt] &\qquad= \frac 1{2\pi i } \int_{(2)} \prod_{\substack{\varpi \equiv 1 \ {\operatorname{mod}}\ (1+i)^{3} \\ (\varpi, 1+i)=1}} \bigg(1+ \frac{\chi_{n}(\varpi)}{N(\varpi)^{s}} \bigg) \mathcal M w (s)X^{s}\, {d} s \\[3pt] &\qquad = \frac 1{2\pi i } \int_{(2)} \frac{L(s, \chi_{n})}{L(2s, \chi^{2}_{n})}\bigg(1+ \frac{\chi_{n}(1+i)}{N(1+i)^{s}} \bigg)^{-1} \mathcal M w (s)X^{s}\, {d} s. \end{align*} $$

Here and henceforth, we write $\int _{(c)}$ for the integral over the vertical line with $\mathrm {Re}(s) = c$ .

We shift the line of integration to $\mathrm {Re}(s)=1/4+\varepsilon $ and we encounter a pole at $s=1$ only when $\chi _{n}$ is a principal character. In that case, the residue is easily seen (recall that the residue of $\zeta _{K}(s)$ at $s = 1$ is $\pi /4$ ) to be

$$ \begin{align*} &\frac {\pi}{6}\zeta^{-1}_{K}(2) \prod_{\varpi | n} \bigg( 1+\frac 1{N(\varpi)} \bigg)^{-1} \mathcal M w (1) X \\[3pt] &\quad = \frac {\pi}{12}\zeta^{-1}_{K}(2) \prod_{\varpi | n} \bigg( 1+\frac 1{N(\varpi)} \bigg)^{-1} \widehat w (0) X, \end{align*} $$

by noting that $\mathcal M w (1)=\widehat w (0)/2 $ when w is even. The remaining integral over the line $\mathrm {Re}(s)=1/4+\varepsilon $ can be estimated by using (2-5) for a suitable E and the bound

(2-9) $$ \begin{align} L^{-1}(2s, \chi_{n}) \ll (N(n)(1+\mathrm{Im}(s))^{\varepsilon}, \end{align} $$

assuming the GRH, which follows from [Reference Iwaniec and Kowalski22, Theorem 5.19]. This gives the result when $\chi _{n}$ is a principal character.

When $\chi _{n}$ is not principal, we apply the convexity bound [Reference Iwaniec and Kowalski22, formula (5.20)] for L-functions attached to nonprincipal characters, such that

$$ \begin{align*} L(s,\chi_{n}) \ll_{\varepsilon} (N(n)(|s|+1)^{2})^{(1-\mathrm{Re}(s))/2+\varepsilon/2} \quad (0\leq \mathrm{Re}(s) \leq 1). \end{align*} $$

Combining this with (2-9) allows us to readily deduce the assertion of the lemma for $\chi _{n}$ being nonprincipal. This completes the proof.

By taking $n=1$ in Lemma 2.3, we immediately obtain that

(2-10) $$ \begin{align} W(X)= \frac{\pi X}{3\zeta_{K}(2)} \widehat w(0) +O ( X^{1/4+\varepsilon} ). \end{align} $$

Lemma 2.4. Suppose that the GRH is true. For any even, nonzero and nonnegative Schwartz function w, we have

$$ \begin{align*} &\frac {1}{ W(X)}\underset{(c, 1+i) =1 }{\sum \nolimits^{*}} w\bigg( \frac {N(c)}X \bigg) \log N(c) \\[3pt] &\quad = \log X + \frac 2{\widehat w(0)}\int_{0}^{\infty} w(x) \log x \, {d} x +O ( X^{-1/2 +\varepsilon} ). \end{align*} $$

Proof. We have

$$ \begin{align*} \underset{(c, 1+i) =1 }{\sum \nolimits^{*}} & w\bigg( \frac {N(c)}X \bigg) \log N(c) \\[3pt] =& -\frac 4{2\pi i}\int_{(2)} \frac{{d}}{{d} s} \bigg( \sum_{ \substack{c \equiv 1 \ {\operatorname{mod}}\ (1+i)^{3}}} \frac{\mu^{2}_{[i]}(c)}{N(c)^{s}} \bigg) \mathcal M w(s)X^{s} \, {d} s \\[3pt] =& -\frac 4{2\pi i}\int_{(2)} \frac{{d}}{{d} s} \bigg( \frac{\zeta_{K}(s)}{\zeta_{K}(2s) ( 1+2^{-s} )} \bigg) \mathcal M w(s)X^{s} \, {d} s. \end{align*} $$

We shift the contour of integration to the line $\mathrm {Re}(s)=1/4+\varepsilon $ . Note that under the GRH, the only poles of the function

$$ \begin{align*} \frac{{d}}{{d} s} & \bigg( \frac{\zeta_{K}(s)}{\zeta_{K}(2s) ( 1+2^{-s} )} \bigg) \\[3pt] & = \frac{\zeta^{\prime}_{K}(s)}{\zeta_{K}(2s) ( 1+2^{-s} )}- \frac{2\zeta^{\prime}_{K}(2s)\zeta_{K}(s)}{\zeta^{2}_{K}(2s) ( 1+2^{-s} )}+\frac{\log 2 \cdot \zeta_{K}(s)}{2^{s} \zeta_{K}(2s) ( 1+2^{-s} )^{2} } \end{align*} $$

in the region $1/4+\varepsilon \leq \mathrm {Re}(s)\leq 2$ are at $s=1$ and $s=1/2$ . Only the contribution of the residue at $s=1$ is much greater than X. It is then easy to compute the contribution of the residues to be

$$ \begin{align*} \frac{2\pi }{3\zeta_{K}(2)} \mathcal M w(1)X \log X+\frac{2\pi }{3\zeta_{K}(2)} (\mathcal M w)^{\prime}(1)X+O(X^{1/2+\varepsilon}). \end{align*} $$

The assertion of the lemma follows from this and (2-10), by noting that $\mathcal M w(1)=\widehat {w}(0)/2$ .

2.4 The explicit formula

Let f be an even, positive Schwartz function whose Fourier transform $\hat {f}$ is a smooth function with compact support. Let $\chi $ be a primitive Hecke character $\chi ({\operatorname {mod}}\,\,m)$ of trivial infinite type. In this section we derive an explicit formula which allows us to convert the evaluation of f at the nontrivial zeros of $L(s,\chi )$ to a sum over powers of prime ideals. Note that the nontrivial zeros of $L(s,\chi )$ are precisely those of the corresponding $\Lambda (s, \chi )$ . For some $c>1$ , consider the integral

$$ \begin{align*} \frac 1{2 \pi i}\int_{(c)}\frac {\Lambda^{\prime}}{\Lambda}(s, \chi)f \bigg( \frac {s-1/2}{2 \pi i} \bigg)\, {d} s. \end{align*} $$

By moving the line of integration to $1-c$ , we obtain

$$ \begin{align*} &\frac 1{2 \pi i}\int_{(c)} \frac {\Lambda^{\prime}}{\Lambda}(s, \chi)f \bigg( \frac {s-1/2}{2 \pi i} \bigg)\, {d} s \\[3pt] &\quad = \sum_{j}f \bigg( \frac {\gamma_{\chi,j}}{2 \pi} \bigg)+ \frac 1{2 \pi i}\int_{(1-c)}\frac {\Lambda^{\prime}}{\Lambda}(s, \chi)f \bigg( \frac {s-1/2}{2 \pi i} \bigg)\, {d} s. \end{align*} $$

Now the functional equation $\Lambda (s, \chi )=W(\chi )(N(m))^{-1/2}\Lambda (1-s, \overline {\chi })$ implies

$$ \begin{align*} \frac {\Lambda^{\prime}}{\Lambda}(s, \chi)=-\frac {\Lambda^{\prime}}{\Lambda}(1-s, \overline{\chi}). \end{align*} $$

It follows that

$$ \begin{align*} \sum_{j}f \bigg( \frac {\gamma_{\chi,j}}{2 \pi} \bigg)&=\frac 1{2 \pi i}\int_{(c)}\frac {\Lambda^{\prime}}{\Lambda} (s, \chi) f \bigg( \frac {s-1/2}{2 \pi i} \bigg)\, {d} s \\[3pt] &\quad +\frac 1{2 \pi i}\int_{(c)}\frac {\Lambda^{\prime}}{\Lambda}(s, \overline{\chi}) f\bigg( \frac {1/2-s}{2 \pi i} \bigg)\, {d} s. \end{align*} $$

Using

$$ \begin{align*} \frac {\Lambda^{\prime}}{\Lambda}(s, \chi)=\frac 1{2}\log \frac {|D_{K}|N(m)}{(2\pi)^{2}}+\frac {\Gamma^{\prime}}{\Gamma}(s)+\frac {L^{\prime}}{L}(s, \chi), \end{align*} $$

we obtain that

$$ \begin{align*} &\frac 1{2 \pi i}\int_{(c)}\frac {\Lambda^{\prime}}{\Lambda}(s, \chi) f \bigg( \frac {s-1/2}{2 \pi i} \bigg)\, {d} s +\frac 1{2 \pi i}\int_{(c)}\frac {\Lambda^{\prime}}{\Lambda}(s, \overline{\chi}) f\bigg( \frac {1/2-s}{2 \pi i} \bigg) \,{d} s =T_{1}+T_{2}, \end{align*} $$

where

$$ \begin{align*} T_{1} &= \frac 1{2 \pi i}\int_{(c)} \bigg( \frac 1{2}\log \frac {|D_{K}|N(m)}{(2\pi)^{2}}+\frac {\Gamma^{\prime}}{\Gamma}(s)\bigg) f \bigg( \frac {s-1/2}{2 \pi i} \bigg) \,{d} s \\[3pt] &\quad +\frac 1{2 \pi i}\int_{(c)}\bigg( \frac 1{2}\log \frac {|D_{K}|N(m)}{(2\pi)^{2}}+\frac {\Gamma^{\prime}}{\Gamma}(s)\bigg) f \bigg( \frac {1/2-s}{2 \pi i} \bigg) \,{d} s \end{align*} $$

and

$$ \begin{align*} T_{2} = \frac 1{2 \pi i}\int_{(c)}\frac {L^{\prime}}{L}(s, \chi)f \bigg( \frac {s-1/2}{2 \pi i} \bigg) \,{d} s+\frac 1{2 \pi i}\int_{(c)}\frac {L^{\prime}}{L}(s, \overline{\chi})f \bigg( \frac {1/2-s}{2 \pi i} \bigg) \,{d} s. \end{align*} $$

For $T_{1}$ , we move the line of integration to $1/2$ and apply the change of variables $s=1/2+2\pi i t$ . In so doing, we obtain

(2-11) $$ \begin{align} T_{1} &=\int^{\infty}_{-\infty}\bigg( \log \frac {|D_{K}|N(m)}{(2\pi)^{2}}+ \frac {\Gamma^{\prime}}{\Gamma} \bigg( \frac{1}{2}+2\pi i t \bigg)+ \frac {\Gamma^{\prime}}{\Gamma} \bigg( \frac{1}{2}-2\pi i t \bigg) \bigg) f(t) \,{d} t \nonumber\\[3pt] &=\int^{\infty}_{-\infty}\bigg( \log\frac {|D_{K}|N(m)}{(2\pi)^{2}}+ 2\frac {\Gamma^{\prime}}{\Gamma} \bigg( \frac 12 \bigg) \bigg )f(t) \,{d} t \nonumber\\[3pt] & \quad+\int^{\infty}_{0}\frac {e^{-t/2}}{1-e^{-t}} ( 2\hat{f}(0)-\hat{f}(t)-\hat{f}(-t) ) \,{d} t, \end{align} $$

where the second equality above follows from [Reference Montgomery and Vaughan33, Lemma 12.14].

We express $T_{2}$ as

$$ \begin{align*} T_{2}&=-\sum_{\substack{(n) \\ n \in \mathcal{O}_{K}}} \chi(n)\Lambda(n) \frac 1{2 \pi i}\int_{(c)}\frac 1{N(n)^{s}}f \bigg( \frac {s-1/2}{2 \pi i} \bigg)\,{d} s \\[3pt] &\quad -\sum_{\substack{(n) \\ n \in \mathcal{O}_{K}}} \overline{\chi}(n) \Lambda(n)\frac 1{2 \pi i}\int_{(c)}\frac 1{N(n)^{s}}f \bigg( \frac {1/2-s}{2 \pi i} \bigg) \,{d} s. \end{align*} $$

Moving the lines of integration for $I_{2}$ to $\mathrm {Re}(s)=1/2$ and setting $s=1/2+2\pi i t$ , the integrations become

$$ \begin{align*} \int^{\infty}_{-\infty}\frac 1{N(n)^{1/2 \pm 2 \pi i t}}f(t)\,dt=\frac 1{\sqrt{N(n)}}\hat{f}(\mp \log N(n)). \end{align*} $$

Thus,

(2-12) $$ \begin{align} T_{2}=-\sum_{\substack{(n) \\ n \in \mathcal{O}_{K}}} \frac {\chi(n)\Lambda(n)}{\sqrt{N(n)}}\hat{f}(-\log N(n))-\sum_{\substack{(n) \\ n \in \mathcal{O}_{K}}} \frac {\overline{\chi}(n)\Lambda(n)}{\sqrt{N(n)}}\hat{f}( \log N(n)). \end{align} $$

We then derive from (2-11) and (2-12) that

$$ \begin{align*} \sum_{j}f \bigg( \frac {\gamma_{\chi,j}}{2 \pi} \bigg) &= \int^{\infty}_{-\infty}\bigg( \log\frac {|D_{K}|N(m)}{(2\pi)^{2}}+ 2\frac {\Gamma^{\prime}}{\Gamma}\bigg( \frac 12 \bigg) \bigg)f(t) \,{d} t \\[3pt] &\quad +\int^{\infty}_{0}\frac {e^{-t/2}}{1-e^{-t}} ( 2\hat{f}(0)-\hat{f}(t)-\hat{f}(- t) ) \,{d} t \\[3pt] &\quad -\sum_{\substack{(n) \\ n \in \mathcal{O}_{K}}} \frac {\chi(n)\Lambda(n)}{\sqrt{N(n)}}\hat{f}(-\log N(n))-\sum_{\substack{(n) \\ n \in \mathcal{O}_{K}}} \frac {\overline{\chi}(n)\Lambda(n)}{\sqrt{N(n)}}\hat{f}( \log N(n)). \end{align*} $$

Now an easy computation in Fourier transforms gives

$$ \begin{align*} \sum_{j}f \bigg( \frac {\log X \gamma_{\chi,j}}{2 \pi} \bigg) &= \frac 1{\log X} \int^{\infty}_{-\infty}\bigg( \log\frac {|D_{K}|N(m)}{(2\pi)^{2}}+ 2\frac {\Gamma^{\prime}}{\Gamma} \bigg( \frac 12 \bigg) \bigg )f(t) \,{d} t \\[1pt] &\quad +\frac 1{\log X}\int^{\infty}_{0}\frac {e^{-t/2}}{1-e^{-t}} \bigg( 2\hat{f}(0)-\hat{f} \bigg( \frac {t}{\log X} \bigg)-\hat{f}\bigg( -\frac {t}{\log X} \bigg) \bigg) \,{d} t \\[1pt] &\quad -\frac 1{\log X}\sum_{\substack{(n) \\ n \in \mathcal{O}_{K}}} \frac {\chi(n)\Lambda(n)}{\sqrt{N(n)}}\hat{f} \bigg( -\frac {\log N(n)}{\log X} \bigg) \\[1pt] &\quad -\frac 1{\log X}\sum_{\substack{(n) \\ n \in \mathcal{O}_{K}}} \frac {\overline{\chi}(n)\Lambda(n)}{\sqrt{N(n)}}\hat{f} \bigg( \frac {\log N(n)}{\log X} \bigg). \end{align*} $$

Recall that $\mathcal {L}=\log X$ and note that as f is even, so is $\hat {f}$ . Thus, when $\chi $ is a quadratic Hecke character, we can simplify the above expression as

(2-13) $$ \begin{align} \sum_{j}f\bigg( \frac {\mathcal{L} \gamma_{\chi,j}}{2 \pi} \bigg) &= \frac 1{\mathcal{L}}\bigg( \log\frac {|D_{K}|N(m)}{(2\pi)^{2}}+ 2\frac {\Gamma^{\prime}}{\Gamma}\bigg( \frac 12 \bigg) \bigg )\hat{f}(0) \nonumber\\[1pt] &\quad +\frac 2{\mathcal{L}}\int^{\infty}_{0}\frac {e^{-t/2}}{1-e^{-t}}\bigg( \hat{f}(0)-\hat{f}\bigg(\frac {t}{\mathcal{L}}\bigg) \bigg) \,{d} t \nonumber\\[1pt] &\quad -\frac 2{\mathcal{L}}\sum_{\substack{(n) \\ n \in \mathcal{O}_{K}}} \frac {\chi(n)\Lambda(n)}{\sqrt{N(n)}}\hat{f} \bigg( \frac {\log N(n)}{\mathcal{L}} \bigg). \end{align} $$

Recall that every odd prime $\varpi \in \mathcal {O}_{K}$ has a primary generator. In our paper, we work explicitly with the Hecke characters $\chi _{i(1+i)^{5}c}$ for odd, square-free c. Our choice for such characters is inspired by the Dirichlet characters $\chi _{8d}$ for odd, square-free rational integers d considered by Soundararajan [Reference Soundararajan40] in his work on nonvanishing of quadratic Dirichlet L-functions at the central value. The advantage of using the characters $\chi _{i(1+i)^{5}c}$ is that, besides their primitivity, the presence of the factors $i(1+i)^{3}$ makes the resulting expression much neater after applying the Poisson summation formula given in Lemma 2.1. We now apply the formula given in (2-13) to the special case $\chi _{i(1+i)^{5}c}$ for odd, square-free c to arrive at the following lemma.

Lemma 2.5 (Explicit formula)

Let $\phi (x)$ be an even Schwartz function whose Fourier transform $\hat {\phi }(u)$ has compact support. Let c be an odd and square-free element of $\mathcal{O}_K$ satisfying $(c, 1+i)=1$ .We have

$$ \begin{align*} S(\kern1.5pt\chi_{i(1+i)^{5}c}, \phi) &= \frac 1{\mathcal{L}} \bigg( \log\frac {32N(c)}{\pi^{2}}+ 2\frac {\Gamma^{\prime}}{\Gamma} \bigg( \frac 12 \bigg) \bigg )\hat{\phi}(0) \\[1pt] &\quad +\frac 2{\mathcal{\mathcal{L}}}\int^{\infty}_{0}\frac {e^{-t/2}}{1-e^{-t}}\bigg( \hat{\phi}(0)-\hat{\phi}\bigg( \frac {t}{\mathcal{\mathcal{L}}} \bigg) \bigg) \,{d} t -\frac 2{\mathcal{L}}\sum_{j \geq 1} S_{j}(\chi_{i(1+i)^{5}c}, \mathcal{L} ;\hat{\phi}), \end{align*} $$

where

$$ \begin{align*} S_{j}(\chi_{i(1+i)^{5}c}, \mathcal{L};\hat{\phi})=\sum_{\varpi \equiv 1 \bmod {(1+i)^{3}}}\frac { \log N(\varpi)}{\sqrt{N(\varpi^{\kern1.8pt j})}}\chi_{i(1+i)^{5}c} ( \varpi^{\kern1.8pt j} ) \hat{\phi}\bigg( \frac {\log N(\varpi^{\kern1.8pt j})}{\mathcal{L}} \bigg), \end{align*} $$

with the sum over $\varpi $ running over primes in $\mathcal {O}_{K}$ .

Using Lemma 2.5, upon summing over c against the weight function w, we obtain the following result (the formula in (1-4)) for $D(\phi ;w, X)$ .

Lemma 2.6. Assume that $\phi $ is an even Schwartz test function whose Fourier transform has compact support. Then we have

$$ \begin{align*} D (\phi;w, X) &= \frac {\widehat{\phi}(0)}{ \mathcal{L} W(X)}\underset{(c, 1+i) =1 }{\sum \nolimits^{*}} w\bigg( \frac {N(c)}X \bigg) \log N(c) +\frac {\widehat\phi(0)}{\mathcal{L}} \bigg( \log\frac {32}{\pi^{2}}+ 2\frac{\Gamma^{\prime}}{\Gamma} \bigg( \frac 12 \bigg) \bigg ) \\[3pt] &\quad -\frac 2{\mathcal{L}W(X)} \underset{(c, 1+i) =1 }{\sum \nolimits^{*}} w\bigg( \frac {N(c)}X \bigg) \sum_{j \geq 1} S_{j}(\chi_{i(1+i)^{5}c}, \mathcal{L};\hat{\phi}) \\[3pt] &\quad +\frac{2}{\mathcal{L}}\int_{0}^{\infty}\frac{e^{-x/2}}{1-e^{-x}}\bigg(\widehat{\phi}(0)-\widehat{\phi}\bigg(\frac{x}{\mathcal{L}}\bigg)\bigg) \,{d} x. \end{align*} $$

Now let $w(t)$ be an even, nonzero and nonnegative Schwartz function as in the theorems. We define

(2-14) $$ \begin{align} g(y)=\widetilde{w}(\sqrt{2 }y ), \quad g_{1}(y)=\tilde{g}( \sqrt{y} ), \end{align} $$

where we recall that for any function W, the definition of $\widetilde {W}$ is given in (2-4).

Our next lemma establishes a relation between the Mellin transforms of g and $g_{2}$ . This is a generalization of [Reference Fiorilli, Parks and Södergren11, formula (2.3)].

Lemma 2.7. For any $z \in \mathbb C, z \neq 0, -1$ , we have

$$ \begin{align*} \zeta_{K}(z+1)\mathcal M g_{1}(z+1)=\zeta_{K}(-z) \mathcal M g(-z). \end{align*} $$

Proof. Our proof of this lemma is motivated by Riemann’s proof of the functional equation of the Riemann zeta function $\zeta (s)$ (see [Reference Davenport6, Section 8]). We first note that, for $\mathrm {Re}(z)>1$ ,

$$ \begin{align*} \zeta_{K}(z) \mathcal M g(z) &= \frac 14\sum_{\substack{ k \in \mathcal O_{K} \\ k \neq 0}}\int^{\infty}_{0}g(t)\bigg( \frac {t}{N(k)} \bigg)^{z} \frac {{d} t}{t} = \frac 14\sum_{\substack{ k \in \mathcal O_{K} \\ k \neq 0}}\int^{\infty}_{0}g(N(k) t)t^{z} \frac {{d} t}{t} \\[3pt] &= -\frac 14\frac {g(0)}z+ \frac 14 \int^{\infty}_{1} \sum_{\substack{ k \in \mathcal O_{K} \\ k \neq 0}} g(N(k) t) t^{z} \frac {{d} t}{t}+ \frac 14\int^{1}_{0} \sum_{\substack{ k \in \mathcal O_{K} }}g(N(k) t) t^{z} \frac {{d} t}{t}. \end{align*} $$

Now applying (2-3) to the second of the above sums and another change of variables, we get

$$ \begin{align*} \zeta_{K}(z) \mathcal M g(z) &= -\frac 14\frac {g(0)}z +\frac 14\frac {\widetilde g(0)}{z-1}+ \frac 14\int^{\infty}_{1} \sum_{\substack{ k \in \mathcal O_{K} \\ k \neq 0}}g(N(k) t) t^{z} \frac {{d} t}{t} \\[3pt] &\quad + \frac 14\int^{\infty}_{1} \sum_{\substack{ k \in \mathcal O_{K} \\ k \neq 0}}\widetilde g(\sqrt{N(k) t}) t^{-z+1} \frac {{d} t}{t}. \end{align*} $$

Note that the last two integrals converge absolutely for all $z \in \mathbb C$ , by applying estimate (2-6) to both g and $\widetilde g$ . The last expression above thus gives an analytical extension of $\zeta _{K}(z) \mathcal M g(z)$ to all $z \in \mathbb C, z \neq 0$ , 1.

Similarly, we also deduce from (2-3) that for $\mathrm {Re}(z)>0$ ,

$$ \begin{align*} \zeta_{K}(z+1) \mathcal M g_{1}(z+1)&= \frac 14\sum_{\substack{ k \in \mathcal O_{K} \\ k \neq 0}}\int^{\infty}_{0} \widetilde g (\sqrt{t}) \bigg( \frac {t}{N(k)} \bigg)^{z+1} \frac {{d} t}{t} \\[3pt] &= \frac 14\frac {g(0)}z-\frac 14\frac {\widetilde g(0)}{z+1}+\frac 14\int^{\infty}_{1} \sum_{\substack{ k \in \mathcal O_{K} \\ k \neq 0}}\widetilde g (\sqrt{N(k)t}) t^{z+1} \frac {{d} t}{t} \\[3pt] &\quad +\frac 14\int^{\infty}_{1} \sum_{\substack{ k \in \mathcal O_{K} \\ k \neq 0}} g(N(k)t) t^{-z} \frac {{d} t}{t}. \end{align*} $$

Once again by applying estimate (2-6) to both g and $\widetilde g$ , we see that the last two integrals above converge absolutely for all $z \in \mathbb C$ , so the last expression above gives an analytical extension of $\zeta _{K}(z+1) \mathcal M g_{1}(z+1)$ to all $z \in \mathbb C, z \neq 0$ , 1. Now, by comparing the above expressions for $\zeta _{K}(z) \mathcal M g(z)$ and $\zeta _{K}(z+1) \mathcal M g_{1}(z+1)$ , we readily deduce the assertion of the lemma.

3.1 Estimation of $S_{\text {even}}$

We first expand $S_{\mathrm {even}}$ into descending powers of $\mathcal {L}$ . We generalize [Reference Fiorilli, Parks and Södergren10, Lemma 3.7] to obtain the following result.

Lemma 3.1. Suppose that $\sigma =\sup (\mathrm {supp }\,\widehat {\phi } )<\infty $ . Then, for any integer $M \geq 1$ , we have the expansion

(3-3) $$ \begin{align} S_{\mathrm{even}} =-\frac{\phi(0)}2 + \sum_{m=1}^{M} \frac{d_{m} \widehat{\phi}^{(m-1)}(0)}{\mathcal{L}^{m}} +O\bigg( \frac 1{\mathcal{L}^{M+1}} \bigg), \end{align} $$

where the coefficients $d_{m}$ are real numbers that can be given explicitly.

Proof. It suffices to show that the expansion given in (3-3) is valid if we ignore the $O (X^{-3/4+\varepsilon } )$ term in (3-2). As $\sigma $ is finite, the sum in (3-2) is finite as we must have $N(\varpi )^{2j} \leq X^{\sigma }$ . It follows that the sum of the terms with $ j \geq 2$ can be expanded as follows:

(3-4) $$ \begin{align} &-\frac 2{\mathcal{L}} \sum_{\substack{\varpi \equiv 1 \bmod {(1+i)^{3}} \\ j\geq 2 , \ N(\varpi)^{2j} \leq X^{\sigma}}} \frac{\log N(\varpi)}{N(\varpi)^{j}} \bigg( 1+\frac 1{N(\varpi)} \bigg)^{-1} \nonumber\\[6pt] &\quad\quad \times \bigg(\sum_{m=0}^{M} \frac{\widehat{\phi}^{(m)}(0)}{m!} \bigg(\frac{2j \log N(\varpi)}{\mathcal{L}} \bigg)^{m}+O\bigg( \bigg(\frac{ 2j\log N(\varpi)}{\mathcal{L}}\bigg)^{M+1} \bigg) \bigg) \nonumber\\[6pt] &\quad= - \frac {2 }{\mathcal{L}} \sum_{m=0}^{M} \frac{\widehat{\phi}^{(m)}(0)}{m!\mathcal{L}^{m}}\sum_{\substack{\varpi \equiv 1 \bmod {(1+i)^{3}} \\ j\geq 2 , \ N(\varpi)^{2j} \leq X^{\sigma}}} \frac{\log N(\varpi)(2j \log N(\varpi))^{m}}{N(\varpi)^{j}} \nonumber\\[6pt] &\quad\quad \times \bigg( 1+\frac 1{N(\varpi)}\bigg)^{-1} +O( \mathcal{L}^{-M-2} ) \nonumber\\[6pt] &\quad= - \frac {2 }{\mathcal{L}} \sum_{m=0}^{M} \frac{\widehat{\phi}^{(m)}(0)}{m!\mathcal{L}^{m}}\sum_{\substack{\varpi \equiv 1 \bmod {(1+i)^{3}} \\ j\geq 2 }} \frac{\log N(\varpi)(2j \log N(\varpi))^{m}}{N(\varpi)^{j}} \nonumber\\[6pt] &\quad\quad \times \bigg( 1+\frac 1{N(\varpi)}\bigg)^{-1} +O( \mathcal{L}^{-M-2} ), \end{align} $$

by noting that the inner sum of the last expression above converges.

It remains to expand the terms with $j=1$ . For this, we first note that, using the Taylor expansion of $\hat {\phi }$ around the origin and rewriting $(1+N(\varpi )^{-1})^{-1}$ as a geometric series,

(3-5) $$ \begin{align} & -\frac 2{\mathcal{L}}\sum_{\substack{\varpi \equiv 1 \bmod {(1+i)^{3}}}} \frac{\log N(\varpi)}{N(\varpi)} \bigg( 1+\frac 1{N(\varpi)} \bigg)^{-1} \widehat{\phi}\,\bigg( \frac{2 \log N(\varpi)}{\mathcal{L}} \bigg) \nonumber\\[3pt] &\quad=-\frac 2{\mathcal{L}}\sum_{\substack{\varpi \equiv 1 \bmod {(1+i)^{3}}}} \frac{\log N(\varpi)}{N(\varpi)} \bigg( 1+\frac 1{N(\varpi)} \bigg)^{-1} \nonumber\\[3pt] &\quad\quad \times \bigg(\sum_{m=0}^{M} \frac{\widehat{\phi}^{(m)}(0)}{m!} \bigg(\frac{2 \log N(\varpi)}{\mathcal{L}} \bigg)^{m}+O\bigg( \bigg(\frac{ \log N(\varpi)}{\mathcal{L}}\bigg)^{M+1} \bigg) \bigg) \nonumber\\[3pt] &\quad=-\frac 2{\mathcal{L}}\sum_{\substack{\varpi \equiv 1 \bmod {(1+i)^{3}}}} \frac{\log N(\varpi)}{N(\varpi)} \bigg(\sum_{m=0}^{M} \frac{\widehat{\phi}^{(m)}(0)}{m!} \bigg(\frac{2 \log N(\varpi)}{\mathcal{L}} \bigg)^{m} \nonumber\\[3pt] &\quad\quad +O\bigg( \bigg(\frac{ \log N(\varpi)}{\mathcal{L}}\bigg)^{M+1} \bigg) \bigg) -\sum_{m=0}^{M} \frac{\widehat{\phi}^{(m)}(0)C_{1}(m)}{m!\mathcal{L}^{m+1}}+O(\mathcal{L}^{-M-1} ) \nonumber\\[3pt] &\quad=-\frac 2{\mathcal{L}}\sum_{\substack{\varpi \equiv 1 \bmod {(1+i)^{3}}}} \frac{\log N(\varpi)}{N(\varpi)}\widehat{\phi}\bigg( \frac{2 \log N(\varpi)}{\mathcal{L}} \bigg) -\sum_{m=0}^{M} \frac{\widehat{\phi}^{(m)}(0)C_{1}(m)}{m!\mathcal{L}^{m+1}} +O_{K}(\mathcal{L}^{-M-1}) \end{align} $$

where

$$ \begin{align*} & C_{1}(m) =\sum_{\substack{\varpi \equiv 1 \bmod {(1+i)^{3}}}} \sum_{l\geq 1} (-1)^{l} \frac{(2\log N(\varpi))^{m+1}}{N(\varpi)^{l+1}} < \infty. \end{align*} $$

Next, we note that

(3-6) $$ \begin{align} E(t) := \sum_{\substack{ N(\varpi) \leq t \\ \varpi \equiv 1 \bmod {(1+i)^{3}}}} \log N(\varpi)- t \ll t^{1/2+\varepsilon} \end{align} $$

from (2-7).

We then apply partial summation to arrive at

(3-7) $$ \begin{align} &-\frac 2{\mathcal{L}} \sum_{\substack{\varpi \equiv 1 \bmod {(1+i)^{3}}}} \frac{\log N(\varpi)}{N(\varpi)}\widehat{\phi}\bigg( \frac{2 \log N(\varpi)}{\mathcal{L}} \bigg) \nonumber\\[3pt] &\quad= -\frac 2{\mathcal{L}}\int_{1}^{\infty} \frac 1t \widehat{\phi}\bigg( \frac{2 \log t}{\mathcal{L}} \bigg) \,{d} ( t+E(t)) \nonumber\\[3pt] &\quad= -\int_{0}^{\infty} \widehat{\phi}(u)\, {d} u +\frac 2{\mathcal{L}}\int_{1}^{\infty} E(t) \frac{{d}}{{d} t} \bigg(\frac{1}{t} \widehat{\phi}\bigg( \frac{2 \log t}{\mathcal{L}} \bigg) \bigg) \, {d} t \nonumber\\[3pt] &\quad= -\frac 12 \phi(0)+\frac 2{\mathcal{L}}\int_{1}^{\infty} E(t) \frac{{d}}{{d} t} \bigg( \frac{1}{t} \widehat{\phi}\bigg( \frac{2 \log t}{\mathcal{L}} \bigg) \bigg) \, {d} t , \end{align} $$

with $E(t)$ defined in (3-6). We can now expand the derivative in the last integrand in (3-7) into Taylor expansions involving powers of $2\log t/\mathcal {L}$ and use the corresponding series to calculate the last integral above. Note that the new integrals emerging from this process are all convergent because of the bound (3-6). The assertion of the lemma now follows by combining (3-4), (3-5) and (3-7).

3.2 Estimation of $S_{\text {odd}}$ : Poisson summation

Starting from this section, we concentrate on the estimation of $S_{\mathrm {odd}}$ . First note that the contribution from the terms with $j\geq 3$ in (3-1) is $O(X^{-3/4+\varepsilon })$ by Lemma 2.3. It thus remains to treat the case for $j=1$ . For this case, we use the Möbius function to detect the condition that c is square-free to get

$$ \begin{align*} S_{\mathrm{odd}}&= -\frac 2{\mathcal{L} W(X) }\sum_{\substack {l \equiv 1 \ {\operatorname{mod}}\ (1+i)^{3}}}\mu_{[i]}(l) \sum_{\substack{ \varpi \equiv 1 \bmod {(1+i)^{3}} }} \frac {\log N(\varpi)}{\sqrt{N(\varpi)}} \hat{\phi} \bigg( \frac {\log N( \varpi)}{\log X} \bigg) \\[3pt] &\quad\times \sum_{(c, 1+i)=1} \bigg(\frac{i(1+i)cl^{2}}{\varpi}\bigg) w \bigg( \frac {N(cl^{2})}{X} \bigg) +O(X^{-3/4+\varepsilon}). \end{align*} $$

We divide the sum over l above into two parts, one over $l \leq Z$ and the other over $l>Z$ , with Z to be chosen optimally later. Note that if c is odd, then $i(1+i)^{5}cl^{2}$ is never a square. Similarly to the treatment of $S_{R}(X,Y; \hat {\phi }, \Phi )$ in [Reference Gao and Zhao15, Section 3.3] (except that we use Lemma 2.2 here instead of [Reference Gao and Zhao15, Lemma 2.5]), we get that the terms with $l> Z$ are

$$ \begin{align*} \ll X^{\varepsilon}(\log Z)^{3}Z^{-1}.\end{align*} $$

For the terms with $l \leq Z$ , we apply the Poisson summation (2-2) given in Lemma 2.1 and argue as in [Reference Gao and Zhao15] (the treatment here is essentially the same as that of $S_{M}(X,Y; \hat {\phi }, \Phi )$ in [Reference Gao and Zhao15, Section 3.2]) to arrive at the following lemma.

Lemma 3.2. Suppose that the GRH is true. We have for any $Z \geq 1$ and any $\epsilon>0$ ,

(3-8) $$ \begin{align} S_{\mathrm{odd}} &= -\frac X{\mathcal{L} W(X)} \sum_{\substack {N(l) \leq Z\\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})} \sum_{k \in \mathbb Z[i]}(-1)^{N(k)} \nonumber\\[3pt] &\quad \times \sum_{\varpi \equiv 1 \bmod {(1+i)^{3}}} \frac {\log N(\varpi)}{N(\varpi)}\bigg(\frac{kl^{2}}{\varpi}\bigg)\hat{\phi}\bigg( \frac {\log N( \varpi)}{\log X} \bigg) \widetilde{w}\bigg(\sqrt{\frac {N(k)X}{2N(l^{2}\varpi)}}\bigg) \nonumber\\[3pt] &\quad +O( X^{-3/4+\varepsilon}+X^{\varepsilon}(\log Z)^{3}Z^{-1} ). \end{align} $$

We now generalize [Reference Fiorilli, Parks and Södergren10, Lemma 2.7] to further analyze the sums in (3-8), obtaining the following result.

Lemma 3.3. Suppose that the GRH is true and that $\sigma =\sup (\mathrm {supp }\,\widehat {\phi })<\infty $ . Then, we have for any $1 \leq Z \leq X^{2}$ and any $\varepsilon>0$ ,

(3-9) $$ \begin{align} S_{\mathrm{odd}} &= \frac X{ W(X) } \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})} \bigg (\frac 12 I_{(1+i)l}(X)-I_{l}(X) \bigg ) \nonumber\\[3pt] &\quad+O(X^{-3/4+\varepsilon}+X^{\varepsilon}(\log Z)^{3}Z^{-1}+ZX^{\sigma/2-1+\varepsilon}+X^{-1/2+\varepsilon}Z^{\varepsilon} ), \end{align} $$

where

(3-10) $$ \begin{align} I_{l}(X) = \int_{0}^{\infty} \widehat{\phi}( u ) \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}} \widetilde{w}\bigg(2N(k)\sqrt{\frac {X^{1-u}}{2N(l^{2})}}\bigg) \,{d} u. \end{align} $$

Proof. Note that as in [Reference Gao and Zhao15, Section 3.4] the inner sum in (3-8) corresponding to $k=0$ is zero. It also follows from the treatment of [Reference Gao and Zhao15, Section 3.5] by setting $U=1$ and dividing the estimation obtained in [Reference Gao and Zhao15, formula (3.9)] by X (since our definition of $S_{\mathrm {odd}}$ differs from $S_{M}(X,Y; \hat {\phi }, \Phi )$ defined in [Reference Gao and Zhao15] by an extra factor $W^{-1}(X)$ ) that the contribution of $k \neq \square $ (k is not a square) to the expression for $S_{\mathrm {odd}}$ given in (3-8) is

$$ \begin{align*} \ll ZX^{\sigma/2-1+\varepsilon}. \end{align*} $$

We are left to consider the contribution from $k=\square $ (k is a square), $k \neq 0$ , to the expression for $S_{\mathrm {odd}}$ given in (3-8). For this, we make a change of variables $k \mapsto k^{2}$ , while noting that $k^{2}_{1}=k^{2}_{2}$ if and only if $k_{1} = \pm k_{2}$ , and deduce that

$$ \begin{align*} S_{\mathrm{odd}} & = -\frac X{2 \mathcal{L} W(X) } \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})} \sum_{\substack {(\varpi , l )=1 \\ \varpi \equiv 1 \bmod {(1+i)^{3}} }} \frac {\log N(\varpi)}{N(\varpi)}\hat{\phi} \bigg( \frac {\log N( \varpi)}{\log X} \bigg) \\[3pt] & \quad \times \sum_{\substack {k \in \mathbb Z[i] , k \neq 0 \\ (k, \varpi)=1}}(-1)^{N(k)}\widetilde{w}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2}\varpi)}}\bigg) \\[3pt] &\quad +O(X^{-3/4+\varepsilon}+X^{\varepsilon}(\log Z)^{3}Z^{-1}+ZX^{\sigma/2-1+\varepsilon} ). \end{align*} $$

In view of the rapid decay property of $\widetilde {w}$ implied by (2-6), we now remove the condition that $(\varpi , l)=1$ at the cost of an error

$$ \begin{align*} & \ll \frac 1{\mathcal{L}}\sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {1}{N(l^{2})} \sum_{\substack {\varpi | l \\ \varpi \equiv 1 \bmod {(1+i)^{3}} }} \frac {\log N(\varpi)}{N(\varpi)}\sum_{\substack {k \in \mathbb Z[i] ,\ k \neq 0 \\ N(k) \leq \sqrt{2N(l^{2}\varpi)/X}}}1 \\[3pt] & \ll X^{-1/2}\sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {1}{N(l)} \sum_{\substack {\varpi | l \\ \varpi \equiv 1 \bmod {(1+i)^{3}} }} \frac {\log N(\varpi)}{\sqrt{N(\varpi)}} \ll X^{-1/2}Z^{\varepsilon}, \end{align*} $$

where we use the well-known fact that for $N(l) \geq 3$ , the number, $\omega (l)$ , of distinct primes in $\mathbb Z[i]$ dividing l can be bounded as

$$ \begin{align*} \omega(l) \ll \frac {\log N(l)}{\log \log N(l)}. \end{align*} $$

One can show similarly that removing the condition $(k,\varpi )=1$ introduces an error of size much less than $ X^{-1/2}Z^{\varepsilon }$ . We thus derive, using (2-7), that

$$ \begin{align*} S_{\mathrm{odd}} &= -\frac X{2 \mathcal{L} W(X)}\sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})} \sum_{\substack {\varpi \equiv 1 \bmod {(1+i)^{3}} }} \frac {\log N(\varpi)}{N(\varpi)} \\[3pt] & \quad \times \hat{\phi} \bigg( \frac {\log N( \varpi)}{\log X} \bigg) \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}(-1)^{N(k)}\widetilde{w}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2}\varpi)}}\bigg)\\[3pt] &\quad +O(X^{-3/4+\varepsilon}+X^{\varepsilon}(\log Z)^{3}Z^{-1}+ZX^{\sigma/2-1+\varepsilon}+X^{-1/2+\varepsilon}Z^{\varepsilon} ) \\[3pt] &= -\frac {X}{2 \mathcal{L} W(X) }\sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})} \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}(-1)^{N(k)} \\[3pt] & \quad \times \int^{\infty}_{1}\frac 1y \hat{\phi} \bigg( \frac {\log y}{\mathcal{L}} \bigg)\widetilde{w}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2})y}}\bigg)\,{d} (y+ h(y) ) \\[3pt] & \quad +O(X^{-3/4+\varepsilon}+X^{\varepsilon}(\log Z)^{3}Z^{-1}+ZX^{\sigma/2-1+\varepsilon}+X^{-1/2+\varepsilon}Z^{\varepsilon} ) \\[3pt] &= S_{1}+S_{2}+O(X^{-3/4+\varepsilon}+X^{\varepsilon}(\log Z)^{3}Z^{-1}+ZX^{\sigma/2-1+\varepsilon}+X^{-1/2+\varepsilon}Z^{\varepsilon} ), \end{align*} $$

where h is a function satisfying $h(y) = O( y^{1/2} \log ^{2} (2y) )$ ,

$$ \begin{align*} S_{1}&= -\frac{X}{2 \mathcal{L}W(X)} \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})} \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}(-1)^{N(k)} \\[3pt] &\quad \times \int^{\infty}_{1}\frac 1y \hat{\phi} \bigg( \frac {\log y}{\mathcal{L}} \bigg)\widetilde{w}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2})y}}\bigg)\,{d} y \end{align*} $$

and

$$ \begin{align*} S_{2} &= -\frac{X}{2 \mathcal{L}W(X)} \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})} \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}(-1)^{N(k)} \\[3pt] &\quad \times \int^{\infty}_{1} \frac 1y \hat{\phi}\bigg( \frac {\log y}{\mathcal{L}} \bigg)\widetilde{w}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2})y}}\bigg)\,{d} h(y). \end{align*} $$

Note that

$$ \begin{align*} S_{2} &\ll \frac 1{\mathcal{L}} \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {1}{N(l^{2})} \\[3pt] &\quad \times \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}\int^{\infty}_{1} y^{1/2} \log^{2} (2y) \frac{{d}}{{d} y} \bigg(\frac 1y \hat{\phi} \bigg( \frac {\log y}{\mathcal{L}} \bigg)\widetilde{w}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2})y}}\bigg) \bigg ) \,{d} y \\[3pt] & \ll S_{2,1}+S_{2,2}, \end{align*} $$

where

$$ \begin{align*} S_{2,1} = \frac 1{\mathcal{L}} \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {1}{N(l^{2})} \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}\bigg | \widetilde{w}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2})}}\bigg)\bigg | \end{align*} $$

and

$$ \begin{align*} S_{2,2} &= \frac 1{\mathcal{L}} \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {1}{N(l^{2})} \int^{\infty}_{1} \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}} y^{1/2} \log^{2} (2y) \\[3pt] &\quad \times \bigg( \bigg |\bigg( \frac 1{y^{2}\mathcal{L}}\hat{\phi}^{\prime} \bigg( \frac {\log y}{\mathcal{L}} \bigg)\widetilde{w}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2})y}}\bigg)\bigg | +\bigg |\frac 1{y^{2}} \hat{\phi} \bigg( \frac {\log y}{\mathcal{L}} \bigg)\widetilde{w}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2})y}}\bigg)\bigg | \\[3pt] &\quad +\bigg | \hat{\phi} \bigg( \frac {\log y}{\mathcal{L}} \bigg)N(k)\sqrt{\frac {X}{8N(l^{2})}}y^{-5/2}\widetilde{w}^{\prime}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2})y}}\bigg)\bigg | \bigg ) \,{d} y. \end{align*} $$

Using (2-6), we deduce that

$$ \begin{align*} \sum_{\substack {k \in\mathbb Z[i] \\ k \neq 0}}\bigg |\widetilde{w}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2})y}}\bigg)\bigg | \ll \sqrt {\frac {N(l^{2})y}{X}} \end{align*} $$

and

$$ \begin{align*} \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}\bigg | N(k)\widetilde{w}^{\prime}\bigg(N(k)\sqrt{\frac {X}{2N(l^{2})y}}\bigg)\bigg | \ll \frac {N(l^{2})y}{X}. \end{align*} $$

Thus, it follows that

$$ \begin{align*} S_{2,2} &\ll \frac 1{\mathcal{L}} \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {1}{N(l^{2})}\int^{\infty}_{1} y^{1/2} \log^{2} (2y) \\[3pt] & \quad \times \bigg( \frac 1{y^{2}\mathcal{L}}\bigg |\hat{\phi} ^{\prime}\bigg( \frac {\log y}{\mathcal{L}} \bigg)\bigg |\sqrt {\frac {N(l^{2})y}{X}}+\frac 1{y^{2}}\bigg | \hat{\phi}\bigg( \frac {\log y}{\mathcal{L}} \bigg)\sqrt {\frac {N(l^{2})y}{X}}\bigg | \\[3pt] & \quad + \bigg |\hat{\phi} \bigg( \frac {\log y}{\mathcal{L}} \bigg)\bigg | \sqrt{\frac {X}{8N(l^{2})}}y^{-5/2}\frac {N(l^{2})y}{X}\bigg )\,{d} y \\[3pt] & \ll X^{-1/2}Z^{\varepsilon}\int^{\infty}_{1} \log^{2} (2y)\bigg ( \bigg | \hat{\phi}^{\prime}\bigg( \frac {\log y}{\mathcal{L}} \bigg)\bigg |+ \bigg |\hat{\phi} \bigg( \frac {\log y}{\mathcal{L}} \bigg)\bigg | \bigg) \,{d} \bigg( \frac {\log y}{\mathcal{L}} \bigg) \\[3pt] & \ll X^{-1/2+\varepsilon}Z^{\varepsilon}, \end{align*} $$

where the last estimation above follows by using a change of variable $u = \log y/\mathcal {L}$ to evaluate the proceeding integral and noting that the integrand has compact support. Similarly, we have that $S_{2,1} \ll X^{-1/2+\varepsilon }Z^{\varepsilon }$ so that

$$ \begin{align*} S_{2} \ll& X^{-1/2+\varepsilon}Z^{\varepsilon}. \end{align*} $$

It follows from the above bound that $S_{2}$ can be absorbed into the O-term in (3-9).It remains to evaluate $S_{1}$ , and applying exactly the same change of variable as above now leads to

(3-11) $$ \begin{align} S_{1} &= -\frac X{ 2W(X) } \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})} \nonumber\\[3pt] &\quad \times \int_{0}^{\infty} \widehat{\phi}( u ) \sum_{\substack {k \in \mathbb Z[i] , k \neq 0}}(-1)^{N(k)} \widetilde{w}\bigg(N(k)\sqrt{\frac {X^{1-u}}{2N(l^{2})}}\bigg) \,{d} u. \end{align} $$

We note that, for any $l \in \mathcal O_{K}$ , we have

$$ \begin{align*} &\sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}(-1)^{N(k)} \widetilde{w}\bigg(N(k)\sqrt{\frac {X^{1-u}}{2N(l^{2})}}\bigg) \\[3pt] &\quad= \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0 \\ 1+i|k}}\widetilde{w}\bigg(N(k)\sqrt{\frac {X^{1-u}}{2N(l^{2})}}\bigg)-\sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0 \\ (1+i,k)=1}}\widetilde{w}\bigg(N(k)\sqrt{\frac {X^{1-u}}{2N(l^{2})}}\bigg) \nonumber \\&\quad= 2\sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0 \\ 1+i|k}}\widetilde{w}\bigg(N(k)\sqrt{\frac {X^{1-u}}{2N(l^{2})}}\bigg)-\sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}\widetilde{w}\bigg(N(k)\sqrt{\frac {X^{1-u}}{2N(l^{2})}}\bigg) \\[3pt] &\quad= 2\sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0 }}\widetilde{w}\bigg(2 N(k)\sqrt{\frac {X^{1-u}}{2N(l^{2})}}\bigg)-\sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}\widetilde{w}\bigg(N(k)\sqrt{\frac {X^{1-u}}{2N(l^{2})}}\bigg). \end{align*} $$

Applying the above in (3-11), we derive that

$$ \begin{align*} S_{1} = \frac X{ W(X) } \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})} \bigg (\frac 12 I_{(1+i)l}(X)-I_{l}(X) \bigg ). \end{align*} $$

The above gives precisely the main term in (3-9) for $S_{\mathrm {odd}}$ , and this completes the proof.

3.3 Estimation of $S_{\text {odd}}$ : small support

In this section, we apply Lemma 3.3 to show that there are no new lower-order terms in powers of $\mathcal {L}^{-1}$ when $\sigma =\sup (\mathrm {supp }\,\widehat {\phi })\,{<}\,1$ . We generalize [Reference Fiorilli, Parks and Södergren10, Proposition 3.1] and state our result in the following proposition.

Proposition 3.4. Suppose that the GRH is true and that $\sigma =\sup (\mathrm {supp }\,\widehat {\phi })<1$ . Then, we have for any $\varepsilon>0$ ,

$$ \begin{align*} S_{\mathrm{odd}} \ll X^{\sigma/4 - 1/2+\varepsilon} + X^{3\sigma/4-3/4+\varepsilon}. \end{align*} $$

Proof. We let

(3-12) $$ \begin{align} \Phi(X)= \sum_{\substack{N(l) \leq X \\ l \equiv 1 \bmod {(1+i)^{3}} }} \frac{\mu_{[i]}(l)}{N(l)^{2}}= \frac 4{3\zeta_{K}(2)} +O(X^{-3/2 + \varepsilon}), \end{align} $$

where the last equality above follows from the observation that under the GRH we have

$$ \begin{align*}\sum_{\substack{N(l) \leq X \\ (l, 1+i)=1 }} \mu_{[i]}(l) \ll X^{1/2+\varepsilon}.\end{align*} $$

We then deduce that for $0\leq u \leq 1$ ,

$$ \begin{align*} &\sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})}\widetilde{w}\bigg(N(k)\sqrt{\frac {X^{1-u}}{2N(l^{2})}}\bigg) = \int_{0^{+}}^{Z} \widetilde{w}\bigg(\frac {N(k)}{t}\sqrt{\frac {X^{1-u}}{2}}\bigg) \,{d} \Phi(t) \\[3pt] &\qquad =\int_{0^{+}}^{Z} \widetilde{w}\bigg(\frac {N(k)}{t}\sqrt{\frac {X^{1-u}}{2}}\bigg) \,{d} \bigg(\frac 4{3\zeta_{K}(2)}+O(t^{-3/2 + \epsilon})\bigg ) \nonumber \\&\qquad = \widetilde{w}\bigg(\frac {N(k)}{Z}\sqrt{\frac {X^{1-u}}{2}}\bigg) O(Z^{-3/2 + \varepsilon})+ N(k)\sqrt{\frac {X^{1-u}}{2}}\int_{0^{+}}^{Z} \widetilde{w}^{\prime}\bigg(\frac {N(k)}{t}\sqrt{\frac {X^{1-u}}{2}}\bigg) O (t^{-3/2 + \varepsilon})\frac{ {d} t}{t^{2}} \\[3pt] & \qquad\ll Z^{-3/2 +\varepsilon} \bigg|\widetilde{w}\bigg(\frac {N(k)}{Z}\sqrt{\frac {X^{1-u}}{2}}\bigg)\bigg| +N(k)X^{(1-u)/2}\int_{0^{+}}^{Z} \bigg|\widetilde{w}^{\prime}\bigg(\frac {N(k)}{t}\sqrt{\frac {X^{1-u}}{2}}\bigg) \bigg| \frac{{d} t}{t^{7/2-\varepsilon}}. \end{align*} $$

Note that the part of the last integral for $ t \in (0,X^{(1-u)/2-\varepsilon }]$ is $O((N(k)X)^{-A})$ for any $A\geq 1$ , by the rapid decay of $\widetilde w^{\prime }$ . Summing over k and integrating over u, we obtain that

$$ \begin{align*} & \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})}I_{l}(X) \\[3pt] &\qquad \ll \int_{0}^{\infty} |\widehat{\phi}( u )| \sum_{\substack{k \in \mathbb Z[i] \\ k \neq 0}} \bigg(Z^{-3/2 +\varepsilon} \bigg|\widetilde{w}\bigg(\frac {N(k)}{Z}\sqrt{\frac {X^{1-u}}{2}}\bigg)\bigg|\\[3pt] &\qquad \quad+N(k)X^{(1-u)/2}\int_{0^{+}}^{Z} \bigg|\widetilde{w}^{\prime}\bigg(\frac {N(k)}{t}\sqrt{\frac {X^{1-u}}{2}}\bigg) \bigg| \frac{{d} t}{t^{7/2-\varepsilon}} \bigg)\,{d} u +X^{-1} \\[3pt] &\qquad \ll \int_{0}^{\infty} |\widehat{\phi}( u )| \bigg(Z^{-3/2 +\varepsilon} \frac{Z}{X^{(1-u)/2}}+\frac 1{X^{(1-u)/2}}\int_{X^{(1-u)/2-\varepsilon}}^{Z} \frac{{d} t}{t^{3/2-\varepsilon}} \bigg)\, {d} u + X^{-1}\\[3pt] &\qquad \ll \frac{X^{(\sigma-1)/2}}{Z^{1/2 -\varepsilon}} + X^{3\sigma/4 -3/4+\varepsilon}. \end{align*} $$

Similarly, we have

$$ \begin{align*} \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})}I_{(1+i)l}(X) \ll \frac{X^{(\sigma-1)/2}}{Z^{1/2 -\varepsilon}} + X^{3\sigma/4 -3/4+\varepsilon}. \end{align*} $$

Hence, it follows from Lemma 3.3 that for $Z\leq X^{2}$ ,

$$ \begin{align*} S_{\mathrm{odd}} &\ll \frac{X^{(\sigma-1)/2}}{Z^{1/2 -\varepsilon}} +X^{3\sigma/4 -3/4+\varepsilon} +X^{-3/4+\varepsilon}+X^{\varepsilon}Z^{-1} +ZX^{\sigma/2-1+\varepsilon}+X^{-1/2+\varepsilon}Z^{\varepsilon}. \end{align*} $$

The result follows by taking $Z=X^{1/2-\sigma /4}$ .

3.5 Estimation of $S_{\text {odd}}$ : extended support

In this section, we analyze the lower-order terms of $S_{\mathrm {odd}}$ when $\sigma =\sup (\mathrm {supp }\,\widehat {\phi }) \geq 1$ .

Lemma 3.5. Suppose that $\sigma =\sup (\mathrm {supp }\,\widehat {\phi })<\infty $ . Then, concerning the function $I_{l}(X)$ defined in (3-10), we have

(3-13) $$ \begin{align} I_{l}(X)&= -\widetilde w (0)\int_{1}^{\infty} \widehat{\phi}(u)\,du +\frac {\tilde{g}(0)}{\mathcal{L}} \int^{\infty}_{0}\phi( 1+\tau/\mathcal{L} ) e^{\tau/2}N(l) \, {d} \tau \nonumber\\[3pt] &\quad +\frac 1{\mathcal{L}} \int_{0}^{\infty} \widehat{\phi}\bigg( 1 + \frac{\tau}{\mathcal{L}} \bigg) e^{\tau/2}N(l)\sum_{\substack{ j \in \mathbb Z[i] \\ j \neq 0}}\tilde{g}(\lower 2pt\sqrt {N(j) e^{\tau/2}N(l)}) \, {d} \tau \nonumber\\[3pt] &\quad +\frac 1{\mathcal{L}} \int_{0}^{\infty} \widehat{\phi} \bigg( 1-\frac{\tau}{\mathcal{L}} \bigg) \sum_{\substack {k \in \mathbb Z[i]\\ k \neq 0}}g\bigg(N(k)\sqrt{\frac {e^{\tau}}{N(l^{2})}}\bigg) \,{d} \tau+O(N(l)X^{-1/2}) \end{align} $$

and

(3-14) $$ \begin{align} I_{(1+i)l}(X)&= -\widetilde w (0)\int_{1}^{\infty} \widehat{\phi}(u)\,du +\frac {2\tilde{g}(0)}{\mathcal{L}} \int^{\infty}_{0}\phi\bigg( 1 + \frac{\tau}{\mathcal{L}} \bigg)e^{\tau/2}N(l) \, {d} \tau \nonumber\\[3pt] & \quad+\frac 1{\mathcal{L}} \int_{0}^{\infty} \widehat{\phi}\bigg( 1 + \frac{\tau}{\mathcal{L}} \bigg) 2e^{\tau/2}N(l)\sum_{\substack{ j \in \mathbb Z[i] \\ j \neq 0}}\tilde{g}(2\lower 2pt\sqrt {N(j) e^{\tau/2}N(l)}) \, {d} \tau \nonumber\\[3pt] &\quad +\frac 1{\mathcal{L}} \int_{0}^{\infty} \widehat{\phi}\bigg( 1 - \frac{\tau}{\mathcal{L}} \bigg) \sum_{\substack {k \in \mathbb Z[i]\\ k \neq 0}}g\bigg(\frac {N(k)}{2}\sqrt{\frac {e^{\tau}}{N(l^{2})}}\bigg) \,{d} \tau, \end{align} $$

where $g(y)$ is given as in (2-14).

Proof. We first extend the integral in (3-10) to $\mathbb R$ and make the substitution $\tau ~=~\mathcal {L}(u-~1)$ to obtain that

(3-15) $$ \begin{align} I_{l}(X)&=\frac 1{\mathcal{L}} \int_{-\infty}^{\infty} \widehat{\phi} \bigg( 1+ \frac{\tau}{\mathcal{L}} \bigg) \sum_{\substack {k \in \mathbb Z[i]\\ k \neq 0}} \widetilde{w}\bigg(2N(k)\sqrt{\frac {e^{-\tau}}{2N(l^{2})}}\bigg) \,{d} \tau \nonumber\\[3pt] &\quad +O(N(l)X^{-1/2}), \end{align} $$

since for $u \leq 0$ , we have

$$ \begin{align*} &\int^{0}_{-\infty} \widehat{\phi}( u ) \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}} \widetilde{w}\bigg(2N(k)\sqrt{\frac {X^{1-u}}{2N(l^{2})}}\bigg) \,{d} u \\[3pt] &\quad \ll \int^{0}_{-\infty} \widehat{\phi}( u ) \sqrt{\frac {N(l^{2})}{X^{1-u}}} \,{d} u \ll N(l)X^{-1/2}. \end{align*} $$

We then break the integral in (3-15) into integrals over $(-\infty ,0]$ and $[0,\infty )$ and denote them, respectively, by $I_{l}^{-}(X)$ and $I_{l}^{+}(X)$ . By applying the Poisson summation formula (2-3) given in Lemma 2.1, we see that

$$ \begin{align*} I_{l}^{+}(X)&= \frac 1{\mathcal{L}}\int_{0}^{\infty} \widehat{\phi}\bigg( 1+ \frac{\tau}{\mathcal{L}} \bigg) \bigg(-\widetilde{w}(0) + \sum_{k \in \mathbb Z[i]} g\bigg(N(k)\sqrt{\frac {e^{-\tau}}{N(l^{2})}}\bigg) \bigg)\, {d} \tau \\[3pt] &= \frac 1{\mathcal{L}} \int_{0}^{\infty} \widehat{\phi}\bigg( 1 + \frac{\tau}{\mathcal{L}} \bigg) \bigg(-\widetilde{w}(0) + e^{\tau/2}N(l)\sum_{\substack{ j \in \mathbb Z[i] }}\tilde{g}(\lower 2pt\sqrt{N(j) e^{\tau/2}N(l)})\bigg) \, {d} \tau \\[3pt] &= \frac 1{\mathcal{L}} \int_{0}^{\infty} \widehat{\phi}\bigg( 1 + \frac{\tau}{\mathcal{L}} \bigg) \bigg(-\widetilde{w}(0) + e^{\tau/2}N(l)\tilde{g}(0)+ e^{\tau/2}N(l)\sum_{\substack{ j \in \mathbb Z[i] \\ j \neq 0}}\tilde{g}(\lower 2pt\sqrt{N(j) e^{\tau/2}N(l)})\bigg) \, {d} \tau. \end{align*} $$

Moreover, substituting $\tau $ by $-\tau $ in $I_{l}^{-}(X)$ , we obtain that

$$ \begin{align*}I^{-}_{l}(X)=\frac 1{\mathcal{L}} \int_{0}^{\infty} \widehat{\phi}\bigg( 1 + \frac{\tau}{\mathcal{L}} \bigg) \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}g \bigg(N(k)\sqrt{\frac {e^{\tau}}{N(l^{2})}}\bigg) \,{d} \tau. \end{align*} $$

Combining the above expressions for $I_{l}^{-}(X)$ and $I_{l}^{+}(X)$ , we readily derive the expression for $I_{l}(X)$ in (3-13). The expression for $I_{(1+i)l}(X)$ in (3-14) can be similarly obtained via the expression for $I_{l}(X)$ with the function $g(y)$ replaced by $g(y/2)$ , and this completes the proof.

We define the functions

$$ \begin{align*} h_{1}(x) = \frac{3\zeta_{K}(2)}{\pi \widehat w(0)}\sum_{\substack{l \equiv 1 \bmod {(1+i)^{3}}}}\frac {\mu_{[i]}(l)}{N(l)} ( \tilde{g}(\!\sqrt{2 N(l)x } )- \tilde{g}(\!\sqrt{ N(l)x } ) ) \end{align*} $$

and

$$ \begin{align*} h_{2}(x)= \frac{3\zeta_{K}(2)}{\pi \widehat w(0)}\sum_{\substack{ l \equiv 1 \bmod {(1+i)^{3}}}}\frac {\mu_{[i]}(l)}{N(l^{2})}\bigg (\frac 12 g\bigg(\frac {x}{2N(l)}\bigg)- g\bigg(\frac {x}{N(l)}\bigg)\bigg). \end{align*} $$

It is easy to see that $h_{1}(x)$ and $h_{2}(x)$ are smooth on $(0, \infty )$ and $[0, \infty )$ , respectively. Moreover, we have the bounds $h_{1}(x)\ll x^{-A}$ for any $A\geq 1$ and $h_{2}(x)\ll x^{-3/2+\varepsilon }$ for any $\varepsilon>0$ under the GRH. We point out here that the above notation and bounds are inspired by the corresponding notation introduced on page 1206 of [Reference Fiorilli, Parks and Södergren10].

We now apply Lemma 3.4 to derive the following generalization of [Reference Fiorilli, Parks and Södergren10, Corollary 3.4].

Lemma 3.6. Suppose that the GRH is true. Then, we have for $\sigma <2$ ,

$$ \begin{align*} S_{\mathrm{odd}} = \int_{1}^{\infty} \widehat{\phi}(u)\, {d} u+ J(X)+O(X^{\sigma/6-1/3+\varepsilon}), \end{align*} $$

where

(3-16) $$ \begin{align} J(X) &= \frac 1{\mathcal{L}} \int_{0}^{\infty} \bigg( \widehat{\phi} \bigg( 1+ \frac{\tau}{\mathcal{L}} \bigg) e^{\tau/2} \sum_{\substack{ j \in \mathbb Z[i] \\ j \neq 0}} h_{1}(N(j)e^{\tau/2}) +\widehat{\phi} \bigg( 1- \frac{\tau}{\mathcal{L}} \bigg) \sum_{\substack {k \in \mathbb Z[i]\\ k \neq 0}} h_{2}(N(k)e^{\tau/2})\bigg)\, {d} \tau. \end{align} $$

Proof. First note that by following the arguments that lead to [Reference Gao and Zhao15, formula (2.13)], we have

(3-17) $$ \begin{align} \widetilde{w}(0) = \frac {\pi }{2}\widehat{w}(0). \end{align} $$

It follows from (2-10), (3-12) and (3-17) that

(3-18) $$ \begin{align} \frac{X}{W(X)} \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}}\frac {\mu_{[i]}(l)}{N(l^{2})} \widetilde{w}(0) \int_{1}^{\infty} \widehat{\phi}(u)\, {d} u = 2 \int_{1}^{\infty} \widehat{\phi}(u)\, {d} u +O(Z^{-3/2+\varepsilon}). \end{align} $$

We now combine Lemmas 3.3, 3.4 and (3-18) to obtain that if $Z \leq X^{2}$ , then

(3-19) $$ \begin{align} S_{\mathrm{odd}} &= \frac{X}{W(X)} \sum_{\substack{N(l) \leq Z \\ l \equiv 1 \bmod {(1+i)^{3}}}} \frac {\mu_{[i]}(l)}{N(l^{2})} \bigg( \int_{0}^{\infty} \widehat \phi \bigg( 1+ \frac{\tau}{\mathcal{L}} \bigg) e^{\tau/2}N(l)\ \nonumber\\[3pt] &\quad \times \sum_{\substack{ j \in \mathbb Z[i] \\ j \neq 0}} ( \tilde{g}(2\lower 2pt\sqrt{N(\,j) e^{\tau/2}N(l)} )-\tilde{g}(\lower 2pt\sqrt {N(\,j) e^{\tau/2}N(l)}) ) \, {d} \tau\nonumber\\[3pt] &\quad +\int_{0}^{\infty} \widehat{\phi}\, \bigg( 1 -\frac{\tau}{\mathcal{L}} \bigg) \sum_{\substack {k \in \mathbb Z[i]\\ k \neq 0}}\bigg ( \frac 12 g\bigg(\frac {N(k)}{2}\sqrt{\frac {e^{\tau}}{N(l^{2})}}\bigg) -g\bigg(N(k)\sqrt{\frac {e^{\tau}}{N(l^{2})}}\bigg)\bigg ) \,{d} \tau \bigg ) \nonumber\\[3pt] &\quad +\int_{1}^{\infty} \widehat{\phi}(u)\, {d} u +O(X^{-3/4+\varepsilon}+X^{\varepsilon}Z^{-1}+ZX^{\sigma/2-1+\varepsilon}+X^{-1/2+\varepsilon}Z^{\varepsilon} ). \end{align} $$

Note that for the first two integrals in the above expression, we have

$$ \begin{align*} &\frac 1{\mathcal{L}} \int_{0}^{\infty} \widehat{\phi}\bigg( 1+ \frac{\tau}{\mathcal{L}} \bigg) e^{\tau/2}N(l) \sum_{\substack{ j \in \mathbb Z[i] \\ j \neq 0}} ( \tilde{g}(2\lower 2pt\sqrt {N(\,j) e^{\tau/2}N(l)}) -\tilde{g}(\lower 2pt\sqrt {N(\,j) e^{\tau/2}N(l)}\,))\, {d} \tau \\[3pt] &\quad \ll \frac {1}{\mathcal{L}} \int_{0}^{\infty} \widehat{\phi}\, \bigg( 1+ \frac{\tau}{\mathcal{L}} \bigg) \, {d} \tau \ll 1 \end{align*} $$

and

$$ \begin{align*} &\frac {1}{\mathcal{L}} \int_{0}^{\infty} \widehat{\phi} \bigg( 1 - \frac{\tau}{\mathcal{L}} \bigg) \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}}\bigg ( \frac 12 g\bigg(\frac {N(k)}{2}\sqrt{\frac {e^{\tau}}{N(l^{2})}}\bigg)-g\bigg(N(k)\sqrt{\frac {e^{\tau}}{N(l^{2})}}\bigg)\bigg) \,{d} \tau \\[3pt] &\quad \ll \frac {1}{\mathcal{L}} \int_{0}^{\infty} \widehat{\phi} \bigg( 1 - \frac{\tau}{\mathcal{L}} \bigg) N(l) e^{-\tau/2} \,{d} \tau \ll N(l). \end{align*} $$

We can therefore use $\Phi (X)$ defined in (3-12) and partial summation to extend the sum over l to all odd elements in $\mathcal {O}_{K}$ by introducing an extra error term of size $O(X^{\varepsilon }Z^{-3/2})$ . We now set $Z=X^{2/3 - \sigma /3}$ , change the order of summation in (3-19) and apply (2-10) to derive the desired result.

6.1 Initial treatment

In this section, we consider the expansions of the $D(\phi ;w, X)$ given in (1-3) as powers of $1/\mathcal {L}$ with $\mathcal {L}=\log X$ . Recall from (5-9) that, up to an error term of size $O(\mathcal {L}^{-2})$ , we have

(6-1) $$ \begin{align} D (\phi;w, X) &=\frac{1}{W(X)} \sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) \frac{1}{2\pi i}\int_{(a^{\prime})} \bigg(2 \dfrac{\zeta^{\prime}_{K}(1+2r)}{\zeta_{K}(1+2r)}+2A_{\alpha}(r,r) \nonumber\\[3pt] & \quad -\frac{X_{c}^{\prime}(1/2+r)}{X_{c}(1/2+r)} -\frac {8}{\pi} X_{c}\bigg(\dfrac{1}{2}+r\bigg)\zeta_{K}(1-2r)A(-r,r)\bigg) \phi\bigg(\frac{i \mathcal{L} r}{2\pi}\bigg) \,{d} r , \end{align} $$

where $\mathcal {L}^{-1}<a^{\prime }< 1/4$ .

We set

(6-2) $$ \begin{align} &I=-\frac {8}{\pi} \frac{1}{W(X)} \sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) \frac{1}{2\pi i} \nonumber\\[3pt] &\quad \int_{(a^{\prime})} \bigg( X_{c}\bigg(\frac{1}{2}+r\bigg)\zeta_{K}(1-2r)A(-r,r)\bigg) \phi\bigg(\frac{i \mathcal{L} r}{2\pi}\bigg) \,{d} r. \end{align} $$

We postpone the evaluation of I to the next section and proceed here with the treatment of the other terms on the right-hand side of (6-1).

We deduce first from Lemma 2.3 and (2-10), after partial summation, that

(6-3) $$ \begin{align} &\frac{1}{W(X)} \sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) \frac{1}{2\pi} \int_{\mathbb R} \log\bigg(\frac{32N(c)}{\pi^{2}}\bigg) \phi\bigg(\frac{t \mathcal{L}}{2\pi}\bigg) \, {d} t \nonumber\\[3pt] &\quad= \frac{\widehat{\phi}(0)}{\mathcal{L}}\frac{1}{W(X)}\sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) \log\bigg(\frac{32N(c)}{\pi^{2}}\bigg) \nonumber\\[3pt] &\quad= \frac{\widehat{\phi}(0)}{\mathcal{L}}\bigg ( \log \frac{32}{\pi^{2}}+\mathcal{L}+ \frac 2{\widehat w(0)}\int_{0}^{\infty} w(x) \log x \, {d} x \bigg )+O ( \mathcal{L}^{-2} ). \end{align} $$

Next note that, similarly to [Reference Gao and Zhao16, formula (4.33)], we have

(6-4) $$ \begin{align} &\frac{1}{{W}(X)} \sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) \frac{1}{2\pi} \int_{\mathbb R}\bigg( \frac{\Gamma^{\prime}}{\Gamma}\bigg(\frac 12-it\bigg) +\frac{\Gamma^{\prime}}{\Gamma}\bigg(\frac 12+it \bigg)\bigg) \phi\bigg(\frac{t\mathcal{L} }{2\pi}\bigg) \, {d} t\nonumber\\[3pt] &\quad= 2\frac{\Gamma^{\prime}}{\Gamma}\bigg(\frac 12\bigg)\frac{\widehat{\phi}(0)}{\mathcal{L}} +\frac{2}{\mathcal{L}} \int_{0}^{\infty}\frac{e^{-t/2}}{1-e^{-t}}\bigg(\widehat{\phi}(0)-\widehat{\phi}\bigg(\frac{t}{\mathcal{L}}\bigg)\bigg) \,{d} t. \end{align} $$

Furthermore, we follow the treatment of [Reference Fiorilli, Parks and Södergren11, Lemma 4.1] to obtain via a direct calculation (noting that $A(r,r)=1$ ) that

$$ \begin{align*} A_{\alpha}(r,r)+\frac{\zeta^{\prime}_{K}(1+2r)}{\zeta_{K}(1+2r)}=\frac{-1}{N(\varpi) +1} \sum_{\varpi \equiv 1 \bmod {(1+i)^{3}}} \frac{N(\varpi)\log N(\varpi)}{N(\varpi)^{1+2r}-1}. \end{align*} $$

It follows from this, after the substitution $u=-i\mathcal {L} r/(2\pi )$ and interchanging the summations and the integral, that

(6-5) $$ \begin{align} &\frac{1}{2\pi i}\int_{(a^{\prime})} \bigg(2 \frac{\zeta^{\prime}_{K}(1+2r)}{\zeta_{K}(1+2r)}+2A_{\alpha}(r,r) \bigg) \phi\bigg(\frac{i\mathcal{L} r}{2\pi}\bigg) \,{d} r \nonumber\\[3pt] &\quad= -\frac{2}{\mathcal{L}}\sum_{\varpi \equiv 1 \bmod {(1+i)^{3}}} \frac{N(\varpi)\log N(\varpi)}{N(\varpi)+1} \nonumber\\[3pt] & \qquad \times \sum_{j=1}^{\infty} \frac{1}{N(\varpi)^{\,j}}\int_{\mathcal C^{\prime}}\phi(u)\exp \bigg( -2\pi i u\bigg(\frac{2j\log N(\varpi)}{\mathcal{L}}\bigg) \bigg) \,{d} u, \end{align} $$

where $\mathcal C^{\prime }$ denotes the horizontal line $\mathrm {Im}(u)=-\mathcal {L} a^{\prime }/(2\pi )$ .

As $\widehat {\phi }$ is compactly supported and $\phi (z)=\int _{\mathbb {R}}\widehat \phi (x)e^{2\pi ixz}\, {d} x$ , it follows from integration by parts that uniformly for $-\mathcal {L} c^{\prime }/(2\pi ) \leq t\leq 0$ ,

$$ \begin{align*} |\phi(T+it)| \ll \frac1{|T|+1}. \end{align*} $$

In view of this, we can shift the contour of the last integration in (6-5) from $\mathcal C^{\prime }$ to $\mathrm {Im}(u) =0$ to deduce that

(6-6) $$ \begin{align} &\frac{1}{2\pi i}\int_{(a^{\prime})} \bigg(2 \frac{\zeta^{\prime}_{K}(1+2r)}{\zeta_{K}(1+2r)}+2A_{\alpha}(r,r)\bigg) \phi\bigg(\frac{i\mathcal{L} r}{2\pi}\bigg) \,{d} r \nonumber\\[3pt] &\quad =- \frac 2{\mathcal{L}}\sum_{\substack{\varpi \equiv 1 \bmod {(1+i)^{3}} \\ j\geq 1}} \frac{\log N(\varpi)}{N(\varpi)^{\,j}} \bigg( 1+\frac 1{N(\varpi)} \bigg)^{-1} \widehat{\phi}\bigg( \frac{2j \log N(\varpi)}{\mathcal{L}} \bigg). \end{align} $$

6.2 Evaluation of I

In this section, we evaluate I, defined in (6-2). Our treatment here follows from the proof of [Reference Fiorilli, Parks and Södergren11, Lemma 4.6]. We deduce from (5-7) that

$$ \begin{align*} A(-\gamma,\gamma)=\frac {3(2-2^{2r})}{4-2^{2r}}\frac {\zeta_{K}(2)}{\zeta_{K}(2-2r)}. \end{align*} $$

Substituting the above in the right-hand side of (6-2), we deduce from the definitions of $X_{c}$ given in (5-3) and a change of variable $r=2\pi i \tau / \mathcal {L}$ that

(6-7) $$ \begin{align} I&= - \frac {8}{\pi} \frac{\zeta_{K}(2)}{\mathcal{L}}\int_{\mathcal C^{\prime}} \bigg(\frac{\Gamma(1/2 - 2\pi i \tau/\mathcal{L})}{\Gamma(1/2 +2\pi i \tau/\mathcal{L} )} \bigg)\bigg(\frac{\pi^{2}}{32}\bigg)^{2 \pi i \tau/\mathcal{L}}\bigg(1+\frac{2-2^{4\pi i \tau/\mathcal{L} +1}}{4-2^{4\pi i \tau/\mathcal{L}}}\bigg) \nonumber\\[3pt] &\quad\times \frac{\zeta_{K}(1-4\pi i \tau/\mathcal{L} )}{\zeta_{K}(2- 4\pi i \tau/\mathcal{L})} \phi(\tau) \frac{1}{W(X)}\sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) N(c)^{- 2\pi i \tau/\mathcal{L}} \,{d} \tau, \end{align} $$

where we also use $\mathcal C^{\prime }$ to denote the horizontal line $\mathrm {Im}(\tau )=- \mathcal {L} a^{\prime }/(2\pi )$ .

Applying Mellin inversion to the last sum in (6-7), we obtain for ${0\leq \mathrm{Re}(r) \leq 1/2}$ ,

$$ \begin{align*} \sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) N(c)^{-r}= \frac 4{2\pi i} \int_{(2)}\frac{2^{s+r}}{2^{s+r}+1} \frac{\zeta_{K}(s+r)}{\zeta_{K}(2(s+r))} X^{s} \mathcal M w(s) \, {d} s. \end{align*} $$

We shift the contour of integration to the line $\mathrm {Re}(s)=1/2 -\mathrm {Re}(r)+\varepsilon $ to encounter a simple pole at $s=1-r$ . On the new line of integration, the convexity bound (see [Reference Iwaniec and Kowalski22, Exercise 3, page 100]), together with the rapid decay of $\mathcal M w$ , gives

$$ \begin{align*} \zeta_{K}(s) \ll (1+|s|^{2})^{1/4+\varepsilon}. \end{align*} $$

With this and recalling that the residue of $\zeta _{K}(s)$ at $s=1$ is $\pi /4$ , we get

$$ \begin{align*} \sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) N(c)^{-r} = \frac{2\pi}{3\zeta_{K}(2)}X^{1-r}\mathcal M w(1-r) +O_{\varepsilon,w}((|\mathrm{Im}(r)|+1)^{1/2+\varepsilon}X^{1/2-\mathrm{Re}(r)+\varepsilon}). \end{align*} $$

Combining the above with (2-10), we deduce that for any $\varepsilon>0$ and $0\leq \mathrm {Re}(r) \leq 1/2$ ,

$$ \begin{align*} \begin{split} & \frac{1}{W(X)}\sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) N(c)^{-r} \\[3pt] & \quad= \frac{2}{\widehat{w}(0)}X^{-r}\mathcal M w(1-r) +O_{\varepsilon,w}((|\mathrm{Im}(r)|+1)^{1/2+\varepsilon}X^{-1/2-\mathrm{Re} (r)+\varepsilon}). \end{split} \end{align*} $$

For small $\varepsilon , \eta>0$ , we change the contour $\mathcal C^{\prime }$ in (6-7) to the path

$$ \begin{align*} C=C_{0}\cup C_{1}\cup C_{2}, \end{align*} $$

where

$$ \begin{align*} C_{0}= \{ \mathrm{Im}(\tau)= 0 , |\mathrm{Re}(\tau)| \geq \mathcal{L}^{\varepsilon} \}, \quad C_{1}= \{ \mathrm{Im}(\tau) = 0 , \eta \leq |\mathrm{Re}(\tau)| \leq \mathcal{L}^{\varepsilon} \} \end{align*} $$

and

$$ \begin{align*} C_{2}= \{ |\tau| = \eta, \mathrm{Im}(\tau) \leq 0 \}. \end{align*} $$

As $\phi $ decays rapidly, the integration of I over $C_{0}$ can be shown to be negligible. We now apply the Taylor expansion to treat the integration of I over $C_{1}\cup C_{2}$ by noting that

$$ \begin{align*} \frac{\Gamma(1/2 - 2\pi i \tau/\mathcal{L} )}{\Gamma(1/2 + 2\pi i \tau/\mathcal{L} )} =1-2\frac {\Gamma^{\prime}(1/2)}{\Gamma(1/2)} \frac{2\pi i \tau}{\mathcal{L}} +O\bigg( \frac{|\tau|^{2}}{\mathcal{L}^{2}}\bigg), \end{align*} $$

and that (see [Reference Gradshteyn and Ryzhik17, Formula 2, Section 8.366])

$$ \begin{align*} -\frac {\Gamma^{\prime}(1/2)}{\Gamma(1/2)}=2\log 2+\gamma. \end{align*} $$

Using Taylor expansion and (1-6), we get

$$ \begin{align*} & \bigg(1+\frac{2-2^{4\pi i \tau/\mathcal{L}+1}}{4-2^{4\pi i \tau/\mathcal{L}}}\bigg)\frac{1}{\zeta_{K}(2- 4\pi i \tau/\mathcal{L})} \\[3pt] &\quad = \frac {1}{\zeta_{K}(2)}+\bigg ( \frac {1}{\zeta_{K}(2)} \bigg(-\frac {2\log 2}{3} \bigg) +\frac {\zeta^{\prime}_{K}(2)}{\zeta_{K}^{2}(2)} \bigg)\frac{4\pi i \tau}{\mathcal{L}}+O\bigg( \frac{|\tau|^{2}}{\mathcal{L}^{2}}\bigg) \end{align*} $$

and

$$ \begin{align*} \zeta_{K}\bigg(1-\frac{4\pi i \tau}{\mathcal{L}}\bigg)= -\frac {\pi}{4}\cdot \frac{\mathcal{L}}{4\pi i \tau}+\gamma_{K} +O\bigg( \frac{|\tau|}{\mathcal{L}}\bigg). \end{align*} $$

Using the above formulas, we get, after a short computation,

$$ \begin{align*} & -\dfrac {8}{\pi} \dfrac{\zeta_{K}(2)}{\mathcal{L}} \dfrac{\Gamma ( 1/2 -2\pi i \tau/\mathcal{L} )}{\Gamma(1/2 + 2\pi i \tau/\mathcal{L})} \bigg(\dfrac{\pi^{2}}{32}\bigg)^{{2 \pi i \tau}/{\mathcal{L}}}\bigg(1+\dfrac{2-2^{({4\pi i \tau}/\mathcal{L})+1}}{4-2^{{4\pi i \tau}/\mathcal{L}}}\bigg) \\[3pt] &\qquad \times \dfrac{\zeta_{K}(1-4\pi i \tau/\mathcal{L})}{\zeta_{K}(2-4\pi i \tau/\mathcal{L})} \phi(\tau)\dfrac{1}{W(X)}\sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) N(c)^{-2\pi i \tau/\mathcal{L}} \\[3pt] &\quad= \dfrac 1{2\pi i\tau} \bigg( 1+ \dfrac{2\pi i \tau }{\mathcal{L}}\bigg( 2\gamma +2 \log 4 + \log \bigg( \dfrac {\pi^{2}}{32} \bigg)+2\dfrac{\zeta^{\prime}_{K}(2)}{\zeta_{K}(2)} -\dfrac{4}{3} \log 2 -\dfrac {8}{\pi} \gamma_{K} \\[3pt] &\qquad -\dfrac{\mathcal M w^{\prime}(1)}{\mathcal Mw(1)} \bigg) +O\bigg( \dfrac{|\tau|^{2}}{\mathcal{L}^{2}}\bigg)\bigg)\phi(\tau) e^{-2\pi i \tau} +O_{\varepsilon,w}(X^{-1/2+\varepsilon}). \end{align*} $$

We then deduce that

$$ \begin{align*} I = \frac{1}{2\pi i} \int_{C_{1}\cup C_{2}} \frac {\phi(\tau)}{\tau} e^{-2\pi i \tau} \,{d} \tau+I^{\prime}+ O_{w}(\mathcal{L}^{-2}),\end{align*} $$

where, combining the logarithm terms,

$$ \begin{align*} I^{\prime} =&\frac 1{\mathcal{L}} \bigg( 2\gamma+ \log \bigg( \frac {\pi^{2}}{2^{7/3}} \bigg)+2\frac{\zeta^{\prime}_{K}(2)}{\zeta_{K}(2)}-\frac {8}{\pi} \gamma_{K} -\frac{\mathcal M w^{\prime}(1)}{\mathcal Mw(1)} \bigg) \int_{C_{1}\cup C_{2}}\phi(\tau) e^{-2\pi i \tau} \,{d} \tau \\[3pt] =& \frac 1{\mathcal{L}} \bigg( 2\gamma+\log \bigg( \frac {\pi^{2}}{2^{7/3}} \bigg)+2\frac{\zeta^{\prime}_{K}(2)}{\zeta_{K}(2)} -\frac {8}{\pi} \gamma_{K} -\frac{\mathcal M w^{\prime}(1)}{\mathcal Mw(1)} \bigg) \nonumber \\& \times \int_{\mathbb R}\phi(\tau) e^{-2\pi i \tau} \,{d} \tau+O(\mathcal{L}^{-2}) \\[3pt] =& \frac {\widehat{\phi}(1)}{\mathcal{L}} \bigg( 2\gamma+\log \bigg( \frac {\pi^{2}}{2^{7/3}} \bigg) +2\frac{\zeta^{\prime}_{K}(2)}{\zeta_{K}(2)} -\frac {8}{\pi} \gamma_{K} -\frac{\mathcal M w^{\prime}(1)}{\mathcal Mw(1)} \bigg) +O(\mathcal{L}^{-2}). \end{align*} $$

Similarly to the treatment of $I_{1}$ in the proof of [Reference Fiorilli, Parks and Södergren11, Lemma 4.6], we have

$$ \begin{align*} \int_{C_{1}\cup C_{2}} \frac 1{2\pi i\tau} \phi(\tau) e^{-2\pi i \tau} \,{d} \tau & =\int^{\infty}_{1} \widehat{\phi}(\tau) \,{d} \tau +O(\mathcal{L}^{-2}). \end{align*} $$

Thus, we conclude that

$$ \begin{align*} I &= \int^{\infty}_{1} \widehat{\phi}(\tau) \,{d} \tau +\frac {\widehat{\phi}(1)}{\mathcal{L}} \bigg( 2\gamma+\log \bigg( \frac {\pi^{2}}{2^{7/3}} \bigg)+2\frac{\zeta^{\prime}_{K}(2)}{\zeta_{K}(2)} -\frac {8}{\pi} \gamma_{K} -\frac{\mathcal M w^{\prime}(1)}{\mathcal Mw(1)} \bigg) \\[3pt] &\quad +O(\mathcal{L}^{-2}). \end{align*} $$

Combining the above expression for I with (6-1)–(6-4) and (6-6), we deduce that expression (1-7) is valid.

6.3 Comparing terms

In this section we show that the expression given in (1-7) is in agreement with that given in (1-4) when $\sigma =\mathrm {sup}(\mathrm {supp }\,\widehat {\phi })<2$ . In fact, applying (6-3) and (6-4) in (1-3) and comparing it with (1-4), we see that, with the help of Lemma 2.3, it suffices to show that

(6-8) $$ \begin{align} &-\frac 2{\mathcal{L} W(X)} \underset{(c, 1+i) =1 }{\sum \nolimits^{*}} w\bigg( \frac {N(c)}X \bigg) \sum_{j \geq 1} S_{j}(\chi_{i(1+i)^{5}c},\mathcal{L};\hat{\phi}) \nonumber\\[3pt] &\quad= \frac{1}{W(X)}\sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) \frac{1}{2\pi} \int_{\mathbb R} \bigg(2 \frac{\zeta^{\prime}_{K}(1+2it)}{\zeta_{K}(1+2it)} + 2A_{\alpha}(it,it) \nonumber\\[3pt] &\quad\quad -\frac {8}{\pi} X_{c}\bigg(\frac{1}{2}+it\bigg)\zeta_{K}(1-2it)A(-it,it) \bigg) \, \phi\bigg(\frac{t\mathcal{L}}{2\pi}\bigg) \, {d} t \nonumber\\[3pt] &\quad= \frac{1}{W(X)}\sideset{}{^{*}}\sum_{(c,1+i)=1} w\bigg( \frac {N(c)}X \bigg) \frac{1}{2\pi i}\int_{(a^{\prime})} \bigg(2 \frac{\zeta^{\prime}_{K}(1+2r)}{\zeta_{K}(1+2r)}+2A_{\alpha}(r,r) \nonumber\\[3pt] &\quad\quad -\frac {8}{\pi} X_{c}\bigg(\frac{1}{2}+r\bigg)\zeta_{K}(1-2r)A(-r,r)\bigg) \phi\bigg(\frac{i\mathcal{L} r}{2\pi}\bigg) \,{d} r, \end{align} $$

where $(\log X)^{-1}<a^{\prime }< 1/4$ .

Now, similarly to the treatment in Section 3, we write

(6-9) $$ \begin{align} -\frac 2{\mathcal{L} W(X)} \underset{(c, 1+i) =1 }{\sum \nolimits^{*}} w\bigg( \frac {N(c)}X \bigg) \sum_{j \geq 1}S_{j}(\chi_{i(1+i)^{5}c},\mathcal{L};\hat{\phi}) = S_{\mathrm{odd}}+S_{\mathrm{even}}, \end{align} $$

where

$$ \begin{align*} S_{\mathrm{odd}}=-\frac 2{\mathcal{L} W(X)} \underset{(c, 1+i) =1 }{\sum \nolimits^{*}} w\bigg( \frac {N(c)}X \bigg) \sum_{\substack{j \geq 1 \\ j \,\equiv 1 ({\operatorname{mod}}\,\,2)}} S_{j}(\chi_{i(1+i)^{5}c},\mathcal{L} ;\hat{\phi}) \end{align*} $$

and

$$ \begin{align*} S_{\mathrm{even}}= -\frac 2{\mathcal{L} W(X)} \underset{(c, 1+i) =1 }{\sum \nolimits^{*}} w\bigg( \frac {N(c)}X \bigg) \sum_{\substack{j \geq 1 \\ j \,\equiv 0 ({\operatorname{mod}}\,\,2)}} S_{j}(\chi_{i(1+i)^{5}c},\mathcal{L} ;\hat{\phi}). \end{align*} $$

We then deduce from (6-6), (6-8), (6-9) and (3-2) that it remains to show that

(6-10) $$ \begin{align} &S_{\mathrm{odd}} =I +O(\mathcal{L}^{-2}), \end{align} $$

where I is defined in (6-2).

6.4 Evaluation of $S_{{\bf odd}}$ , first order term

In this section we evaluate $S_{\mathrm {odd}}$ to the first lower-order term. Our treatment here largely follows the approach in the proof of [Reference Fiorilli, Parks and Södergren11, Theorem 1.1]. We recall from (3.5) that

(6-11) $$ \begin{align} S_{\mathrm{odd}} = \int_{1}^{\infty} \widehat{\phi}(u)\, {d} u+ J(X)+O(\mathcal{L}^{-2} ), \end{align} $$

where

$$ \begin{align*} J(X) &= \frac 1{\mathcal{L}} \int_{0}^{\infty} \bigg( \widehat{\phi} \bigg( 1+ \frac{\tau}{\mathcal{L}} \bigg) e^{\tau/2} \sum_{\substack{ j \in \mathbb Z[i] \\ j \neq 0}} h_{1}(N(j)e^{\tau/2}) \\[3pt] &\quad +\widehat{\phi} \bigg( 1- \frac{\tau}{\mathcal{L}} \bigg) \sum_{\substack {k \in \mathbb Z[i] \\ k \neq 0}} h_{2}(N(k)e^{\tau/2})\bigg)\, {d} \tau. \end{align*} $$

We now evaluate $h_{1}(x)$ by applying the Mellin inversion to recast it as

$$ \begin{align*} h_{1}(x) &= \frac{3\zeta_{K}(2)}{\pi \widehat w(0)}\frac 1{2 \pi i} \int_{(5/2)}\sum_{\substack{ l \,\equiv 1 \bmod {(1+i)^{3}}}}(2^{-z}-1)\frac {\mu_{[i]}(l)}{N(l)^{1+z}} \mathcal M g_{1}(z)\frac{{d} z}{x^{z}} \\[3pt] &= \frac{3\zeta_{K}(2)}{\pi \widehat w(0)}\frac 1{2 \pi i} \int_{(5/2)} \frac{2^{-z}-1}{( 1- 2^{-1-z})\zeta_{K}(1+z)} \mathcal M g_{1}(z)\frac{{d} z}{x^{z}}. \end{align*} $$

Similarly, we have, with a change of variables $z \to -z$ ,

$$ \begin{align*} h_{2} (x) &= \frac{3\zeta_{K}(2)}{\pi \widehat w(0)}\frac 1{2 \pi i} \int_{(1/2)}\sum_{\substack{l \equiv 1 \bmod {(1+i)^{3}}}}(2^{z-1}-1)\frac {\mu_{[i]}(l)}{N(l)^{2-z}} \mathcal M g(z)\frac{{d} z}{x^{z}} \\[3pt] &= \frac{3\zeta_{K}(2)}{\pi \widehat w(0)}\frac 1{2 \pi i} \int_{(-1/2)} \frac{2^{-z-1}-1}{( 1- 2^{-2-z})\zeta_{K}(2+z)} \mathcal M g(-z)x^{z} \,{d} z \\[3pt] &= \frac{3\zeta_{K}(2)}{\pi \widehat w(0)}\frac 1{2 \pi i} \int_{(-5/4)} \frac{2^{-z-1}-1}{( 1- 2^{-2-z})\zeta_{K}(2+z)} \mathcal M g(-z)x^{z} \,{d} z. \end{align*} $$

With the above expressions for $h_{1}(x)$ and $h_{2}(x)$ , we can write $J(x)$ given in (3-16) as

(6-12) $$ \begin{align} J(X) &= \frac{Y}{\mathcal{L}}\int_{0}^{\infty} \bigg( \widehat{\phi} \bigg( 1+ \frac{\tau}{\mathcal{L}} \bigg) \int_{(5/2)} \frac{4(2^{-z}-1)\zeta_{K}(z)}{( 1- 2^{-1-z})\zeta_{K}(1+z)} \mathcal Mg_{1}(z)\frac{{d} z}{e^{(z-1)\tau/2}} \nonumber\\[3pt] &\quad +\widehat{\phi} \bigg( 1- \frac{\tau}{\mathcal{L}} \bigg) \int_{(-5/4)} \frac{ 4(2^{-z-1}-1)\zeta_{K}(-z)}{( 1- 2^{-2-z})\zeta_{K}(2+z)} \mathcal M g(-z)e^{z\tau/2} \,{d} z \bigg) \, {d} \tau \nonumber\\[3pt] &= \frac{Y}{\mathcal{L}}\bigg( \int_{(3/2)} \frac{4(2^{-z}-1)\zeta_{K}(z)}{( 1- 2^{-1-z})\zeta_{K}(1+z)} \mathcal M g_{1} (z) \int_{0}^{\infty} \widehat{\phi} \bigg( 1+\frac{\tau}{\mathcal{L}} \bigg)e^{-(z-1)\tau/2} \,{d} \tau \,{d} z \nonumber\\[3pt] &\quad + \int_{(-5/4)} \frac{4(2^{-z-1}-1)\zeta_{K}(-z)}{( 1- 2^{-2-z})\zeta_{K}(2+z)} \mathcal M g(-z) \int_{0}^{\infty}\widehat\phi \bigg( 1- \frac{\tau}{\mathcal{L}} \bigg) e^{z\tau/2} \,{d} \tau \,{d} z \bigg)\, , \end{align} $$

where

$$ \begin{align*} Y = \frac{3\zeta_{K}(2)}{\pi \widehat w(0)}\frac 1{2 \pi i}. \end{align*} $$

Now we consider the Taylor expansions, centered at $1$ , of the terms $\widehat \phi ( 1+ \tau /\mathcal {L} )$ and ${\widehat \phi ( 1- \tau /\mathcal {L} )}$ in (6-12). By keeping only the constant terms, we see that their contribution to $J(X)$ equals, with another change of variables $z \to z+1$ in the first integral,

(6-13) $$ \begin{align} &\frac{3\zeta_{K}(2)}{\mathcal{L} \pi \widehat w(0)} \frac{ 8\widehat{\phi}( 1)}{2 \pi i}\bigg( \int_{(1/2)} \frac{(2^{-z-1}-1)\zeta_{K}(z+1)}{( 1- 2^{-2-z})\zeta_{K}(2+z)} \mathcal M g_{1}(z+1) \frac{ {d} z}{z} \nonumber\\[3pt] &\quad - \int_{(-5/4)} \frac{(2^{-z-1}-1)\zeta_{K}(-z)}{( 1- 2^{-2-z})\zeta_{K}(2+z)} \mathcal M g(-z) \frac{{d} z}z \bigg). \end{align} $$

We now shift the contour of the last integration to the line $\mathrm {Re}(z) = 1/2$ . We apply Lemma 2.7 to see that the quantity in (6-13) equals

$$ \begin{align*} \frac{3\zeta_{K}(2)}{\mathcal{L} \pi \widehat w(0)}\frac{ 8\widehat{\phi}( 1)}{2 \pi i}R, \end{align*} $$

where R is the residue of the function

$$ \begin{align*} \frac{(2^{-z-1}-1)\zeta_{K}(-z)}{( 1- 2^{-2-z})\zeta_{K}(2+z)} \frac{\mathcal M g(-z)}z \end{align*} $$

at $z=0$ . We observe via integration by parts that

$$ \begin{align*} \mathcal M g(-z) =\int^{\infty}_{0} g(t)t^{-z}\frac {{d} t}{t}=\int^{\infty}_{0} g(t) \,{d} \bigg ( \frac {t^{-z}}{-z} \bigg )=\frac 1{z}\int^{\infty}_{0}t^{-z} g^{\prime}(t) \,{d} t. \end{align*} $$

As

$$ \begin{align*} \lim_{z \rightarrow 0}\int^{\infty}_{0}t^{-z} g^{\prime}(t) \,{d} t=g(0)=\widehat{w}(0)>0, \end{align*} $$

it follows that $\mathcal M g(-z)$ has a pole at $z=0$ . We apply (3-17) and get that around $z=0$ ,

(6-14) $$ \begin{align} \mathcal M g(-z) &= \frac {-g(0)}{z}-\int^{\infty}_{0}(\log t) g^{\prime}(t) \,{d} t+O(z^{2}) \nonumber\\[3pt] &= \frac {-\widetilde{w}(0)}{z}-\int^{\infty}_{0}(\log t) g^{\prime}(t) \,{d} t+O(z^{2}) \nonumber\\[3pt] &= -\frac {\pi}{2}\frac {\widehat{w}(0)}{z}-\int^{\infty}_{0}(\log t) g^{\prime}(t) \,{d} t+O(z^{2}). \end{align} $$

Moreover, we have around $z=0$ ,

(6-15) $$ \begin{align} \frac{2^{-z-1}-1}{1- 2^{-2-z}}= -\frac 23+\frac{-\log 2 \big(\tfrac 38\big)+\tfrac 12 (\log 2) \tfrac 14}{(1- 2^{-2})^{2}}z=-\frac 23-\frac 49 (\log 2)z \end{align} $$

and

(6-16) $$ \begin{align} \frac{\zeta_{K}(-z)}{\zeta_{K}(2+z)}= \frac{\zeta_{K}(0)}{\zeta_{K}(2)}+\frac{-\zeta^{\prime}_{K}(0)\zeta_{K}(2)-\zeta_{K}(0)\zeta^{\prime}_{K}(2)}{\zeta^{2}_{K}(2)}z+O(z^{2}). \end{align} $$

Using (6-14), (6-15) and (6-16), we get

(6-17) $$ \begin{align} R&=-\frac {\pi}{2} \widehat{w}(0) \bigg( -\frac 23 \bigg) \frac{-\zeta^{\prime}_{K}(0)\zeta_{K}(2)-\zeta_{K}(0)\zeta^{\prime}_{K}(2)}{\zeta^{2}_{K}(2)} \nonumber\\[3pt] &\quad - \frac {\pi}{2}\widehat{w}(0) \bigg( -\frac 49 \log 2 \bigg) \frac{\zeta_{K}(0)}{\zeta_{K}(2)} -\frac 23\frac{\zeta_{K}(0)}{\zeta_{K}(2)} \bigg( -\int^{\infty}_{0}(\log t) g^{\prime}(t) \,{d} t \bigg). \end{align} $$

To further simplify R, we use the fact that $s\Gamma (s) = 1$ (see [Reference Davenport6, Section 10]) when $s = 0$ and the functional equation for $\zeta _{K}(s)$ (see [Reference Iwaniec and Kowalski22, Theorem 3.8]),

$$ \begin{align*} \pi^{-s}\Gamma(s)\zeta_{K}(s) = \pi^{-(1-s)}\Gamma(1 -s)\zeta_{K}(1 -s), \end{align*} $$

to obtain that $\zeta _{K}(0) = -1/4$ .

We further use the relation (see [Reference Davenport6, Section 10])

$$ \begin{align*} \Gamma(s)\Gamma(1-s)=\frac {\pi}{\sin (\pi s)} \end{align*} $$

to derive that

(6-18) $$ \begin{align} \zeta_{K}(1 -s)=\pi^{-2s}\Gamma(s)^{2}\sin(\pi s)\zeta_{K}(s). \end{align} $$

Applying (1-6), we see that around $s=1$ , we have

$$ \begin{align*} \zeta_{K}(s)\sin (\pi s)=-\frac {\pi^{2}}{4}-\pi \gamma_{K}(s-1)+O((s-1)^{2}). \end{align*} $$

Using the above expansion and the fact that $\Gamma ^{\prime }(1)=-\gamma $ (see [Reference Gradshteyn and Ryzhik17, Formula 1, Section 8.366] by noting also that $\Gamma (1)=1$ ), we take the derivative on both sides of (6-18) to see that

$$ \begin{align*} -\zeta^{\prime}_{K}(0)=-\frac {1}{\pi}\gamma_{K}+\frac {\gamma}{2}+\frac {\log \pi}{2}. \end{align*} $$

Now, inserting the values of $\zeta _{K}(0), -\zeta ^{\prime }_{K}(0)$ into (6-17), together with a short calculation, we obtain that

(6-19) $$ \begin{align} J(X)&= \frac{3\zeta_{K}(2)}{\mathcal{L} \pi \widehat w(0)}\frac{ 8\widehat{\phi}( 1)}{2 \pi i}R+O(\mathcal{L}^{-2}) \nonumber\\[3pt] &= \frac{8\widehat{\phi}( 1)}{\mathcal{L}}\bigg(-\frac {\gamma_{K}}{\pi}+\frac {\gamma}{2}+\frac {\log \pi}{2}+\frac 14\frac {\zeta^{\prime}_{K}(2)}{\zeta_{K}(2)}\nonumber\\[3pt] &\quad+ \bigg( \frac 23 \log 2 \bigg)\zeta_{K}(0) -\frac {1}{2\pi \widehat{w}(0)} \bigg( \int^{\infty}_{0}(\log t) g^{\prime}(t) \,{d} t \bigg) \bigg)+O(\mathcal{L}^{-2}). \end{align} $$

We evaluate the last integral above by noticing that for small $\eta>0$ , we have

$$ \begin{align*} \int_{0}^{\infty} (\log x) g^{\prime}(x) \,{d} x &=\int_{0}^{\infty} \sqrt{2} \log x \widetilde w^{\prime}(\sqrt{2}x) \,{d} x = \int_{0}^{\infty} \log \bigg(\frac x{\sqrt{2}}\bigg) \widetilde w^{\prime}(x) \,{d} x \\[3pt] &= \widetilde w(0) \log (\sqrt{2}) +\int_{\eta}^{\infty} (\log x) \widetilde w^{\prime}(x) \,{d} x +O(\eta \log (\eta^{-1})). \end{align*} $$

Now the above expression becomes, after integration by parts,

$$ \begin{align*} \widetilde w(0) \log (\sqrt{2}) -\int_{\eta}^{\infty} \frac{\widetilde w(x)-\widetilde w(0)I_{[0,1]}(x)}x \,{d} x +O(\eta \log (\eta^{-1})), \end{align*} $$

where $I_{[0,1]}$ is the characteristic function of the interval $[0,1]$ .

By evaluating $\widetilde {w}(x)$ in polar coordinates, we see that

$$ \begin{align*} \widetilde{w}(x) =& 4\int^{\pi/2}_{0}\int^{\infty}_{0}\cos (2\pi r x \sin \theta)w(r^{2}) \ r \,{d} r \,{d} \theta. \end{align*} $$

It follows from this, and by letting $\eta \rightarrow 0^{+}$ and using [Reference Vladimirov and Jeffrey41, Example (e) on page 132], that

(6-20) $$ \begin{align} \int_{0}^{\infty} (\log x )g^{\prime}(x) \,{d} x &= \widetilde w(0) \log (\!\sqrt{2}) -\int_{0}^{\infty} \frac{\widetilde w(x)-\widetilde w(0)I_{[0,1]}(x)}x \,{d} x \nonumber\\[3pt] &= \widetilde w(0) \log (\sqrt{2}) -4\int^{\infty}_{0}w(r^{2})r\int^{\pi/2}_{0} \mathcal{I}(\theta) \,{d} \theta \,{d} r , \end{align} $$

where

$$ \begin{align*} \mathcal{I} (\theta) = \int^{1}_{0} \frac {\cos (2\pi r x \sin \theta )-1}{x} \,{d} x + \int^{\infty}_{1} \frac {\cos (2\pi r x \sin \theta )}{x} \,{d} x. \end{align*} $$

Now, with some changes of variables,

$$ \begin{align*} \mathcal{I}(\theta) = \int^{2\pi r \sin \theta }_{0} \frac {\cos (u)-1}{u} \,{d} u + \int^{\infty}_{2\pi r \sin \theta} \frac {\cos (u)}{u} \,{d} u = \gamma +\log ( 2\pi r \sin \theta ). \end{align*} $$

Hence, the expression in (6-20) is

$$ \begin{align*} \widetilde w(0) \log (\sqrt{2}) & +\frac {\pi \gamma \widehat{w}(0)}{2} +\frac {\pi \log \pi \widehat{w}(0)}{2}+\frac {\pi \log 2 \widehat{w}(0)}{2} \\[3pt] & +\frac {\pi}{2} \int^{\infty}_{0}w(r)\log r \,{d} r+ 2\int^{\infty}_{0}w(r) \,{d} r\int^{\pi/2}_{0} \log ( \sin \theta ) \,{d} \theta. \end{align*} $$

As we have (see [Reference Gradshteyn and Ryzhik17, Formula 3, Section 4.224])

$$ \begin{align*} \int^{\pi/2}_{0} \log ( \sin \theta ) \,{d} \theta=-\frac {\pi}{2}\log 2. \end{align*} $$

We thus conclude that

$$ \begin{align*} \int_{0}^{\infty} (\log x )g^{\prime}(x) \,{d} x &= \frac {\pi \widehat{w}(0)}{4} \log 2 +\frac {\pi \gamma \widehat{w}(0)}{2} +\frac {\pi \log \pi \widehat{w}(0)}{2}+ \frac {\pi}{2} \mathcal M w^{\prime}(1). \end{align*} $$

Applying this to (6-19), we see that

$$ \begin{align*} J(X)&=\frac {\widehat{\phi}(1)}{\mathcal{L}} \bigg( 2\gamma +2\log 4+\log \bigg( \frac {\pi^{2}}{32} \bigg)+2\frac{\zeta^{\prime}_{K}(2)}{\zeta_{K}(2)} \\[3pt] &\quad - \frac 43 \log 2 -\frac {8}{\pi} \gamma_{K} -\frac{\mathcal M w^{\prime}(1)}{\mathcal Mw(1)} \bigg) +O(\mathcal{L}^{-2}). \end{align*} $$

With the above expression for $J(X)$ and (6-11), we conclude that the expression given in (6-10) is valid, and this completes the proof of Theorem 1.3.