Hardy's work

Hardy and Mellin transform

In a previous post (link) we used the following identity. We repost here the proof based on the work of Hardy: a mix of the Mellin transform and theta functions properties. \[\mathop{\sum_{n=-\infty }^{\infty}\sum_{m=-\infty}^{\infty}}_{n\neq 0 \wedge m\neq 0}\frac{1}{(m^2+n^2)^s} = 4\zeta(s)\beta(s)\]
Proof
Let $n,m \neq 0$ then, \[ \int_{0}^{\infty} t^{s-1} e^{-(n^2+m^2)t}dt = \frac{1}{(n^2+m^2)} \int_{0}^{\infty} w^{s-1}e^{-w}dw \quad \left(w \mapsto (n^2+m^2)t\right) \] Hence \[\int_{0}^{\infty} t^{s-1} e^{-(n^2+m^2)t}dt = \frac{1}{(n^2+m^2)} \Gamma(s) \Longrightarrow \frac{1}{(n^2+m^2)}= \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1} e^{-(n^2+m^2)t}dt \] Therefore \begin{align*} \mathop{\sum_{n=-\infty}^{\infty}\sum_{m=-\infty }^{\infty}}_{n\neq 0 \wedge m\neq 0} \frac{1}{(m^2+n^2)^s} =& \mathop{\sum_{n=-\infty }^{\infty}\sum_{\substack{m=-\infty}}^{\infty}}_{n\neq 0 \wedge m\neq 0}\frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1} e^{-(n^2+m^2)t}dt \\ =& \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1}\mathop{\sum_{n=-\infty }^{\infty}\sum_{m=-\infty }^{\infty}}_{n\neq 0 \wedge m\neq 0} e^{-(n^2+m^2)t}dt \\ =& \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1}\left[\sum_{\substack{n=-\infty }}^{\infty}\sum_{\substack{m=-\infty }}^{\infty} e^{-(n^2+m^2)t} -1\right]dt \\ =& \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1}\left[\sum_{\substack{n=-\infty}}^{\infty} e^{-n^2t}\sum_{\substack{m=-\infty}}^{\infty}e^{-m^2t}-1\right]dt \\ =& \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1}\left[\left(\sum_{\substack{n=-\infty}}^{\infty} e^{-n^2t}\right)^2-1\right]dt \\ \end{align*} Now recall the definition of the theta function $\displaystyle \theta_{3}(q,z)=\sum_{n=-\infty}^{\infty} q^{n^2}e^{2inz}$
If $\displaystyle q=e^{-t}$ and $z=0$ then:
\[\displaystyle \theta_{3}(e^{-t},0)=\sum_{n=-\infty}^{\infty} e^{-n^2t}\Longrightarrow \theta^2_{3}(e^{-t},0)=\left(\sum_{n=-\infty}^{\infty} e^{-n^2t} \right)^2\] Therefore \[\left(\sum_{\substack{n=-\infty}}^{\infty} e^{-n^2t}\right)^2 -1 = \theta^2_{3}(e^{-t},0)-1\] Now recall the following identity for $\displaystyle \theta_{3}(q,0)$: \[ \theta^2_{3}(q,0) =1+4\sum_{n=1}^{\infty} \frac{q^n}{(1+q^{2n})}\]
In this case with $q=e^{-t}$ \[\theta^2_{3}(e^{-t},0) -1=4\sum_{n=1}^{\infty} \frac{e^{-tn}}{(1+e^{-2tn})} = 4 \sum_{n=1}^{\infty}\sum_{m=0}^{\infty}e^{-tn}(-1)^m e^{-2tnm}\] Therefore \begin{align*}\left\{\mathscr{M} \left[\theta^2_{3}(e^{-t},0) -1\right] \right\}(s) =& \int_{0}^{\infty} t^{s-1} \left[\theta^2_{3}(e^{-t},0) -1\right] dt\\ =& \int_{0}^{\infty} t^{s-1}4 \sum_{n=1}^{\infty}\sum_{m=0}^{\infty}(-1)^m e^{-n(2m+1)t} dt\\ =& \Gamma(s) 4\sum_{n=1}^{\infty}\frac{1}{n^s}\sum_{n=0}^{\infty}(-1)^m\frac{1}{(2m+1)^s}\\ =& 4\Gamma(s)\zeta(s)\beta(s) \end{align*} Therefore \[\boxed{\mathop{\sum_{n=-\infty }^{\infty}\sum_{m=-\infty}^{\infty}}_{n\neq 0 \wedge m\neq 0}\frac{1}{(m^2+n^2)^s} = 4\zeta(s)\beta(s)} \]