Is it possible to express this sum exactly in closed form?

Question

Is it possible to express the following sum in closed form? $$S=\sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}e^{-\sqrt{n^2+m^2}}$$ I can compute this numerically to find $S\approx 6.5072$, but I can't think of any way to perform the sum analytically.

Answer 1

We can rewrite the sum by $$S= \sum_{m=1}^\infty r(m) e^{-\sqrt{m}} +1,$$ where $r(n)$ denotes the number of representations of $n$ as a sum of two squares as usal. By Gauß-formula we know that $$r(n) = \begin{cases} 4 (l_1+1) \cdot \ldots \cdot (l_r+1) & \text{if } m_1 \equiv \ldots \equiv m_s \equiv 0 \ \mathrm{mod} \ 2 \\ 0 & \text{otherwise} \end{cases}$$ if $n = 2^l p_1^{l_1} \cdot \ldots \cdot p_r^{l_r} q_1^{m_1} \cdot \ldots \cdot q_s^{m_s}$, where $p_i$ primes with $p_i \equiv 1 \ \mathrm{mod} \ 4$ and $q_i$ are primes with $q_i \equiv 3 \ \mathrm{mod} \ 4$. This gives a identity in terms of the prime numbers.

Maybe it is easier to study the function $$S(z)= \sum_{m=1}^\infty r(m) z^{\sqrt{m}} +1.$$ That is something like a modifcated Theta-series.

Answer 2

The proof of the integral expression can be found here. \begin{align*} &\;\;\,\int_{0}^{+\infty}e^{-t^{2}z^{2}}\vartheta_{3}\left(\frac{x}{2},e^{-\frac{1}{4t^{2}}}\right)\vartheta_{3}\left(\frac{y}{2},e^{-\frac{1}{4t^{2}}}\right)\mathrm{d}t\\ &=\frac{\sqrt{\pi}}{2z}\sum_{n=-\infty}^{+\infty}\sum_{m=-\infty}^{+\infty}e^{-z\sqrt{m^{2}+n^{2}}}\cos(mx)\cos(ny) \end{align*}