# Residues of Zeta-like Function

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?

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.
• 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 $f(0,0)=0$? Sep 8, 2018 at 17:55
• 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). Sep 8, 2018 at 19:33