# 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

In this post the following is proved

$$\small \sum _{n=1}^{\infty } \frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}=\frac{1}{2} \pi J_0(a x)-\frac{\sin (a x)}{2 a},\ \ \sum _{n=1}^{\infty } \frac{\cos \left(x \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}=-\frac{1}{2} \pi Y_0(a x)-\frac{\cos (a x)}{2 a}$$

But how to established the harder one

$$\small\sum _{n=1}^{\infty } \frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}=-\frac{\sin (a x)}{2 a^3}+\frac{\pi x \coth (\pi a)}{2 a}+\frac{1}{4} \pi ^2 x^2 (\pmb{H}_1(a x) J_0(a x)-\pmb{H}_0(a x) J_1(a x))-\frac{1}{2} \pi x^2 J_0(a x)+\frac{\pi x J_1(a x)}{2 a}$$

Here $$J, \pmb{H}$$ denotes Bessel and Struve functions.

Update: By M-L theorem and repeated integration one have

$$\small \sum _{n=1}^{\infty } \frac{\cos \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^2}=-\frac{\cos (a x)}{2 a^4}+\frac{\pi \coth (\pi a)}{4 a^3}+\frac{1}{8} \pi ^2 x^3 \left(1-\frac{1}{a^2 x^2}\right) F(a x)+\frac{\pi ^2 \text{csch}^2(\pi a)}{4 a^2}+\frac{1}{4} \pi x^3 J_0(a x)-\frac{\pi x^2 \coth (\pi a)}{4 a}-\frac{\pi x^2 J_1(a x)}{4 a}$$

Where $$F(t)=\pmb{H}_0(t) J_1(t)-\pmb{H}_1(t) J_0(t)$$. Analytic continuation allows us to extend the range to $$|a|<1$$, $$x\in (0,2\pi)$$, for instance

$$\small\sum _{n=1}^{\infty } \frac{\cos \left(\sqrt{4 n^2-1}\right)}{\left(n^2-\frac{1}{4}\right)^2}=2 \pi ^2 \pmb{L}_1(1) I_0(1)-2 \pi ^2 \pmb{L}_0(1) I_1(1)+\pi ^2-8 \cosh (1)+2 \pi I_0(1)-2 \pi I_1(1)$$

Moreover, differentiating closed-form of $$\sum _{n=1}^{\infty } \left(\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}-\frac{\sin (n x)}{n}\right)$$ w.r.t $$x$$ yields:

$$\small\sum _{n=1}^{\infty } \left(\cos \left(\pi \sqrt{n^2+1}\right)-(-1)^n\right)=1-\frac{\pi J_1(\pi )}{2}$$

Using the representation $$$$\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}=x\int_0^1 \frac{\cos \left(tx \sqrt{a^2+n^2}\right)}{a^2+n^2}\,dt\\$$$$ and integrating by parts, $$$$\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}=x\frac{\cos \left(x \sqrt{a^2+n^2}\right)}{a^2+n^2}+x^2\int_0^1 t\frac{\sin \left(tx \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}\,dt$$$$ We have then to evaluate \begin{align} S(x)&=\sum_{n\ge1}\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}\\ &=xA(x)+x^2B(x)\\ A(x)&=\sum_{n\ge1}\frac{\cos \left(x \sqrt{a^2+n^2}\right)}{a^2+n^2}\\ B(x)&=\sum_{n\ge1}\int_0^1 t\frac{\sin \left(tx \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}\,dt \end{align} We have \begin{align} A'(x)&=-\sum_{n\ge1}\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}\\ &=-\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a} \end{align} And thus, considering that $$$$A(0)=\sum_{n\ge1}\frac{1}{n^2+a^2}=\frac{\pi}{2a}\coth \pi a-\frac{1}{2a^2}$$$$ we deduce \begin{align} &A(x)=\frac{\pi}{2a}\coth \pi a-\frac{1}{2a^2}+\int_0^x\left[\frac{\sin (a t)}{2 a}-\frac{1}{2} \pi J_0(a t)\right]\,dt\\ &=\frac{\pi}{2a}\coth \pi a-\frac{1}{2a^2}+\frac{x\pi^2}{4}\left( \pmb{H}_1(a x) J_0(a x)-\pmb{H}_0(a x) J_1(a x) \right)-\frac{x\pi}{2}J_0(ax)+\frac{1-\cos ax}{2a^2} \end{align} Now, \begin{align} B(x)&=\int_0^1 \left[\frac{1}{2} \pi J_0(a xt)-\frac{\sin (a xt)}{2 a}\right]t\,dt\\ &=\frac{1}{x^2}\int_0^x\left[\frac{1}{2} \pi J_0(a u)-\frac{\sin (a u)}{2 a}\right]u\,du\\ &=\frac{\pi }{2ax}J_1(ax)+\frac{\cos ax}{2xa^2}-\frac{\sin xa}{2x^2a^3}\\ \end{align} Finally, as expected, \begin{align} S(x)=&\frac{\pi x}{2a}\coth \pi a-\frac{x}{2a^2}+\frac{x^2\pi^2}{4}\left( \pmb{H}_1(a x) J_0(a x)-\pmb{H}_0(a x) J_1(a x) \right)-\frac{x^2\pi}{2}J_0(ax)\\ &+\frac{x}{2a^2}+ \frac{x\pi }{2a}J_1(ax)-\frac{\sin ax}{2a^3} \end{align} The series with cosines could be evaluated in the same way. In fact working directly with complex numbers $$\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.
• I might suggest not using $I$ and $J$ for your intermediate integrals, but looks good otherwise. Commented Sep 21, 2019 at 11:51