Evaluate $\sum _{n=1}^{\infty } \frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}$ and generalize it

Solution 1:

Using the representation \begin{equation} \frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}=x\int_0^1 \frac{\cos \left(tx \sqrt{a^2+n^2}\right)}{a^2+n^2}\,dt\\ \end{equation} and integrating by parts, \begin{equation} \frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}=x\frac{\cos \left(x \sqrt{a^2+n^2}\right)}{a^2+n^2}+x^2\int_0^1 t\frac{\sin \left(tx \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}\,dt \end{equation} We have then to evaluate \begin{align} S(x)&=\sum_{n\ge1}\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}\\ &=xA(x)+x^2B(x)\\ A(x)&=\sum_{n\ge1}\frac{\cos \left(x \sqrt{a^2+n^2}\right)}{a^2+n^2}\\ B(x)&=\sum_{n\ge1}\int_0^1 t\frac{\sin \left(tx \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}\,dt \end{align} We have \begin{align} A'(x)&=-\sum_{n\ge1}\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}\\ &=-\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a} \end{align} And thus, considering that \begin{equation} A(0)=\sum_{n\ge1}\frac{1}{n^2+a^2}=\frac{\pi}{2a}\coth \pi a-\frac{1}{2a^2} \end{equation} we deduce \begin{align} &A(x)=\frac{\pi}{2a}\coth \pi a-\frac{1}{2a^2}+\int_0^x\left[\frac{\sin (a t)}{2 a}-\frac{1}{2} \pi J_0(a t)\right]\,dt\\ &=\frac{\pi}{2a}\coth \pi a-\frac{1}{2a^2}+\frac{x\pi^2}{4}\left( \pmb{H}_1(a x) J_0(a x)-\pmb{H}_0(a x) J_1(a x) \right)-\frac{x\pi}{2}J_0(ax)+\frac{1-\cos ax}{2a^2} \end{align} Now, \begin{align} B(x)&=\int_0^1 \left[\frac{1}{2} \pi J_0(a xt)-\frac{\sin (a xt)}{2 a}\right]t\,dt\\ &=\frac{1}{x^2}\int_0^x\left[\frac{1}{2} \pi J_0(a u)-\frac{\sin (a u)}{2 a}\right]u\,du\\ &=\frac{\pi }{2ax}J_1(ax)+\frac{\cos ax}{2xa^2}-\frac{\sin xa}{2x^2a^3}\\ \end{align} Finally, as expected, \begin{align} S(x)=&\frac{\pi x}{2a}\coth \pi a-\frac{x}{2a^2}+\frac{x^2\pi^2}{4}\left( \pmb{H}_1(a x) J_0(a x)-\pmb{H}_0(a x) J_1(a x) \right)-\frac{x^2\pi}{2}J_0(ax)\\ &+\frac{x}{2a^2}+ \frac{x\pi }{2a}J_1(ax)-\frac{\sin ax}{2a^3} \end{align} The series with cosines could be evaluated in the same way. In fact working directly with complex numbers $\exp\left(i x\sqrt{n^2+a^2} \right)$ and Hankel functions may simplify, but I didn't try. Indefinite integrals of $H_0^{1}(z)$ and $zH_0^{1}(z)$ are indeed tabulated DLMF.