2. Notation

Throughout the paper, we shall adopt the following list of notation:

1. (a) we say $f(x)=O(g(x))$ or $f(x) \ll g(x)$ if there exists a constant $C>0$ which might depend on $k,q, \boldsymbol{\alpha}, \boldsymbol{\beta}, \boldsymbol{c}, \boldsymbol{\delta}$ such that $|f(x)| \leqslant C \cdot g(x)$ whenever $x >x_0$ for some $x_0>0;$

2. (b) $[n]\;:\!=\;\{1,2,\ldots, n\};$

3. (c) $\tau_k(n)\;:\!=\;|\{(d_1,\ldots, d_k) \in \mathbb{N}^k \, : \, n=d_1\cdots d_{k} \} |$ and $\tau(n)\;:\!=\;\tau_2(n);$

4. (d) $\mathcal{M}_q\;:\!=\;\{F \in \mathbb{F}_q[x] \, : \, F \text{ is monic}\};$

5. (e) $\mathcal{M}_q(n)\;:\!=\;\{F \in \mathcal{M}_q \, : \, \deg{F}=n\};$

6. (f) $\tau_k(F)\;:\!=\;|\{(D_1,\ldots, D_k) \in \mathcal{M}_q^k \, : \, F=D_1\cdots D_{k} \} |;$

7. (g) $S_n$ denotes the group of permutations on $[n];$

8. (h) $c(\sigma)$ denotes the number of disjoint cycles of the permutation $\sigma;$

9. (i) $\tau_{\alpha}(\sigma)\;:\!=\;{\alpha}^{c(\sigma)};$

10. (j) $\left[{n \atop k}\right]\;:\!=\;|\{\sigma \in S_n \, : \, c(\sigma)=k\}|$ denote (unsigned) Stirling numbers of the first kind.

3. Properties of $\mathcal{D}(s_1,\ldots, s_k)$

Both [ Reference Deshouillers, Dress and Tenenbaum5, Reference Nyandwi and Smati16 ] deal with the divisors one by one using [ Reference Deshouillers, Dress and Tenenbaum5 , théorème T] followed by partial summation. However, as k gets larger, it is increasingly laborious to achieve full uniformity as well as the optimal rate of convergence, especially when one of the u’s is small or $u_1+\cdots+u_{k-1}$ is close to 1. Instead, we would like to apply Mellin’s inversion formula to (the second derivative of) the multiple Dirichlet series $\mathcal{D}(s_1,\ldots, s_k)$ defined below, which allows a more symmetric approach to the problem so that all the divisors can be handled simultaneously. We first establish a few properties of the multiple Dirichlet series that are essential to the proof of Theorem 1·1.

Lemma 3·1. Let $\mathcal{D}(s_1,\ldots, s_k)$ denote the multiple Dirichlet series

\begin{align*}\sum_{n_1=1}^{\infty}\cdots\sum_{n_k=1}^{\infty}\frac{\tau_k(n_1\cdots n_k)^{-1}}{n_1^{s_1}\cdots n_k^{s_k}}.\end{align*}

Then $\mathcal{D}(s_1,\ldots, s_k)$ converges absolutely in the domain

\begin{align*}\Omega\;:\!=\;\{(s_1,\dots,s_k)\in \mathbb{C}^k \, : \, \text{$\mathrm{Re}(s_j)>1$ for $j=1,\dots,k$}\}\end{align*}

and uniformly on any compact subset of $\Omega.$ In particular, $\mathcal{D}(s_1,\dots,s_k)$ is an analytic function of k variables in $\Omega.$

Proof. Let $\sigma_j\;:\!=\;\mathrm{Re}(s_j)$ for $j=1,\ldots,k.$ Then, since

\begin{align*}\sum_{n_1=1}^{\infty}\cdots\sum_{n_k=1}^{\infty}\left|\frac{\tau_k(n_1\cdots n_k)^{-1}}{n_1^{s_1}\cdots n_k^{s_k}}\right| \leqslant \zeta(\sigma_1)\cdots\zeta(\sigma_k) <\infty,\end{align*}

the lemma follows.

Lemma 3·2. The multiple Dirichlet series $D(s_1,\dots,s_k)$ can be expressed as the Euler product

\begin{align*}\prod_p\sum_{v_1=0}^{\infty}\cdots\sum_{v_k=0}^{\infty}\left(\substack{v_1+\cdots+v_k+k-1 \\[2pt] {k-1}}\right)^{-1}p^{-(v_1s_1+\cdots+v_ks_k)}\end{align*}

in the domain $\Omega$ defined above.

Proof. Let $y \geqslant 2$ and $\sigma_j\;:\!=\;\mathrm{Re}(s_j)$ for $j=1,\dots,k.$ Then, since

\begin{align*}\sum_{v_1=0}^{\infty}\cdots\sum_{v_k=0}^{\infty}\left(\substack{v_1+\cdots+v_k+k-1 \\[2pt] {k-1}}\right)^{-1}p^{-(v_1\sigma_1+\cdots+v_k\sigma_k)}\leqslant \left(1-\frac{1}{p^{\sigma_1}} \right)^{-1}\cdots \left(1-\frac{1}{p^{\sigma_k}} \right)^{-1}<\infty,\end{align*}

the finite product

\begin{align*}\prod_{p \leqslant y}\sum_{v_1=0}^{\infty}\cdots\sum_{v_k=0}^{\infty}\left(\substack{v_1+\cdots+v_k+k-1 \\[2pt] {k-1}}\right)^{-1}p^{-(v_1s_1+\cdots+v_ks_k)}\end{align*}

is well-defined.

Let $S(y)\;:\!=\;\{n\geqslant 1 \,: \, p|n \text{ implies } p \leqslant y\}$ be the set of y-smooth numbers. Then, since

\begin{align*}\tau_k(p^v)=\left(\substack{v+k-1 \\[2pt] k-1}\right),\end{align*}

we have

\begin{align*}&\left|\prod_{p \leqslant y}\sum_{v_1=0}^{\infty}\cdots\sum_{v_k=0}^{\infty}\left(\substack{v_1+\cdots+v_k+k-1 \\[2pt] {k-1}}\right)^{-1}p^{-(v_1s_1+\cdots+v_ks_k)}- \mathcal{D}(s_1,\dots,s_k) \right|\\[5pt] &\quad =\left|\sum_{n_1 \in S(y)}\cdots\sum_{n_k \in S(y)}\frac{\tau_k(n_1\cdots n_k)^{-1}}{n_1^{s_1}\cdots n_k^{s_k}} - \mathcal{D}(s_1,\dots,s_k)\right|\\[5pt] & \qquad \leqslant \sum_{j=1}^k \prod_{\substack{i=1\\[2pt] i\neq j}}^k \zeta(\sigma_i)\sum_{n_j>y}\frac{1}{n_j^{\sigma_j}}.\end{align*}

The lemma follows by letting $y \to \infty.$

Lemma 3·3. For $j=1,\ldots,k,$ let $R_j \subseteq \left\{ s_j \in \mathbb{C} \,:\, \mathrm{Re}(s_j) > {3}/{4}, \, |\mathrm{Im}(s_j)| > {1}/{4} \right\}$ be a zero-free region for $\zeta(s_j).$ Then the multiple Dirichlet series $\mathcal{D}(s_1,\dots,s_k)$ can be continued analytically to the domain $\prod_{j=1}^k R_j.$ Moreover, we have the bound

(3·1) \begin{align} \mathcal{D}(s_1,\dots,s_k) \ll |\zeta(s_1)|^{\frac{1}{k}}\cdots|\zeta(s_k)|^{\frac{1}{k}}.\end{align}

Proof. Let $(s_1,\dots,s_k)\in \mathbb{C}^k$ with $\sigma_j\;:\!=\;\mathrm{Re}(s_j)>1$ for $j=1,\dots,k.$ Then by Lemma 3·2 we have the Euler product expression

\begin{align*}&\zeta(s_1)^{-\frac{1}{k}}\cdots\zeta(s_k)^{-\frac{1}{k}}\mathcal{D}(s_1,\dots,s_k)\\[5pt] =&\prod_p\left(\prod_{j=1}^{k}\left(1-\frac{1}{p^{s_j}}\right)^{\frac{1}{k}}\right)\sum_{v_1=0}^{\infty}\cdots\sum_{v_k=0}^{\infty}\left(\substack{v_1+\cdots+v_k+k-1 \\[2pt] {k-1}}\right)^{-1}p^{-(v_1s_1+\cdots+v_ks_k)},\end{align*}

where the kth root is understood as its principal branch.

For $j=1,\ldots,k,$ expanding the kth root as

\begin{align*}\sum_{r=0}^{\infty}(\!-\!1)^{r}\left(\substack{\frac{1}{k} \\[2pt] r}\right)p^{-rs_j},\end{align*}

we find that the factors of the Euler product are $1+O\left(\sum_{i=1}^{k}\sum_{j=1}^{k} p^{-(\sigma_i+\sigma_j)} \right) $ by Taylor’s theorem. Therefore, the function

(3·2) \begin{align}\zeta(s_1)^{-\frac{1}{k}}\cdots\zeta(s_k)^{-\frac{1}{k}}\mathcal{D}(s_1,\dots,s_k)\end{align}

can be continued analytically to the domain where $\mathrm{Re}(s_j) > {3}/{4}$ for $j=1,\dots,k,$ in which it is uniformly bounded.

On the other hand, for $(s_1,\dots,s_k) \in \prod_{j=1}^k R_j,$ we can express $\mathcal{D}(s_1,\dots,s_k)$ as

\begin{align*}\zeta(s_1)^{\frac{1}{k}}\cdots\zeta(s_k)^{\frac{1}{k}}\left(\zeta(s_1)^{-\frac{1}{k}}\cdots\zeta(s_k)^{-\frac{1}{k}}\mathcal{D}(s_1,\dots,s_k)\right),\end{align*}

and so the lemma follows.

Lemma 3·4. In the open hypercube

\begin{align*}Q\;:\!=\;\left\{(s_1,\ldots,s_k) \in \mathbb{C}^k \, : \, 1<\mathrm{Re}(s_j)<\frac{7}{4},\, |\mathrm{Im}(s_j)|<\frac{3}{4} \text{ $j=1,\dots,k$}\right\},\end{align*}

we have the estimate

\begin{align*}\frac{\partial^{2k}}{\partial s_1^2\cdots\partial s_k^2}\mathcal{D}(s_1,\dots,s_k)=\left(1+\frac{1}{k}\right)^k\frac{1}{k^k}&(s_1-1)^{-\frac{1}{k}-2}\cdots(s_k-1)^{-\frac{1}{k}-2}\\[5pt] \times &(1+O(|s_1-1|+\cdots+|s_k-1|)).\end{align*}

Proof. By (3·2) and the fact that

(3·3) \begin{align} \zeta(s)=\frac{1}{s-1}+O(1),\end{align}

we have the power series representation

\begin{align*}(s_1-1)^{\frac{1}{k}}\cdots(s_k-1)^{\frac{1}{k}}\mathcal{D}(s_1,\dots,s_k)=\sum_{(i_1,\dots,i_k)\in \mathbb{Z}_{\geqslant 0}^k}a_{i_1,\dots,i_k}(s_1-1)^{i_1}\cdots(s_k-1)^{i_k}\end{align*}

in Q for some constants $a_{i_1,\dots,i_k} \in \mathbb{C}.$ It follows that

\begin{align*}\frac{\partial^{2k}}{\partial s_1^2\cdots\partial s_k^2}\mathcal{D}(s_1,\dots,s_k)=\left(1+\frac{1}{k}\right)^k\frac{1}{k^k}&(s_1-1)^{-\frac{1}{k}-2}\cdots(s_k-1)^{-\frac{1}{k}-2}\nonumber\\[5pt] \times & (a_{\textbf{0}}+O(|s_1-1|+\cdots+|s_k-1|)),\end{align*}

where by (3·3) and Lemma 3·2, the leading coefficient

\begin{align*}a_{\textbf{0}}&=\prod_p\left(1-\frac{1}{p}\right)\sum_{v_1=0}^{\infty}\cdots\sum_{v_k=0}^{\infty}\left(\substack{v_1+\cdots+v_k+k-1 \\[2pt] {k-1}}\right)^{-1}p^{-(v_1+\cdots+v_k)}\\[5pt] &=\prod_p \left(1-\frac{1}{p} \right)\sum_{v=0}^{\infty}\left(\substack{v+k-1 \\[2pt] k-1}\right)\left(\substack{v+k-1 \\[2pt] k-1}\right)^{-1}p^{-v}=1.\end{align*}

4. Proof of Theorem 1·1

We begin with writing

where the main term is

(4·1) \begin{align}\sum_{n \leqslant x}\frac{1}{\tau_k(n)}\underset{d_1\cdots d_{k-1} \mid n}{\sum_{d_1 \leqslant x^{u_1}}\cdots\sum_{d_{k-1} \leqslant x^{u_{k-1}}}}1=\sum_{d_1 \leqslant x^{u_1}}\cdots\sum_{d_{k-1} \leqslant x^{u_{k-1}}}\sum_{d_k \leqslant x/(d_1\cdots d_{k-1})}\frac{1}{\tau_k(d_1\cdots d_{k})},\end{align}

and the error term is

(4·2)

Let us first bound the error term (4·2). For $j=1,\dots,k-1$ with $u_j \neq 0,$ we write $n=d_jm$ for some integer m. Then $d_j>n^{u_j}$ implies $m<d_j^{(1-u_j)/u_j},$ and the number of ways of obtaining m as a product $d_1\cdots d_{j-1}d_{j+1}\cdots d_k$ is bounded by $\tau_{k-1}(m).$ It follows that

(4·3) \begin{align} \sum_{n \leqslant x}\frac{1}{\tau_k(n)}\sum_{n^{u_j}<d_j\leqslant x^{u_j}}\underset{\substack{d_i \leqslant x^{u_i}\text{ for }i \neq j\\[2pt] d_1\cdots d_{k-1} \mid n}}{\sum\cdots\sum}\quad1 \leqslant\sum_{d_j\leqslant x^{u_j}}\sum_{m<d_j^{(1-u_j)/u_j}}\frac{\tau_{k-1}(m)}{\tau_k(d_jm)}.\end{align}

If $u_{j}\leqslant {1}/{2},$ then using [ Reference Koukoulopoulos14 , theorem 14·2], this is bounded by

\begin{align}\sum_{d_{j}\leqslant x^{u_{j}}} \sum_{m < d_{j}^{(1-u_{j})/u_{j}}}\frac{\tau_{k-1}(m)}{\tau_k(m)}&\ll \sum_{2\leqslant d_{j}\leqslant x^{u_{j}}} \frac{d_{j}^{(1-u_{j})/u_{j}}}{\left(\!\log d_{j}^{(1-u_{j})/u_{j}}\right)^{\frac{1}{k}}} \nonumber \\[5pt] &\ll \frac{x}{(\!\log x)^{\frac{1}{k}}}. \nonumber\end{align}

Otherwise, it follows from the simple observation

(4·4) \begin{align}\frac{\tau_{k-1}(m)}{\tau_k(d_jm)} &\leqslant\frac{\tau_{k-1}(d_{j}m)}{\tau_k(d_{j}m)}\leqslant \frac{\tau_{k-1}(d_{j})}{\tau_k(d_{j})}\end{align}

that the expression (4·3) is bounded by

\begin{align}\sum_{d_{j}\leqslant x^{u_{j}}}d_{j}^{(1-u_{j})/u_{j}}\frac{\tau_{k-1}(d_{j})}{\tau_k(d_{j})}&\ll x^{1-u_{j}} \sum_{d_{j}\leqslant x^{u_{j}}} \frac{\tau_{k-1}(d_{j})}{\tau_k(d_{j})} \nonumber\\[5pt] &\ll \frac{x}{(\!\log x)^{\frac{1}{k}}} \nonumber\end{align}

again using [ Reference Koukoulopoulos14 , theorem 14·2].

Now we are left with the main term (4·1). In order to apply Mellin’s inversion formula, we follow the treatment in [ Reference Granville and Koukoulopoulos11 ] and [ Reference Koukoulopoulos14 , chapter 13].

Lemma 4·1. Let $T \geqslant 1.$ Let $\phi, \psi \colon [0, \infty) \to \mathbb{R}$ be smooth functions supported on [0,1] and $\left[0, 1+{1}/{T}\right]$ respectively with

\begin{align*}\begin{cases}\phi(y)=1 & \mbox{if } y \leqslant 1-\frac{1}{T},\\[2pt] \phi(y) \in [0,1] & \mbox{if } 1-\frac{1}{T} < y \leqslant 1,\\[2pt] \phi(y) = 0 & \mbox{if}\; y > 1 \end{cases}\end{align*}

and

\begin{align*}\begin{cases}\psi(y)=1 & \mbox{if } y \leqslant 1,\\[2pt] \psi(y) \in [0,1] & \mbox{if } 1 < y \leqslant 1 + \frac{1}{T}\\[2pt] \phi(y) = 0 & \mbox{if}\; y > 1 + \frac{1}{T} .\end{cases}\end{align*}

Moreover, for each integer $j\geqslant 0$ , their derivatives satisfy the growth condition $\phi^{(j)}(y), \psi^{(j)}(y) \ll_j T^j$ uniformly for $y\geqslant 0.$ Let $\Phi(s), \Psi(s)$ be the Mellin transform of $\phi(y), \psi(y)$ respectively for $1 \leqslant \mathrm{Re}(s) \leqslant 2$ , i.e.

\begin{align*}\Phi(s)=\int_{0}^{\infty}\phi(y)y^s\frac{dy}{y},\end{align*}

and

\begin{align*}\Psi(s)=\int_{0}^{\infty}\psi(y)y^s\frac{dy}{y}.\end{align*}

Then we have the estimates

(4·5) \begin{align} \Phi(s), \Psi(s)=\frac{1}{s}+O\left(\frac{1}{T}\right),\end{align}

and

(4·6) \begin{align} \Phi(s), \Psi(s) \ll_j \frac{T^{j-1}}{|s|^j}\end{align}

for $j\geqslant 1.$

We need the following version of Hankel’s lemma to extract the main contribution from the multidimensional contour integral in the proof of Lemma 4·3.

Lemma 4·2. Let $x>1, \sigma>1$ and $\mathrm{Re}(\alpha)>1.$ Then we have

\begin{align*}\frac{1}{2\pi i}\int_{\mathrm{Re}(s)=\sigma}\frac{x^s}{s(s-1)^{\alpha}}ds=\frac{1}{\Gamma(\alpha)}\int_{1}^x(\!\log y)^{\alpha-1}dy.\end{align*}

We now prove the main lemma.

Lemma 4·3. Let $x_1,\dots,x_k \geqslant e$ and $S(x_1,\dots,x_k)$ denote the weighted sum

\begin{align*}\sum_{d_1 \leqslant x_1}\cdots\sum_{d_k \leqslant x_k}\frac{(\!\log d_1)^2 \cdots (\!\log d_k)^2}{\tau_k(d_1\cdots d_k)}.\end{align*}

Then we have

\begin{align*}S(x_1,\dots,x_k)=\frac{1}{\Gamma\left(\frac{1}{k} \right)^k}\prod_{j=1}^{k}\int_1^{x_j}(\!\log y_j)^{\frac{1}{k}+1}dy_j+R(x_1,\dots,x_k)\end{align*}

with

\begin{align*}R(x_1,\dots,x_k) \ll x_1\cdots x_k\sum_{j=1}^k \left(\prod_{\substack{i=1\\[2pt] i \neq j}}^k (\!\log x_i)^{\frac{1}{k}+1}\right) (\!\log x_j)^{\frac{1}{k}}.\end{align*}

As in [ Reference Granville and Koukoulopoulos11 ] and [ Reference Koukoulopoulos14 , chapter 13], we introduce powers of logarithms to ensure that the major contribution to the multiple Perron integral below comes from $s_1,\ldots,s_k \approx 1.$ Later on, they will be removed by partial summation.

Proof. The proof consists of four steps: Mellin inversion, localisation, approximation and completion. For $j=1,\ldots,k$ , let $T_j=2(\!\log x_j)^{2}$ and $\phi_j, \psi_j$ be any smooth functions coincide with $\phi, \psi$ respectively from Lemma 4·1. Then the weighted sum $S(x_1,\ldots,x_k)$ is bounded between

\begin{align*}\sum_{d_1=1}^{\infty}\cdots\sum_{d_k =1}^{\infty}\frac{(\!\log d_1)^2 \cdots (\!\log d_k)^2}{\tau_k(d_1\cdots d_k)}\phi_1\left(\frac{d_1}{x_1} \right)\cdots\phi_k\left(\frac{d_k}{x_k}\right),\end{align*}

and

(4·7) \begin{align} \sum_{d_1=1}^{\infty}\cdots\sum_{d_k =1}^{\infty}\frac{(\!\log d_1)^2 \cdots (\!\log d_k)^2}{\tau_k(d_1\cdots d_k)}\psi_1\left(\frac{d_1}{x_1} \right)\cdots\psi_k\left(\frac{d_k}{x_k}\right).\end{align}

To avoid repetitions, we only establish the upper bound here. Applying Mellin’s inversion formula, the expression (4·7) becomes

\begin{align*}\sum_{d_1 =1}^{\infty}\cdots\sum_{d_k =1}^{\infty}\frac{(\!\log d_1)^2 \cdots (\!\log d_k)^2}{\tau_k(d_1\cdots d_k)}\prod_{j=1}^{k}\left(\frac{1}{2\pi i}\int_{\mathrm{Re}(s_j)=1+\frac{1}{2\log x_j}}\Psi_j(s_j)\left( \frac{d_j}{x_j}\right)^{-s_j}ds_j \right).\end{align*}

Then, by Lemma 3·1 and Lemma 4·1, it is valid to interchange the order of summation and integration, and so this becomes

(4·8) \begin{align} \frac{1}{(2\pi i)^k}\int_{\mathrm{Re}(s_1)=1+\frac{1}{2\log x_1}}\cdots\int_{\mathrm{Re}(s_k)=1+\frac{1}{2\log x_k}}&\left(\frac{\partial^{2k}}{\partial s_1^2\cdots\partial s_k^2}\mathcal{D}(s_1,\dots,s_k)\right) \nonumber\\[5pt] \times &\Psi_1(s_1)\cdots\Psi_k(s_k)x_1^{s_1}\cdots x_k^{s_k}ds_1\cdots ds_k.\end{align}

For each $j=1,\dots,k,$ we decompose the vertical contour $ I_j\;:\!=\;\left\{s_j \in \mathbb{C}\, : \,\mathrm{Re}(s_j)=1+{1}/{2\log x_j}\right\}$ as $I_{j}^{(1)} \cup I_{j}^{(2)} \cup I_{j}^{(3)}$ (traversed upwards), where

\begin{align*}I_{j}^{(1)}\;:\!=\;\left\{s_j \in I_j \, : \, |\mathrm{Im}(s_j)|\leqslant 1/2\right\},\end{align*}

\begin{align*}I_{j}^{(2)}\;:\!=\;\left\{s_j \in I_j \, : \, 1/2<|\mathrm{Im}(s_j)|\leqslant T_j^2/2 \right\}\end{align*}

and

\begin{align*}I_{j}^{(3)}\;:\!=\;\left\{s_j \in I_j \, : \, |\mathrm{Im}(s_j)|> T_j^2/2\right\}.\end{align*}

To establish an upper bound on the second derivative of the multiple Dirichlet series, we shall apply Cauchy’s integral formula for derivatives of k variables. For this purpose, we invoke Lemma 3·3 with the classical zero-free region

with $c=1/100$ say, for $j=1,\ldots,k.$ Moreover, we introduce the distinguished boundary

\begin{align*}\Gamma_{s_1,\dots,s_k}\;:\!=\;\left\{(w_1,\dots,w_k) \in \mathbb{C}^k \, : \,|w_j-s_j|=\begin{cases}\frac{|s_j-1|}{2} & \mbox{if } s_j \in I_j^{(1)},\\[5pt] \frac{c}{4\log T_j} & \mbox{if } s_j \in I_j^{(2)},\\[5pt] \frac{1}{4\log x_j} & \mbox{if } s_j \in I_j^{(3)}\end{cases}\text{ for $j=1,\dots,k$}\right\}\end{align*}

as there are various bounds on $\zeta(w_j)$ depending on the height. Then, Cauchy’s formula implies

\begin{align*}\frac{\partial^{2k}}{\partial s_1^2\cdots\partial s_k^2}\mathcal{D}(s_1,\dots,s_k)&\ll \frac{\sup_{(w_1,\dots,w_k) \in \Gamma_{s_1,\dots,s_k}}|\mathcal{D}(w_1,\dots,w_k)|}{\!\left(\!\prod_{\substack{j:s_j \in I_{j}^{(1)}}}|s_j-1|^2\!\right)\!\left(\!\prod_{j:s_j \in I_{j}^{(2)}}(\!\log T_j)^{-2}\!\right)\!\left(\!\prod_{j:s_j \in I_{j}^{(3)}}(\!\log x_j)^{-2}\!\right)}\end{align*}

with

\begin{align*}\mathcal{D}(w_1,\dots,w_k)\ll |\zeta(w_1)|^{\frac{1}{k}}\cdots|\zeta(w_k)|^{\frac{1}{k}}\end{align*}

given by (3·1) from Lemma 3·3. Using (3·3), [ Reference Titchmarsh21 , theorem 3·5] that

\begin{align*}\zeta(w_j) \ll \log T_j\end{align*}

whenever ${1}/{4}\leqslant|\mathrm{Im}(w_j)| \leqslant T_j^2,$ and the simple upper bound

\begin{align*}|\zeta(w_j)| \leqslant \sum_{n=1}^{\infty}\frac{1}{n^{1+\frac{1}{4\log x_j}}} \ll \log x_j,\end{align*}

we arrive at the derivative bound

(4·9) \begin{align} \frac{\partial^{2k}}{\partial s_1^2\cdots\partial s_k^2}\mathcal{D}(s_1,\dots,s_k) &\ll\prod_{j:s_j \in I_{j}^{(1)}}|s_j-1|^{-\frac{1}{k}-2}\prod_{j:s_j \in I_{j}^{(2)}}(\!\log T_j)^{\frac{1}{k}+2}\prod_{j:s_j \in I_{j}^{(3)}}(\!\log x_j)^{\frac{1}{k}+2}.\end{align}

Applying (4·6) with $j=1,2$ from Lemma 4·1, for $j=1,\dots,k,$ we have the estimates

\begin{align*}(\!\log x_j)^{\frac{1}{k}+2}\int_{I_{j}^{(3)}}|\Psi(s_j)x_j^{s_j}ds_j| &\ll x_j(\!\log x_j)^{\frac{1}{k}+2}T_j\int_{T_j^2/2}^{\infty}\frac{dt}{t^2}\\[5pt] &\ll \frac{x_j(\!\log x_j)^{\frac{1}{k}+2}}{T_j},\end{align*}

\begin{align*}(\!\log T_j)^{\frac{1}{k}+2}\int_{I_{j}^{(2)}}|\Psi(s_j){x_j}^{s_j}ds_j| &\ll x_j(\!\log T_j)^{\frac{1}{k}+2}\int_{1/2}^{T_j^2/2}\frac{dt}{t}\\[5pt] &\ll x_j(\!\log T_j)^{\frac{1}{k}+3},\end{align*}

and

(4·10) \begin{align} \int_{I_{j}^{(1)}}|(s_j-1)^{-\frac{1}{k}-2}\Psi(s_j){x_j}^{s_j}ds_j|&\ll \int_{I_{j}^{(1)}}\left|(s_j-1)^{-\frac{1}{k}-2}{x_j}^{s_j}\frac{ds_j}{s_j}\right| \nonumber\\[5pt] &\ll x_j\int_{-1/2}^{1/2}\left|\frac{1}{2\log x_j}+it \right|^{-\frac{1}{k}-2}dt \nonumber\\[5pt] &\ll x_j(\!\log x_j)^{\frac{1}{k}+1}.\end{align}

Therefore, combining with (4·5) from Lemma 4·1 and (4·9), the main contribution to (4·8) is

(4·11) \begin{align} \frac{1}{(2\pi i)^k}\int_{I_{1}^{(1)}}\cdots\int_{I_{k}^{(1)}}\left(\frac{\partial^{2k}}{\partial s_1^2\cdots\partial s_k^2}\mathcal{D}(s_1,\dots,s_k)\right)x_1^{s_1}\cdots x_k^{s_k}\frac{ds_1}{s_1}\cdots\frac{ds_k}{s_k}\end{align}

with an error term

(4·12) \begin{align} \ll x_1\cdots x_k\sum_{j=1}^k \left(\prod_{\substack{i=1\\[2pt] i \neq j}}^k (\!\log x_i)^{\frac{1}{k}+1}\right) (\!\log x_j)^{\frac{1}{k}}\end{align}

as $T_j= 2(\!\log x_j)^{2}$ for $j=1,\ldots,k.$

Applying Lemma 3·4, the main contribution to (4·11) is

(4·13) \begin{align} \left(1+\frac{1}{k}\right)^k\frac{1}{k^k}\prod_{j=1}^k\left(\frac{1}{2\pi i}\int_{I_j^{(1)}}(s_j-1)^{-\frac{1}{k}-2}x_j^{s_j}\frac{ds_j}{s_j}\right)\end{align}

with an error term

(4·14) \begin{align} \ll \sum_{j=1}^k \int_{I_{j}^{(1)}}\left|(s_j-1)^{-\frac{1}{k}-1}x_j^{s_j}\frac{ds_j}{s_j} \right| \prod_{\substack{i=1\\[2pt] i \neq j}}^k\int_{I_{i}^{(1)}} \left|(s_i-1)^{-\frac{1}{k}-2}x_i^{s_i}\frac{ds_i}{s_i} \right|.\end{align}

For $j=1,\dots,k,$ we have

\begin{align*}\int_{I_{j}^{(1)}} \left|(s_j-1)^{-\frac{1}{k}-1}x_j^{s_j}\frac{ds_j}{s_j} \right|&\ll x_j\int_{-1/2}^{1/2} \left|\frac{1}{2\log x_j}+it \right|^{-\frac{1}{k}-1}dt\\[5pt] &\ll x_j(\!\log x_j)^{\frac{1}{k}}.\end{align*}

Combining with (4·10), the expression (4·14) is

(4·15) \begin{align} \ll x_1\cdots x_k\sum_{j=1}^k \left(\prod_{\substack{i=1\\[2pt] i \neq j}}^k (\!\log x_i)^{\frac{1}{k}+1}\right)(\!\log x_j)^{\frac{1}{k}}.\end{align}

Since for $j=1,\dots,k$ we have the bound

\begin{align*}\int\limits_{\substack{\mathrm{Re}(s_j)=1+\frac{1}{2\log x_j}\\[2pt] |\mathrm{Im}(s_j)|>\frac{1}{2}}} \left|(s_j-1)^{-\frac{1}{k}-2}x_j^{s_j}\frac{ds_j}{s_j} \right|&\ll x_j\int_{1/2}^{\infty} t^{-\frac{1}{k}-3}dt\\[5pt] &\ll x_j,\end{align*}

it follows from (4·10) that the main contribution to (4·13) is

(4·16) \begin{align} \left(1+\frac{1}{k}\right)^k\frac{1}{k^k} \prod_{j=1}^k \left(\frac{1}{2\pi i} \int_{\mathrm{Re}(s_j)=1+\frac{1}{2\log x_j}}(s_j-1)^{-\frac{1}{k}-2}x_j^{s_j}\frac{ds_j}{s_j}\right)\end{align}

with an error term

(4·17) \begin{align} \ll x_1\cdots x_k\sum_{j=1}^k \prod_{\substack{i=1\\[2pt] i \neq j}}^k (\!\log x_i)^{\frac{1}{k}+1}.\end{align}

Applying Lemma 4·2, for $j=1,\dots,k$ we have

\begin{align*}\frac{1}{2\pi i}\int_{\mathrm{Re}(s_j)=1+\frac{1}{2\log x_j}}(s_j-1)^{-\frac{1}{k}-2}x_j^{s_j}\frac{ds_j}{s_j}&=\frac{1}{\Gamma \left(\frac{1}{k}+2 \right)}\int_{1}^{x_j}(\!\log y_j)^{\frac{1}{k}+1}dy_j\nonumber\\[5pt] &=\left(1+\frac{1}{k}\right)^{-1} \frac{k}{\Gamma \left(\frac{1}{k} \right)}\int_{1}^{x_j}(\!\log y_j)^{\frac{1}{k}+1}dy_j.\end{align*}

Finally, the lemma follows from collecting the main term (4·16) and the error terms (4·12), (4·15) and (4·17).

To proceed to the computation of the main term (4·1), we first show that it suffices to limit ourselves to the region where $u_1+\cdots+u_{k-1} \leqslant 1-{1}/{\log x}.$ Otherwise, if $u_1+\cdots+u_{k-1} > 1-{1}/{\log x},$ then we may assume without loss of generality that $u_{k-1}\geqslant {1}/{2k}.$ We now show that when replacing $u_{k-1}$ by $u_{k-1}-{1}/{\log x},$ both the right-hand sides of (4·1) and (1·1) are changed by a negligible amount. Arguing similarly as before, we have

(4·18) \begin{align} \sum_{d_1 \leqslant x^{u_1}}\cdots\sum_{d_{k-2}\leqslant x^{u_{k-2}}}&\sum_{x^{u_{k-1}-\frac{1}{\log x}}\leqslant d_{k-1} \leqslant x^{u_{k-1}}}\sum_{d_k\leqslant x/(d_1\cdots d_{k-1})}\frac{1}{\tau_k(d_1\cdots d_{k})} \nonumber\\[5pt] \leqslant & \sum_{x^{u_{k-1}-\frac{1}{\log x}}\leqslant d_{k-1} \leqslant x^{u_{k-1}}}\sum_{m\leqslant x/d_{k-1}}\frac{\tau_{k-1}(m)}{\tau_{k}(d_{k-1}m)}.\end{align}

If $u_{k-1}\leqslant {1}/{2},$ then using [ Reference Koukoulopoulos14 , theorem 14·2], this is bounded by

\begin{align*}\ll x^{u_{k-1}}\sum_{m\leqslant x^{1+\frac{1}{\log x}-u_{k-1}} }\frac{\tau_{k-1}(m)}{\tau_{k}(m)}\ll \frac{x}{(\!\log x)^{\frac{1}{k}}}.\end{align*}

Otherwise, again it follows from the observation (4·4) that (4·18) is

\begin{align*}\ll x^{1-u_{k-1}}\sum_{d_{k-1}\leqslant x^{u_{k-1}}}\frac{\tau_{k-1}(d_{k-1})}{\tau_{k}(d_{k-1})}\ll \frac{x}{(\!\log x)^{\frac{1}{k}}}.\end{align*}

On the other hand, by making the change of variables $t_j=(1-t_{k-1})s_j$ for $j=1,\ldots, k-2,$ we have

(4·19) \begin{align} x\int_{u_{k-1}-\frac{1}{\log x}}^{u_{k-1}}&t_{k-1}^{\frac{1}{k}-1}\left(\int_0^{u_1}\cdots\int_0^{u_{k-2}}t_1^{\frac{1}{k}-1}\cdots t_{k-2}^{\frac{1}{k}-1}(1-t_1-\cdots-t_{k-1})^{\frac{1}{k}-1}dt_1\cdots dt_{k-2}\right)dt_{k-1} \nonumber\\[5pt] \leqslant & x \int_{u_{k-1}-\frac{1}{\log x}}^{u_{k-1}}t_{k-1}^{\frac{1}{k}-1}(1-t_{k-1})^{-\frac{1}{k}}dt_{k-1} \nonumber\\[5pt] &\times \int_0^{\frac{u_1}{1-u_{k-1}}}\cdots\int_0^{\frac{u_{k-2}}{1-u_{k-1}}}s_1^{\frac{1}{k}-1}\cdots s_{k-2}^{\frac{1}{k}-1}(1-s_1-\cdots-s_{k-2})^{\frac{1}{k}-1}ds_1\cdots ds_{k-2} \nonumber\\[5pt] \ll & x \int_{0}^{\frac{1}{\log x}}t_{k-1}^{\frac{1}{k}-1}(1-t_{k-1})^{-\frac{1}{k}}dt_{k-1}\ll \frac{x}{(\!\log x)^{\frac{1}{k}}}.\end{align}

Therefore, we can always assume $u_1+\cdots+u_{k-1} \leqslant 1-{1}/{\log x}.$ Arguing similarly, we can further limit ourselves to the smaller region where $u_1,\dots,u_{k-1} \geqslant {1}/{\log x}$ as well.

In order to apply Lemma 4·3, we express the main term (4·1) as

(4·20) \begin{align} \sum_{3 \leqslant d_1 \leqslant x^{u_1}}\cdots&\sum_{3 \leqslant d_{k-1}\leqslant x^{u_{k-1}}}\sum_{3 \leqslant d_k \leqslant x/(d_1\cdots d_{k-1})}\frac{1}{\tau_k(d_1\cdots d_{k})} \nonumber\\[5pt] &+O\left(\sum_{j=1}^{k-1}\sum_{d_j=1,2}\underset{\substack{d_i \leqslant x^{u_i}\\[2pt] i=1,\dots,k-1, i \neq j}}{\sum\cdots\sum}\sum_{d_k \leqslant x/(d_1\cdots d_{k-1})}\frac{1}{\tau_k(d_1\cdots d_{k})}\right) \nonumber \\[5pt] &+O\left(\sum_{d_k=1,2}\sum_{d_1 \leqslant x^{u_1}}\cdots\sum_{d_{k-1}\leqslant x^{u_{k-1}}}\frac{1}{\tau_k(d_1\cdots d_{k})}\right).\end{align}

For $j=1,\dots,k-1,$ again it follows from [ Reference Koukoulopoulos14 , theorem 14·2] that

\begin{align*}\sum_{d_j=1,2}\underset{\substack{d_i \leqslant x^{u_i}\\[2pt] i=1,\dots,k-1, i \neq j}}{\sum\cdots\sum}\sum_{d_k \leqslant x/(d_1\cdots d_{k-1})}\frac{1}{\tau_k(d_1\cdots d_{k})}&\leqslant \sum_{m \leqslant x}\frac{\tau_{k-1}(m)}{\tau_k(m)}+\sum_{m \leqslant x/2} \frac{\tau_{k-1}(m)}{\tau_{k-1}(2m)}\nonumber\\[5pt] &\ll \frac{x}{(\!\log x)^{\frac{1}{k}}},\end{align*}

and similarly

\begin{align*}\sum_{d_{k}=1,2}\sum_{d_1 \leqslant x^{u_1}}\cdots\sum_{d_{k-1}\leqslant x^{u_{k-1}}}\frac{1}{\tau_k(d_1\cdots d_{k})}& \leqslant \sum_{m \leqslant x}\frac{\tau_{k-1}(m)}{\tau_k(m)}+\sum_{m \leqslant x/2} \frac{\tau_{k-1}(m)}{\tau_{k-1}(2m)} \nonumber\\[5pt] &\ll \frac{x}{(\!\log x)^{\frac{1}{k}}}.\end{align*}

By partial summation (or more precisely multiple Riemann–Stieltjes integration) and Lemma 4·3, the main term of (4·20) is

\begin{align}&\int_{e}^{x^{u_1}}\cdots\int_{e}^{x^{u_{k-1}}}\int_{e}^{\frac{x}{x_1\cdots x_{k-1}}}\frac{1}{(\!\log x_1)^2}\cdots\frac{1}{(\!\log x_k)^2}d S(x_1,\dots,x_k) \nonumber\\[5pt] =&\frac{1}{\Gamma\left(\frac{1}{k} \right)^k}\int_{e}^{x^{u_1}}\cdots\int_{e}^{x^{u_{k-1}}}\int_{e}^{\frac{x}{x_1\cdots x_{k-1}}}(\!\log x_1)^{\frac{1}{k}-1}\cdots(\!\log x_k)^{\frac{1}{k}-1} dx_1\cdots dx_k \nonumber\\[5pt] &+\int_{e}^{x^{u_1}}\cdots\int_{e}^{x^{u_{k-1}}}\int_{e}^{\frac{x}{x_1\cdots x_{k-1}}}\frac{1}{(\!\log x_1)^2}\cdots\frac{1}{(\!\log x_k)^2}d R(x_1,\dots,x_k)\nonumber\\[5pt] \;=\!:\;& I_1 + I_2. \nonumber\end{align}

Finally, it remains to compute the integrals $I_1$ and $I_2.$

Lemma 4·4. The first integral $I_1$ equals

\begin{align*}F(u_1,\ldots,u_{k-1})x+O\left(\frac{x}{(\!\log x)^{\frac{1}{k}}} \right).\end{align*}

Proof. Making the change of variables $x_j=x^{t_j}$ for $j=1,\ldots,k-1$ , the integral $I_1$ becomes

\begin{align*}\frac{\log x}{\Gamma\left(\frac{1}{k} \right)^k}\int_{{\frac{1}{\log x}}}^{u_1}\cdots\int_{{\frac{1}{\log x}}}^{u_{k-1}}\int_{{\frac{1}{\log x}}}^{1-t_1-\cdots -t_{k-1}}t_1^{\frac{1}{k}-1}\cdots t_k^{\frac{1}{k}-1}x^{t_1+\cdots+t_k}dt_1\cdots dt_k.\end{align*}

Integrating by parts with respect to $t_k$ gives

(4·21) \begin{align} \int_{\frac{1}{\log x}}^{1-t_1-\cdots-t_{k-1}}t_k^{\frac{1}{k}-1}x^{t_k}dt_k= &(1-t_1-\cdots-t_{k-1})^{\frac{1}{k}-1}\frac{x^{1-t_1-\cdots-t_{k-1}}}{\log x}-\frac{e}{(\!\log x)^{\frac{1}{k}}} \nonumber\\[5pt] &+\left(1-\frac{1}{k}\right)\frac{1}{\log x}\int_{\frac{1}{\log x}}^{1-t_1-\cdots t_{k-1}}t_k^{\frac{1}{k}-2}x^{t_k}dt_k.\end{align}

Therefore, the contribution of the first term of (4·21) to the integral $I_1$ is

\begin{align*}\frac{x}{\Gamma\left(\frac{1}{k} \right)^k}\int_{{\frac{1}{\log x}}}^{u_1}\cdots\int_{{\frac{1}{\log x}}}^{u_{k-1}}t_1^{\frac{1}{k}-1}\cdots t_{k-1}^{\frac{1}{k}-1}(1-t_1-\cdots t_{k-1})^{\frac{1}{k}-1}dt_1\cdots dt_{k-1}.\end{align*}

Note that

(4·22) \begin{equation} \begin{gathered}[b] \!\!F(u_1,\ldots,u_{k-1})x=\frac{x}{\Gamma\left(\frac{1}{k} \right)^k}\int_{{\frac{1}{\log x}}}^{u_1}\cdots\int_{{\frac{1}{\log x}}}^{u_{k-1}}t_1^{\frac{1}{k}-1}\cdots t_{k-1}^{\frac{1}{k}-1}(1-t_1-\cdots t_{k-1})^{\frac{1}{k}-1}dt_1\cdots dt_{k-1} \\[5pt] +O\left(x\sum_{j=1}^{k-1}\int_0^{\frac{1}{\log x}}\underset{\substack{0 \leqslant t_i \leqslant u_i\\[2pt] i \neq j}}{\int\cdots\int}t_1^{\frac{1}{k}-1}\cdots t_{k-1}^{\frac{1}{k}-1}(1-t_1-\cdots t_{k-1})^{\frac{1}{k}-1}dt_1\cdots dt_{k-1} \right). \end{gathered}\end{equation}

Without loss of generality, it suffices to bound the term where $j=k-1.$ Similar to (4·19), we have

(4·23) \begin{align} x\int_0^{\frac{1}{\log x}}t_{k-1}^{\frac{1}{k}-1}&\left(\int_0^{u_1}\cdots\int_0^{u_{k-2}}t_1^{\frac{1}{k}-1}\cdots t_{k-2}^{\frac{1}{k}-1}(1-t_1-\cdots-t_{k-1})^{\frac{1}{k}-1}dt_1\cdots dt_{k-2}\right)dt_{k-1} \nonumber \\[5pt] &\ll x\int_0^{\frac{1}{\log x}}t_{k-1}^{\frac{1}{k}-1}(1-t_{k-1})^{-\frac{1}{k}}dt_{k-1}\ll \frac{x}{(\!\log x)^{\frac{1}{k}}}.\end{align}

On the other hand, the contribution of the second term of (4·21) to the integral $I_1$ is

\begin{align*}\ll (\!\log x)^{1-\frac{1}{k}}\prod_{j=1}^{k-1}\int_{\frac{1}{\log x}}^{u_j}t_j^{\frac{1}{k}-1}x^{t_j}dt_j&\ll (\!\log x)^{1-\frac{1}{k}}\prod_{j=1}^{k-1} u_j^{\frac{1}{k}-1}\frac{x^{u_j}}{\log x}.\end{align*}

If $u_j > {1}/{2k}$ for some $j=1,\ldots, k-1,$ then this is

(4·24) \begin{align} \ll x(\!\log x)^{-(k-1)/k} \leqslant \frac{x}{(\!\log x)^{\frac{1}{k}}},\end{align}

as $k \geqslant 2$ . Otherwise, the contribution is

\begin{align*}\ll x^{u_1+\cdots+u_{k-1}}(\!\log x)^{1-\frac{1}{k}} \ll x^{1/2}.\end{align*}

We also have

\begin{align*}\int_{\frac{1}{\log x}}^{1-t_1-\cdots t_{k-1}}t_k^{\frac{1}{k}-2}x^{t_k}dt_k &\ll (1-t_1-\cdots-t_{k-1})^{\frac{1}{k}-2}\frac{x^{1-t_1-\cdots-t_{k-1}}}{\log x}\end{align*}

so that the contribution of the last term of (4·21) to the integral $I_1$ is

\begin{align*}\ll \frac{x}{\log x}&\int_{\frac{1}{\log x}}^{u_1}\cdots\int_{\frac{1}{\log x}}^{u_{k-1}}t_1^{\frac{1}{k}-1}\cdot st_{k-1}^{\frac{1}{k}-1}(1-t_1-\cdots-t_{k-1})^{\frac{1}{k}-2}dt_1\cdots dt_{k-1}.\end{align*}

Making the change of variables $s=1-t_1-\cdots-t_{k-1},$ this is bounded by

(4·25) \begin{align} \frac{x}{\log x} \int_{\frac{1}{\log x}}^{1-\frac{k-1}{\log x}}s^{\frac{1}{k}-2}&\left(\int_{\frac{1}{\log x}}^{u_1}\cdots\int_{\frac{1}{\log x}}^{u_{k-2}}t_1^{\frac{1}{k}-1}\cdots t_{k-2}^{\frac{1}{k}-1}(1-s-t_1-\cdots-t_{k-2})^{\frac{1}{k}-1}dt_1\cdots dt_{k-2}\right)ds.\end{align}

Similar to (4·19), the integral in the parentheses is $\ll (1-s)^{-\frac{1}{k}},$ and so (4·25) is

(4·26) \begin{align} \ll \frac{x}{\log x} \int_{\frac{1}{\log x}}^{1-\frac{k-1}{\log x}}s^{\frac{1}{k}-2}(1-s)^{-\frac{1}{k}}ds \ll \frac{x}{(\!\log x)^{\frac{1}{k}}}.\end{align}

Collecting the main term of (4·22) and the error terms (4·23), (4·24) and (4·26), the lemma follows.

Lemma 4·5. The second integral $I_2$ is

\begin{align*}\ll \frac{x}{(\!\log x)^{\frac{1}{k}}}.\end{align*}

Proof. The integral $I_2$ is bounded by

\begin{align*}\underset{\substack{l_j \leqslant u_j \log x , j=1,\ldots, k-1\\[2pt] l_{k} \leqslant \log x-l_1-\cdots-l_{k-1} }}{\sum \cdots \sum} I_{2}^{(\boldsymbol{l})},\end{align*}

where

\begin{align*}I_{2}^{(\boldsymbol{l})}\;:\!=\;\underset{x_j \in [e^{l_j}, e^{l_j+1}], j=1,\dots,k}{\int \cdots \int}\frac{1}{(\!\log x_1)^2}\cdots\frac{1}{(\!\log x_k)^2}dR(x_1,\ldots,x_k).\end{align*}

By integration by parts, for each $\boldsymbol{l},$ the integral $I_{2}^{(\boldsymbol{l})}$ is bounded by

\begin{align*}\sum_{J \subseteq [k]}\underset{x_j \in [e^{l_j}, e^{l_j+1}], j\in J}{\int \cdots \int}\!\left(\sum_{x_j \in \{e^{l_j}, e^{l_j+1}\}, j \notin J} |R(x_1,\ldots, x_k)|\!\right)\!\prod_{j \notin J}\frac{1}{{l_j}^2}\prod_{j \in J}\left|d\!\left(\frac{1}{(\!\log x_j)^2} \right) \right|\;=\!:\;\sum_{J \subseteq [k]} I_{2}^{(\boldsymbol{l};J)}.\end{align*}

For each subset $J \subseteq [k],$ the integral $I_{2}^{(\boldsymbol{l};J)}$ is

\begin{align*}\ll \left(\max_{x_j \in [e^{l_j}, e^{l_j+1}], j=1,\ldots,k}|R(x_1,\ldots,x_k)|\right)\underset{x_j \in [e^{l_j}, e^{l_j+1}], j=1,\ldots,k}{\int \cdots \int} \prod_{j \notin J}\frac{1}{x_j (\!\log x_j)^2}\prod_{j \in J} & \frac{1}{x_j (\!\log x_j)^3}\nonumber\\[5pt] \times & dx_1 \cdots dx_k.\end{align*}

Applying Lemma 4·3, this is

\begin{align*}\ll \sum_{i=1}^k\underset{x_j \in [e^{l_j}, e^{l_j+1}], j=1,\ldots,k}{\int \cdots \int} \frac{1}{\log x_i} \prod_{j \notin J}\frac{1}{(\!\log x_j)^{1-\frac{1}{k}}} \prod_{j \in J} \frac{1}{(\!\log x_j)^{2-\frac{1}{k}}} dx_1\cdots dx_k.\end{align*}

Summing over every $\boldsymbol{l},$ we have

\begin{align*}\underset{\substack{l_j \leqslant u_j \log x , j=1,\ldots, k-1\\[2pt] l_k \leqslant \log x -l_1-\cdots-l_{k-1}}}{\sum \cdots \sum} I_2^{(\boldsymbol{l};J)}\ll \sum_{i=1}^k \int_{e}^{x^{u_1}e}\cdots \int_{e}^{x^{u_{k-1}}e}\int_{e}^{\frac{e^kx}{x_1\cdots x_{k-1}}} &\frac{1}{\log x_i} \prod_{j \notin J}\frac{1}{(\!\log x_j)^{1-\frac{1}{k}}} \\[5pt] \times & \prod_{j \in J} \frac{1}{(\!\log x_j)^{2-\frac{1}{k}}}dx_1\cdots dx_k.\end{align*}

To avoid repetitions, we only bound the contribution of the term where $i=k$ here. Making the change of variables $x_j=x^{t_j}$ for $j=1,\ldots,k-1$ , it becomes

\begin{align*}\frac{1}{(\!\log x)^{|J|}} \int_{\frac{1}{\log x}}^{u_1+\frac{1}{\log x}} \cdots \int_{\frac{1}{\log x}}^{u_{k-1}+\frac{1}{\log x}}\int_{\frac{1}{\log x}}^{1+\frac{k}{\log x}-t_1-\cdots-t_{k-1}}t_k^{-1} \prod_{j \notin J}t_j^{\frac{1}{k}-1}\prod_{j \in J}t_j^{\frac{1}{k}-2} x^{t_1+\cdots+t_k} dt_1\cdots dt_k,\end{align*}

which is

(4·27) \begin{align} \ll \int_{\frac{1}{\log x}}^{u_1+\frac{1}{\log x}} \cdots \int_{\frac{1}{\log x}}^{u_{k-1}+\frac{1}{\log x}}\int_{\frac{1}{\log x}}^{1+\frac{k}{\log x}-t_1-\cdots-t_{k-1}}&t_1^{\frac{1}{k}-1}\cdots t_{k-1}^{\frac{1}{k}-1}t_k^{\frac{1}{k}-2}x^{t_1+\cdots+t_k}dt_1\cdots dt_k.\end{align}

Integrating by parts with respect to $t_k$ gives

\begin{align*}\int_{\frac{1}{\log x}}^{1+\frac{k}{\log x}-t_1-\cdots-t_{k-1}} t_k^{\frac{1}{k}-2}x^{t_k}dt_k\ll \left(1+\frac{k}{\log x}-t_1-\cdots-t_{k-1} \right)^{\frac{1}{k}-2}\frac{x^{1-t_1-\cdots-t_{k-1}}}{\log x}.\end{align*}

Therefore, the expression (4·27) is

\begin{align*}\ll \frac{x}{\log x} \int_{\frac{1}{\log x}}^{u_1+\frac{1}{\log x}} \cdots \int_{\frac{1}{\log x}}^{u_{k-1}+\frac{1}{\log x}} t_1^{\frac{1}{k}-1}\cdots t_{k-1}^{\frac{1}{k}-1}\left(1+\frac{k}{\log x}-t_1-\cdots-t_{k-1} \right)^{\frac{1}{k}-2} dt_1\cdots d_{k-1}.\end{align*}

Similar to (4·19), this is

\begin{align*}\ll \frac{x}{\log x}\int_{\frac{2}{\log x}}^{1+\frac{1}{\log x}}s^{\frac{1}{k}-2}\left( 1+\frac{k}{\log x}-s \right)^{-\frac{1}{k}}ds\ll \frac{x}{(\!\log x)^{\frac{1}{k}}}.\end{align*}

Finally, the lemma follows from summing over every subset $J \subseteq [k].$

5. Proof of Theorem 1·2

We first define the function field analogue of the multiple Dirichlet series $\mathcal{D}(s_1,\dots,s_k).$

Definition 5·1. For $(s_1,\dots,s_k) \in \Omega$ , the multiple Dirichlet series $\mathcal{D}_{\mathbb{F}_q[x]}(s_1,\dots,s_k)$ is defined as

\begin{align*}\underset{F_1,\dots,F_k \in \mathcal{M}_q}{\sum\cdots\sum}\frac{\tau_k(F_1\cdots F_k)^{-1}}{q^{s_1\deg F_1+\cdots+s_k \deg F_k}}.\end{align*}

Having the multiple Dirichlet series $\mathcal{D}_{\mathbb{F}_q[x]}(s_1,\dots,s_k)$ in hand, it is now clear that we can follow exactly the same steps as before. Moreover, since the function field zeta function $\zeta_{\mathbb{F}_q}(s)$ never vanishes (see [ Reference Rosen18 , chapter 2]), some of the computations above can be simplified considerably. To avoid repetitions, the complete proof is omitted here.

6. Proof of Theorem 1·3

It is clear that one can argue similarly but a more direct and elementary proof is presented here. We begin with the combinatorial analogue of the mean of divisor functions.

Lemma 6·1. Let $\alpha \in \mathbb{C}\setminus \mathbb{Z}_{\leqslant 0}$ and $n\in \mathbb{Z}_{\geqslant 0}.$ Then we have

(6·1) \begin{align} \frac{1}{n!}\sum_{\sigma \in S_n}\tau_{\alpha}(\sigma)=\left(\substack{n+\alpha-1 \\[2pt] n}\right).\end{align}

Moreover, we have the estimate

(6·2) \begin{align} \left(\substack{n+\alpha-1 \\[2pt] n}\right)=\begin{cases}1 &\mbox{if } n = 0, \\[5pt] \frac{1}{\Gamma \left(\alpha \right)}n^{\alpha-1}\left(1+O_{\alpha}\left(\frac{1}{n}\right) \right) & \mbox{if } n\geqslant1.\end{cases}\end{align}

Proof. Although (6·1) is fairly standard, say for instance one may apply [ Reference Stanley19 , corollary 5·1·9] with $f \equiv \alpha$ , we provide a short proof here for the sake of completeness. Adopting the notations in Section 2, we write

\begin{align*}\sum_{\sigma \in S_n}\tau_{\alpha}(\sigma)&=\sum_{k=0}^{n}\left[{n\atop k}\right]\alpha^k.\end{align*}

Now any permutation $\sigma \in S_n$ with k disjoint cycles can be constructed by the following procedure. To begin with, there are $(n-1)(n-2)\cdots(n-i_1+1)$ ways of choosing $i_1-1$ distinct integers from $[n]\setminus\{1\}$ to form a cycle $C_1$ with length $i_1$ containing $1.$ Then, fix any integer $m \in [n]$ not contained in the cycle $C_1$ . Similarly, there are $(n-i_1-1)\cdots(n-i_1-i_2+1)$ ways of choosing $i_2-1$ integers from $[n]\setminus C_1$ to form another cycle $C_2$ with length $i_2$ containing m. Repeating the same procedure until $i_1+\cdots+i_k=n$ , we arrive at a permutation with k disjoint cycles.

Therefore, the expression (6·1) follows from the explicit formula

\begin{align*}\left[{n\atop k}\right]=\underset{i_1+\cdots+i_k=n}{\sum_{i_1=1}^n\cdots\sum_{i_k=1}^n}\frac{(n-1)!}{\prod_{j=1}^{k}(n-i_1-\cdots-i_j)},\end{align*}

which can be seen as the coefficient of ${\alpha}^k$ in the falling factorial

\begin{align*}(n+\alpha-1)(n+\alpha-2)\cdots(\alpha+1)\alpha.\end{align*}

On the other hand, to prove (6·2), we express the binomial coefficient as a ratio of gamma functions, followed by the application of Stirling’s formula.

Similar to Theorem 1·1, without loss of generality we shall assume $u_1,\ldots u_{k-1}, 1-u_1-\cdots-u_{k-1} \geqslant {1}/{n}.$ Interchanging the order of summation, we have

(6·3) \begin{align} \frac{1}{n!}\sum_{\sigma \in S_n}\frac{1}{\tau_k(\sigma)}\underset{{\substack{[n]=A_1 \sqcup \cdots \sqcup A_k\\[2pt] \sigma(A_i)=A_i, i=1,\dots,k \\[5pt] 0 \leqslant |A_i| \leqslant nu_i, i=1,\dots,k-1}}}{\sum\cdots\sum}1&=\frac{1}{n!}\underset{\substack{0 \leqslant m_i \leqslant nu_i \\[2pt] i=1,\dots,k-1}}{\sum\cdots\sum}\left(\substack{n \\[2pt] m_1,\dots,m_k}\right)\prod_{i=1}^k\left(\sum_{\sigma \in S_{m_i}}\frac{1}{\tau_k(\sigma)} \right) \nonumber\\[5pt] &=\underset{\substack{0 \leqslant m_i \leqslant nu_i \\[2pt] i=1,\dots,k-1}}{\sum\cdots\sum}\prod_{i=1}^k \left(\frac{1}{m_i!}\sum_{\sigma \in S_{m_i}}\frac{1}{\tau_k(\sigma)} \right),\end{align}

where $m_k\;:\!=\;n-m_1-\cdots -m_{k-1}.$

Note that $\tau_k(\sigma)^{-1}=\tau_{1/k}(\sigma).$ Applying (6·1) from Lemma 6·1, the expression (6·3) equals

(6·4) \begin{align} \underset{\substack{0 \leqslant m_i \leqslant nu_i \\[2pt] i=1,\dots,k-1}}{\sum\cdots\sum}\prod_{i=1}^k\left(\substack{m_i+\frac{1}{k}-1 \\[2pt] m_i}\right).\end{align}

Let $I \subseteq [k-1]$ be a nonempty subset. Then using (6·2) from Lemma 6·1, the contribution of $m_i=0$ to (6·4) for $i\in I$ is

\begin{align}\ll \underset{\substack{1 \leqslant m_i \leqslant nu_i\\[2pt] i \notin I}}{\sum\cdots\sum}\left(\prod_{i \notin I}m_i^{\frac{1}{k}-1}\right)m_k^{\frac{1}{k}-1}= n^{-\frac{|I|}{k}}\underset{\substack{1 \leqslant m_i \leqslant nu_i \\[2pt] i \notin I}}{\sum\cdots\sum}&\left(\prod_{i \notin I}\left(\frac{m_i}{n}\right)^{\frac{1}{k}-1}\right)\nonumber \\[5pt] \times & \left(1-\frac{m_1}{n}-\cdots-\frac{m_{k-1}}{n}\right)^{\frac{1}{k}-1}n^{-(k-|I|)}, \nonumber\end{align}

which is

(6·5) \begin{align} \ll n^{-\frac{|I|}{k}} \underset{\substack{0 \leqslant t_i \leqslant u_i \\[2pt] i \notin I}}{\int\cdots\int}\left(\prod_{i \notin I}t_i^{\frac{1}{k}-1}\right)\left(1-\sum_{i \notin I}t_i\right)^{\frac{1}{k}-1}\prod_{i \notin I}dt_i\ll n^{-\frac{|I|}{k}}.\ \end{align}

Also, the contribution of $m_j>nu_j-1$ for some $j=1,\dots,k-1$ to (6·4) given that $m_1,\dots,m_k\geqslant 1$ is

\begin{align}\ll \sum_{j=1}^{k-1}{(nu_j)}^{\frac{1}{k}-1}\underset{\substack{1 \leqslant m_i \leqslant nu_i\\[2pt] i\neq j}}{\sum\cdots\sum}\prod_{\substack{i=1\\[2pt] i\neq j}}^k m_i^{\frac{1}{k}-1}=n^{-\frac{1}{k}} \sum_{j=1}^{k-1}{(nu_j)}^{\frac{1}{k}-1}\underset{\substack{1 \leqslant m_i \leqslant nu_i\\[2pt] i\neq j}}{\sum\cdots\sum}\prod_{\substack{i=1\\[2pt] i\neq j}}^k\left(\frac{m_i}{n}\right)^{\frac{1}{k}-1}. \nonumber\end{align}

Since $u_j \geqslant {1}/{n}$ for $j=1,\dots,k-1,$ this is

(6·6) \begin{align} &\ll n^{-\frac{1}{k}} \sum_{j=1}^{k-1} (nu_j)^{\frac{1}{k}-1}\underset{\substack{0 \leqslant t_i \leqslant u_i, i=1,\dots,k-1\\[2pt] i \neq j}}{\int\cdots\int}\prod_{\substack{i=1\\[2pt] i \neq j}}^{k-1}t_i^{\frac{1}{k}-1}(1-t_1-\cdots-t_{k-1})^{\frac{1}{k}-1}dt_1\cdots dt_{k-1} \nonumber\\[5pt] &\ll n^{-\frac{1}{k}} \sum_{j=1}^{k-1} (nu_j)^{\frac{1}{k}-1}\left(1-u_j\right)^{-\frac{1}{k}}\ll n^{-\frac{1}{k}}.\end{align}

Collecting the error terms (6·5) and (6·6), the expression (6·4) equals

(6·7) \begin{align} \underset{\substack{1 \leqslant m_i \leqslant nu_i-1 \\[2pt] i=1,\dots,k-1}}{\sum\cdots\sum}\prod_{i=1}^k\left(\substack{m_i+\frac{1}{k}-1\\[2pt] m_i}\right)+O(n^{-\frac{1}{k}}).\end{align}

Applying (6·2) from Lemma 6·1, the main term of (6·7) is the Riemann sum

(6·8) \begin{align}\frac{1}{\Gamma \left(\frac{1}{k} \right)^k}\underset{\substack{1 \leqslant m_i \leqslant nu_i-1 \\[2pt] i=1,\dots,k-1}}{\sum\cdots\sum}\prod_{i=1}^k \left(\frac{m_i}{n}\right)^{\frac{1}{k}-1}\frac{1}{n^{k-1}},\end{align}

with an error term

(6·9) \begin{align} \ll \frac{1}{n}\sum_{j=1}^{k}\left(\frac{m_j}{n}\right)^{\frac{1}{k}-2}\underset{\substack{1 \leqslant m_i \leqslant nu_i-1\\[2pt] i=1,\dots,k-1}}{\sum\cdots\sum}\prod_{\substack{i=1\\[2pt] i \neq j}}^k\left(\frac{m_i}{n}\right)^{\frac{1}{k}-1}\frac{1}{n^{k-1}}.\end{align}

Let us first bound the error term (6·9). For each $j=1,\dots,k-1,$ we have

\begin{align}\frac{1}{n}&\sum_{1 \leqslant m_j \leqslant nu_j-1}\left(\frac{m_j}{n} \right)^{\frac{1}{k}-2}\underset{\substack{1 \leqslant m_i \leqslant nu_i-1 \\[2pt] i=1,\dots,k-1, i\neq j}}{\sum\cdots\sum}\prod_{\substack{i=1\\[2pt] i \neq j}}^{k-1}\left(\frac{m_i}{n} \right)^{\frac{1}{k}-1}\frac{1}{n^{k-1}} \nonumber\\[5pt] &\ll \frac{1}{n} \int_{\frac{1}{n}}^{1-\frac{1}{n}} t_j^{\frac{1}{k}-2}\underset{\substack{0 \leqslant t_i \leqslant u_i\\[2pt] i=1,\dots,k-1, i \neq j}}{\int\cdots\int}\prod_{\substack{i=1\\[2pt] i \neq j}}^{k-1}t_i^{\frac{1}{k}-1}(1-t_1-\cdots-t_{k-1})^{\frac{1}{k}-1}dt_1\cdots dt_{k-1} \nonumber\\[5pt] &\ll \frac{1}{n}\int_{\frac{1}{n}}^{1-\frac{1}{n}}t_j^{\frac{1}{k}-2}(1-t_j)^{-\frac{1}{k}}dt_j\ll n^{-\frac{1}{k}}. \nonumber\end{align}

Arguing similarly for $j=k,$ we also have

\begin{align}\frac{1}{n}\underset{\substack{1 \leqslant m_i \leqslant nu_i-1 \\[2pt] i=1,\dots,k-1}}{\sum\cdots\sum}\prod_{i=1}^{k-1}\left(\frac{m_i}{n} \right)^{\frac{1}{k}-1}\left(\frac{m_k}{n} \right)^{\frac{1}{k}-2}\frac{1}{n^{k-1}} \ll n^{-\frac{1}{k}}.\nonumber\end{align}

Therefore, the error term (6·9) is $\ll n^{-\frac{1}{k}}$ and we are left with the main term (6·8).

The distribution function $F(u_1,\ldots,u_{k-1})$ equals

(6·10) \begin{align} \frac{1}{\Gamma \left(\frac{1}{k} \right)^k}&\underset{\substack{1 \leqslant m_i \leqslant nu_i-1 \\[2pt] i=1,\dots,k-1 }}{\sum\cdots\sum}\int_{\frac{m_1}{n}}^{\frac{m_1+1}{n}}\cdots\int_{\frac{m_{k-1}}{n}}^{\frac{m_{k-1}+1}{n}}t_1^{\frac{1}{k}-1}\cdots t_{k-1}^{\frac{1}{k}-1}(1-t_1-\cdots-t_{k-1})^{\frac{1}{k}-1}dt_1\cdots dt_{k-1} \nonumber\\[3pt] &+O\left(\sum_{j=1}^{k-1}\int_{0}^{\frac{1}{n}}t_j^{\frac{1}{k}-1}\underset{\substack{0 \leqslant t_i \leqslant u_i\\[2pt] i\neq j}}{\int\cdots\int}\prod_{\substack{i=1\\[2pt] i \neq j}}^{k-1}t_j^{\frac{1}{k}-1}(1-t_1-\cdots-t_{k-1})^{\frac{1}{k}-1}dt_1\cdots dt_{k-1} \right) \nonumber\\[3pt] &+O\left(\sum_{j=1}^{k-1}\int_{u_j-\frac{1}{n}}^{u_j}t_j^{\frac{1}{k}-1}\underset{\substack{0 \leqslant t_i \leqslant u_i \\[2pt] i\neq j}}{\int\cdots\int}\prod_{\substack{i=1\\[2pt] i \neq j}}^{k-1}t_j^{\frac{1}{k}-1}(1-t_1-\cdots-t_{k-1})^{\frac{1}{k}-1}dt_1\cdots dt_{k-1} \right).\end{align}

The first error term in (6·10) is

(6·11) \begin{align} \ll \sum_{j=1}^{k-1}\int_{0}^{\frac{1}{n}}t_j^{\frac{1}{k}-1}(1-t_j)^{-\frac{1}{k}}dt_j\ll n^{-\frac{1}{k}}.\end{align}

The second error term in (6·10) is

(6·12) \begin{align} \ll \sum_{j=1}^{k-1}\int_{u_j-\frac{1}{n}}^{u_j}t_j^{\frac{1}{k}-1}(1-t_j)^{-\frac{1}{k}}dt_j&\leqslant \sum_{j=1}^{k-1}\int_{0}^{\frac{1}{n}}t_j^{\frac{1}{k}-1}(1-t_j)^{-\frac{1}{k}}dt_j \nonumber\\[3pt] &\ll n^{-\frac{1}{k}}.\end{align}

By Taylor’s theorem, for $(t_1,\dots,t_{k-1}) \in\left[{m_1}/{n}, ({m_1+1})/{n} \right]\times \cdots \times [({m_{k-1}})/{n}, ({m_{k-1}+1})/{n}],$ we have

\begin{align*}t_1^{\frac{1}{k}-1}\cdots t_{k-1}^{\frac{1}{k}-1}(1-t_1-\cdots-t_{k-1})^{\frac{1}{k}-1}=\prod_{i=1}^k\left(\frac{m_i}{n}\right)^{\frac{1}{k}-1}+O\!\left(\!\frac{1}{n}\sum_{j=1}^{k}\left(\frac{m_j}{n}\right)^{\frac{1}{k}-2}\prod_{\substack{i=1\\[2pt] i \neq j}}^k\!\left(\!\frac{m_i}{n}\right)^{\frac{1}{k}-1}\!\right)\!.\end{align*}

Using the approximation, we conclude from (6·10), (6·11) and (6·12) that

\begin{align*}F(u_1,\ldots,u_{k-1})=&\frac{1}{\Gamma \left(\frac{1}{k} \right)^k}\underset{\substack{1 \leqslant m_i \leqslant nu_i-1 \\[2pt] i=1,\dots,k-1}}{\sum\cdots\sum}\prod_{i=1}^k \left(\frac{m_i}{n}\right)^{\frac{1}{k}-1}\frac{1}{n^{k-1}}+O\left(n^{-\frac{1}{k}} \right)\\[5pt] &+O\left(\frac{1}{n}\underset{\substack{1 \leqslant m_i \leqslant nu_i-1 \\[2pt] i=1,\dots,k-1}}{\sum\cdots\sum}\sum_{j=1}^{k}\left(\frac{m_j}{n}\right)^{\frac{1}{k}-2}\prod_{\substack{i=1\\[2pt] i \neq j}}^k\left(\frac{m_i}{n}\right)^{\frac{1}{k}-1}\frac{1}{n^{k-1}} \right),\end{align*}

and the last error term here is exactly the same as (6·9), which is again $\ll n^{-\frac{1}{k}}.$

7. Factorisation into k parts in the general setting

With a view to model Dirichlet distribution with arbitrary parameters, we further explore the factorisation of integers into k parts in the general setting using multiplicative functions of several variables defined below.

Definition 7·1. An arithmetic function of k variables $F \colon \mathbb{N}^k \to \mathbb{C}$ is said to be multiplicative if it satisfies the condition $F(1,\ldots,1)=1$ and the functional equation

\begin{align*}F(m_1n_1,\ldots,m_kn_k)=F(m_1,\ldots,m_k)F(n_1,\ldots,n_k)\end{align*}

whenever $(m_1\cdots m_k, n_1\cdots n_k)=1,$ or equivalently,

\begin{align*}F(n_1,\ldots,n_k)=\prod_{p}F(p^{v_p(n_1)},\ldots,p^{v_p(n_k)}),\end{align*}

where $v_p(n)\;:\!=\;\max\{k \geqslant 0 \, : \, p^k | n\}.$

To adapt the proof of Theorem 1·1, we consider the following class of multiplicative functions.

For instance, the multiplicative function $F(n_1,\ldots,n_k)=\tau_k(n_1\cdots n_k)^{-1}$ belongs to the class

Applying the Mellin transform to (higher derivatives of) the multiple Dirichlet series

\begin{align*}\mathcal{D}_F(s_1,\ldots,s_k)\;:\!=\;\sum_{n_1=1}^{\infty}\cdots\sum_{n_k=1}^{\infty}\frac{F(n_1,\ldots,n_k)}{n_1^{s_1}\cdots n_k^{s_k}}\end{align*}

as before, one can prove the following generalisation of Lemma 4·3.

Lemma 7·1. Given a multiplicative function of k variables $ F \in \mathcal{M}(\boldsymbol{\alpha};\boldsymbol{\beta}, \boldsymbol{c}, \boldsymbol{\delta}).$ Let $m \geqslant 2$ be an integer and $x_1,\dots,x_k \geqslant e.$ We denote by $S_F(x_1,\dots,x_k;m)$ the weighted sum

\begin{align*}\sum_{d_1\leqslant x_1}\cdots\sum_{d_k\leqslant x_k}(\!\log d_1)^m \cdots (\!\log d_k)^m F(d_1,\ldots,d_k).\end{align*}

Then there exists $m_0=m_0(\boldsymbol{\alpha},\boldsymbol{\beta}, \boldsymbol{c}, \boldsymbol{\delta})$ such that for any integer $m\geqslant m_0,$ we have

\begin{align*}S_F(x_1,\ldots,x_k;m) = \prod_{j=1}^k\frac{1}{\Gamma(\alpha_j)}\int_{1}^{x_j}(\!\log y_j)^{\alpha_j+m-1} dy_j + R_F(x_1,\ldots,x_k;m)\end{align*}

with

\begin{align*}R_F(x_1,\dots,x_k;m) \ll x_1\cdots x_k\sum_{j=1}^k \left(\prod_{\substack{i=1\\[2pt] i \neq j}}^k (\!\log x_i)^{\alpha_i+m-1}\right) (\!\log x_j)^{\alpha_j+m-2}.\end{align*}

To model the Dirichlet distribution by factorizing integers into k parts, we consider the following class of pairs of multiplicative functions.

Then, applying Lemma 7·1 followed by partial summation as before, one can prove the following generalisation of Theorem 1·1.

Theorem 7·1. Let $(f;G)$ be a pair of multiplicative functions belonging to the class $\mathcal{M}_{\theta}(\boldsymbol{\alpha}).$ Then uniformly for $x \geqslant 2$ and $u_1,\ldots,u_{k-1} \geqslant 0$ satisfying $u_1+\cdots+u_{k-1}\leqslant 1,$ we have

\begin{align*}\left(\sum_{m \leqslant x}f(m)\right)^{-1}&\sum_{n\leqslant x}f(n)\left( \sum_{n=e_1\cdots e_k}G(e_1,\ldots,e_k) \right)^{-1}\underset{n=d_1\cdots d_{k}}{\sum_{d_1 \leqslant n^{u_1}}\cdots\sum_{d_{k-1} \leqslant n^{u_{k-1}}}\sum_{d_k \leqslant n}}G(d_1,\ldots,d_k)\\[5pt] =&F_{\boldsymbol{\alpha}}(u_1,\ldots, u_{k-1})+O\left(\frac{1}{(\!\log x)^{\min \{1,\alpha_1,\ldots,\alpha_k\}}} \right).\end{align*}

Finally, we conclude with the following generalisation of Corollary 1·1.

Corollary 7·1. Given a pair of multiplicative functions $(f;G)$ belonging to the class $\mathcal{M}_{\theta}(\boldsymbol{\alpha}).$ For $x\geqslant 1,$ let n be a random integer chosen from [1, x] with probability $\left(\sum_{m \leqslant x}f(m)\right)^{-1}f(n)$ and $(d_1,\ldots,d_k)$ be a random k-tuple chosen from the set of all possible factorisation $\{(m_1,\ldots,m_k) \in \mathbb{N}^k\, : \, n=m_1\cdots m_k\}$ with probability $\left( \sum_{n=e_1\cdots e_k}G(e_1,\ldots,e_k) \right)^{-1}G(d_1,\ldots,d_k)$ . Then as $x \to \infty,$ we have the convergence in distribution

\[\left(\frac{\log d_1}{\log n}, \ldots, \frac{\log d_k}{\log n}\right) \xrightarrow[]{\;\;\;\;d\;\;\;\;}\mathrm{Dir}\left(\alpha_1,\ldots,\alpha_k \right).\]

Example 7·1. For $k\geqslant 2,$ let $\theta, \lambda_1, \ldots, \lambda_k>0.$ We consider the pair of multiplicative functions

\begin{align*}f(n)=\tau_{\theta}(n);\quad G(d_1,\ldots,d_k)= \tau_{\lambda_1}(d_1) \cdots\tau_{\lambda_k}(d_k).\end{align*}

Then the Dirichlet distribution of dimension k

\begin{align*}\mathrm{Dir}\left(\frac{\theta\lambda_1}{\lambda_1+\cdots+\lambda_k},\ldots,\frac{\theta\lambda_k}{\lambda_1+\cdots+\lambda_k} \right)\end{align*}

can be modelled in the sense of Corollary 7·1. In particular, when $\theta, \lambda_1,\ldots, \lambda_k=1,$ it reduces to Theorem 1·1.

Example 7·2. For $q\geqslant 3,$ let $\{a_1,\ldots,a_{\varphi(q)}\}$ be a reduced residue system $\ (\mathrm{mod}\ q)$ . We consider the pair of multiplicative functions

Then the Dirichlet distribution of dimension $\varphi(q)$

\begin{align*}\mathrm{Dir}\left(\frac{1}{\varphi(q)},\ldots,\frac{1}{\varphi(q)} \right)\end{align*}

can be modelled in the sense of Corollary 7·1. In particular, when $q=4,$ it reduces to [ Reference Montgomery and Vaughan15 , exercise 6·2·22].

Example 7·3. For $k\geqslant 2,$ we consider the pair of multiplicative functions

Then the Dirichlet distribution of dimension k

\begin{align*}\mathrm{Dir}\left(\frac{1}{2k},\ldots,\frac{1}{2k} \right)\end{align*}

can be modelled in the sense of Corollary 7·1. In particular, when $k=2,$ it reduces to [ Reference Daoud, Hidri and Naimi6 , theorem 2].

Example 7·4. For $k\geqslant 2,$ we consider the pair of multiplicative functions

Then the Dirichlet distribution of dimension k

\begin{align*}\mathrm{Dir}\left(\frac{1}{k},\ldots,\frac{1}{k} \right)\end{align*}

can be modelled in the sense of Corollary 7·1. In particular, when $k=2,$ it reduces to [ Reference Feng and Cui8 , theorem 2] with $y=x.$

Example 7·5. For $k\geqslant 2$ , let $\mathcal{R}$ be a subset of $\{\{i,j\}\,:\, 1 \leqslant i \neq j \leqslant k\}.$ We consider the pair of multiplicative functions

\begin{align*}f(n) \equiv 1;\quad G(d_1,\ldots,d_k)=\begin{cases}1 & \mbox{if $(d_i,d_j)=1$ whenever$\{i,j\}\notin \mathcal{R},$}\\[5pt] 0 & \mbox{otherwise.}\end{cases}\end{align*}

Then the Dirichlet distribution of dimension k

\begin{align*}\mathrm{Dir}\left( \frac{1}{k},\ldots,\frac{1}{k} \right)\end{align*}

can be modelled in the sense of Corollary 7·1. In particular, when $k=2^r$ for $r \geqslant 2,$ it reduces to [ Reference de la Bretèche and Tenenbaum4 , théorème 1·1] with a suitable subset $\mathcal{R}$ via total decomposition sets (see [ Reference Hall12 , theorem 0·20]), which is itself a generalisation of [ Reference Bareikis and Mac̆iulis1 , theorem 2·1] for $r=2.$

Example 7·6. For $k\geqslant 3$ , we consider the pair of multiplicative functions

\begin{align*}f(n) \equiv 1;\quad G(d_1,\ldots,d_k)=\prod_{j=1}^{k-1}\frac{1}{\tau(d_j\cdots d_k)}.\end{align*}

Then the Dirichlet distribution of dimension k

\begin{align*}\mathrm{Dir}\left(\frac{1}{2},\frac{1}{4},\ldots,\frac{1}{2^{k-2}},\frac{1}{2^{k-1}},\frac{1}{2^{k-1}} \right)\end{align*}

can be modelled in the sense of Corollary 7·1. In particular, when $k=3,$ it reduces to [ Reference de la Bretèche and Tenenbaum4 , théorème 1·2].

Unsurprisingly, we expect that Theorem 7·1 should also hold for polynomials or permutations. Specifically, in the realm of permutations, the counterpart to multiplicative functions is the generalised Ewens measure (see [ Reference Elboim and Gorodetsky7 ]). Detailed proofs will be provided in the author’s doctoral thesis.