I'm looking for the residues of the following function $$s \mapsto\sum^\infty_{m,n =1} (m+n) \left[ amn + (m-n)^2 \right]^{-s}$$ at $s=\frac{1}{2}$ and $s=\frac{3}{2}$, where $a$ is some real positive number.

Yet, I have literally no idea how to precedure here. I tried to rearrange the terms in order to get some well known zeta function but it failed. Is there any useful integral representation or any simple trick I'm not aware of?

Thanks in advance!


1 Answer 1


The following is just a sketch, for the detail you can find in the reference of Zagier [http://people.mpim-bonn.mpg.de/zagier/files/scanned/ValeursZeta/ZetaFunctionRQF.pdf]. Let us denote $$f(x,y)=(x+y)e^{-(axy+(x-y)^2)}.$$ Then you can check that $$F(s):=\sum_{m,n\ge 1}\frac{m+n}{(amn+(m-n)^2)^s}=\frac{1}{\Gamma(s)}\int_{0}^{\infty}u^{s-3/2}\left(\sum_{m,n\ge 1}f(\sqrt{u}m,\sqrt{u}n)\right)\,du$$ by direct calculation. Then, from a result of Zagier [page 10th of http://people.mpim-bonn.mpg.de/zagier/files/scanned/ValeursZeta/ZetaFunctionRQF.pdf], actually the Euler–Maclaurin formula, we have the following asymptotic expansion: \begin{align} \sum_{m,n\ge 1}f(mt,nt)\sim&\frac{1}{t^2}\int_{{\mathbb{R}}_+^2}f(x,y)\,dx\,dy+\sum_{r,s\ge 0}\beta_r\beta_sf^{(r,s)}(0,0)t^{r+s}\\ &+\frac{1}{t}\sum_{r\ge 0}\beta_rt^r\left(\int_{\mathbb{R}_+}f^{(0,r)}(x,0)\,dx+\int_{\mathbb{R}_+}f^{(r,0)}(0,y)\,dy\right) \end{align} for $t\rightarrow 0^+$, where $\beta_r=(-1)^rB_{r+1}/(r+1)!, r\in\mathbb{Z}_{\ge 0}$ and $B_r$ denote the $r$-th Bernoulli number. There we have as $|u|\le 1,$ $$\sum_{m,n\ge 1}f(\sqrt{u}m,\sqrt{u}n)=\frac{I_f}{u}+\frac{c_{-1}}{u^{1/2}}+c_0+O(u^{1/2})$$ with $I_f,c_{-1}, c_0$ be defined as above asymptotic expansion. Hence we can obtain that $$F(s)=\frac{1}{\Gamma(s)}\int_{0}^{1}u^{s-3/2}\left(\frac{I_f}{u}+\frac{c_{-1}}{u^{1/2}}+c_0\right)\,du+h(u)$$ with $h(u)$ is an analytic function on $\Re(s)>0$. Moreover, for $\Re(s)>3/2$, $$F(s)=\frac{1}{\Gamma(s)}\left(\frac{I_f}{s-3/2}+\frac{c_{-1}}{s-1}+\frac{c_0}{s-1/2}\right)+h(s).$$ This means that $F(s)$ can analytic continuation to all $\Re(s)>0$. The following is easy and I omit the detail.

  • $\begingroup$ My previous comment is wrong, it is a one-sided double sum so is not directly related to Epstein zeta (I do not know how to delete a comment). I agree with Zhou's answer. $\endgroup$ Sep 8, 2018 at 13:16
  • $\begingroup$ Just to make sure I understand. The residue at $s=\frac{3}{2}$ is given by $I_f = \int_{\mathbb{R}_+^2} f(x,y) dx dy$, and at $s=\frac{1}{2}$ by $c_0=0$ since $f(0,0)=0$? $\endgroup$
    – YoungMath
    Sep 8, 2018 at 17:55
  • $\begingroup$ For $s=3/2$ yes (result $\sqrt{\pi}/a$), for $s=1/2$ no, you must take into account the coefficient of $\beta_1t^1$ (again trivial to compute as Zhou says, but I am too lazy to work it out). $\endgroup$ Sep 8, 2018 at 19:33
  • $\begingroup$ Professor @HenriCohen to delete a comment click the x that appears at the right when you click near your comment. $\endgroup$
    – efs
    Sep 9, 2018 at 1:47
  • $\begingroup$ @YoungMath It is very strage that the same paper has two errors (the other being the computation of your other question). Are you sure you didn't missed something while writing the functions here on MO? If they are actual errors of the paper, then maybe some other articles that cite the one you are studying hve corrected them. $\endgroup$
    – efs
    Sep 9, 2018 at 1:52

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