Evaluate $\sum _{n=1}^{\infty } \frac{J_v(2 \pi n a)}{n^v}$

Question

How to prove $$\sum _{n=1}^{\infty } \frac{J_v(2 \pi n a)}{n^v}=\frac{\pi ^{v-\frac{1}{2}} a^{v-1}}{2 \Gamma \left(v+\frac{1}{2}\right)}+\frac{\pi ^{v-\frac{1}{2}} a^v \sin ^{2 v}\left(\cos ^{-1}\left(\frac{1}{a}\right)\right)}{\sqrt{a^2-1} \Gamma \left(v+\frac{1}{2}\right)}-\frac{(\pi a)^v}{2 \Gamma (v+1)}$$ For $1<a<2$? Any help will be appreciated.

Answer 1

Actually $k=0$ is well-driven this way in the distributional sense. Using Poisson's $$J_\nu(z)=\frac{(z/2)^\nu}{\sqrt\pi\,\Gamma(\nu+1/2)}\int_{-1}^{1}(1-t^2)^{\nu-1/2}\cos zt\,dt$$ restricted to $\Re\nu>1/2$ (for $\delta$'s to "behave good"), and taking any real $a>0$, we find \begin{align} \sum_{n=1}^{\infty}\frac{J_\nu(2na\pi)}{n^\nu}&=\frac{(a\pi)^\nu}{\sqrt\pi\,\Gamma(\nu+1/2)}\int_{-1}^{1}(1-t^2)^{\nu-1/2}\sum_{n=1}^{\infty}\cos 2na\pi t\,dt\\&=\frac{(a\pi)^\nu}{2\sqrt\pi\,\Gamma(\nu+1/2)}\int_{-1}^{1}(1-t^2)^{\nu-1/2}\left(-1+\sum_{n=-\infty}^{\infty}\delta(at-n)\right)dt\\&=\frac{(a\pi)^\nu}{2\sqrt\pi\,\Gamma(\nu+1/2)}\left(-\mathrm{B}\Big(\frac{1}{2},\nu+\frac{1}{2}\Big)+\frac{1}{a}\sum_{|n|\leqslant a}\Big(1-\frac{n^2}{a^2}\Big)^{\nu-1/2}\right),\end{align} generalizing your formula.