Proving an identity for Riemann zeta type sum I have asked here for a closed form of the sum
$$\sum_{n=1}^∞\left(\frac{1}{n^2+\alpha^2}\right)^a.$$
However, it was suggested in the comments that even though a closed form might not be possible, the sum can be related to the Bessel function of the first kind in the following mannner
$$\sum_{n=1}^∞\left(\frac{1}{n^2+\alpha^2}\right)^a=\frac{2^{\frac{1}{2}−a}\sqrt{π}\alpha^{\frac{1}{2}−a}}{\Gamma(a)}\int_0^\infty\frac{x^{-\frac{1}{2}+a}J_{-\frac{1}{2}+a}(x\alpha)}{e^x-1}{\rm d}x.$$
I'm looking for a proof of this identity.
 A: After the straightforward $(e^x-1)^{-1}=\sum_{n=1}^\infty e^{-nx}$ for $x>0$, it reduces to $$\int_0^\infty x^\nu J_\nu(ax)e^{-bx}\,dx=\frac{(2a)^\nu}{\sqrt\pi}\frac{\Gamma(\nu+1/2)}{(a^2+b^2)^{\nu+1/2}}$$ for $a>0$, $\Re b>0$, $\Re\nu>-1/2$ (say). One way to show the latter is to consider $|b|>a$ first:
\begin{align*}
\int_0^\infty x^\nu J_\nu(ax)e^{-bx}\,dx
&=\int_0^\infty x^\nu e^{-bx}\sum_{n=0}^\infty\frac{(-1)^n(ax/2)^{2n+\nu}}{n!\ \Gamma(n+\nu+1)}\,dx
\\&=(a/2)^\nu\sum_{n=0}^\infty\frac{(-1)^n(a/2)^{2n}}{n!\ \Gamma(n+\nu+1)}\int_0^\infty x^{2n+2\nu}e^{-bx}\,dx
\\&=(a/2)^\nu\sum_{n=0}^\infty\frac{(-1)^n(a/2)^{2n}}{n!\ \Gamma(n+\nu+1)}\frac{\Gamma(2n+2\nu+1)}{b^{2n+2\nu+1}}
\\&=\frac{(2a)^\nu}{b^{2\nu+1}\sqrt\pi}\sum_{n=0}^\infty\frac{(-1)^n}{n!}\left(\frac{a}{b}\right)^{2n}\Gamma\left(n+\nu+\frac12\right),
\end{align*}
where $\int\sum\mapsto\sum\int$ is valid because of absolute convergence, and the duplication formula is used for $\Gamma(2n+2\nu+1)$; the final sum is $\Gamma(\nu+1/2)(1+a^2/b^2)^{-\nu-1/2}$ (binomial series).
The full case of $\Re b>0$ follows by analytic continuation (w.r.t. $b$).
