Evaluate the sum $\sum ^{\infty}_{n=1} \frac{1}{n^2+a^4}$ Evaluate the sum $\sum ^{\infty}_{n=1} \frac{1}{n^2+a^4}$
I don't to know how to processed this problem. Can any one help with problem please. thanks
 A: Complex analysis gives a very clear explanation of the identity
$$ \sum_{n=1}^{\infty} \frac{1}{n^2+z^2} = \frac{1}{z}\bigg( \frac{\pi}{2}\coth(\pi z) - \frac{1}{2z} \bigg) $$
where $z = a^2$. This equality follows from the observation that both sides have exactly the same poles and decay to $0$ as $\Re(z) \to \pm \infty$.

If you want to avoid complex-analysis method, there are some alternatives. Let me briefly discuss the idea. Let $b = a^2$ and assume that $b > 0$. Using the following Fourier transform,
$$ \int_{-\infty}^{\infty} \cos(nx) e^{-b|x|} \, dx = \frac{2b}{n^2 + b^2}, $$
we can perform the following heuristic computation:
$$\sum_{n=1}^{\infty} \frac{1}{n^2 + b^2}
= \frac{1}{2b} \sum_{n=1}^{\infty} \int_{-\infty}^{\infty} \cos(nx)e^{-b|x|} \, dx
\color{red}{\stackrel{!?}{=}} \frac{1}{2b} \int_{-\infty}^{\infty} \bigg( \sum_{n=1}^{\infty} \cos(nx) \bigg) e^{-b|x|} \, dx. $$
Now using the (distributional) Poisson summation formula
$$ 2\pi \sum_{n=-\infty}^{\infty} \delta(x - 2\pi n)
= \sum_{n=-\infty}^{\infty} e^{inx}
= 1 + 2 \sum_{n=1}^{\infty} \cos(nx), $$
we find that
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2 + b^2}
&\color{red}{\stackrel{!?}{=}} \frac{1}{2b} \int_{-\infty}^{\infty} \bigg( \pi \sum_{n=-\infty}^{\infty} \delta(x - 2\pi n) - \frac{1}{2} \bigg) e^{-b|x|} \, dx \\
&\color{red}{\stackrel{!?}{=}} \frac{1}{2b} \bigg( \pi \sum_{n=-\infty}^{\infty} e^{-2\pi b |n|} - \frac{1}{b} \bigg) \\
&= \frac{1}{2b}\bigg( \pi\coth(\pi b) - \frac{1}{b} \bigg).
\end{align*}
Up to this point, all the computations are only heuristic and desperately require some rigorous justification. Fortunately not much extra work is required. All this nonsense can be salvaged by adopting an appropriate regularization.
Indeed, using the dominated convergence theorem, we may safely write
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2 + b^2}
&= \lim_{r\uparrow 1} \sum_{n=1}^{\infty} \frac{r^n}{n^2 + b^2} \\
&= \frac{1}{2b} \lim_{r\uparrow 1} \sum_{n=1}^{\infty} r^n \int_{-\infty}^{\infty} \cos(nx)e^{-b|x|} \, dx \\
&= \frac{1}{2b} \lim_{r\uparrow 1} \int_{-\infty}^{\infty} \bigg( \sum_{n=1}^{\infty} r^n \cos(nx) \bigg) e^{-b|x|} \, dx
\end{align*}
Then from the formula of the 2-dimensional Poisson kernel
$$ \frac{1 - r^2}{1 - 2r\cos x + r^2} = 1 + 2 \sum_{n=1}^{\infty} r^n \cos(nx), \qquad |r| < 1$$
we have
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2 + b^2}
&= \frac{1}{2b} \lim_{r\uparrow 1} \int_{-\infty}^{\infty} \bigg( \frac{1}{2} \frac{1 - r^2}{1 - 2r\cos x + r^2} - \frac{1}{2} \bigg) e^{-b|x|} \, dx \\
&= \frac{1}{2b} \bigg( \frac{1}{2} \lim_{r\uparrow 1} \int_{-\pi}^{\pi} \frac{1 - r^2}{1 - 2r\cos x + r^2} \bigg( \sum_{n=-\infty}^{\infty}  e^{-b|x+2\pi n|} \bigg) \, dx - \frac{1}{b} \bigg).
\end{align*}
Notice that $x \mapsto \sum_{n=-\infty}^{\infty}  e^{-b|x+2\pi n|}$ defines a continuous function. Thus by appealing to the approximation-to-the-identity property of the Poisson kernel, we obtain the same answer as in the heuristics.
A: Of course, Sangchul's answer above is great and satisfying, but From here, we get $$\sum_{n=1}^{\infty} \frac{1}{k + n^{2}}=- \frac{\sqrt{- k}}{2 k} (- \psi{ (0,- \sqrt{- k} + 1  )} + \psi{ (0,\sqrt{- k} + 1  )})=- \frac{0.5}{k}(- k)^{0.5} (-\psi{(0,- \sqrt{- k} + 1 )} + \psi{(0,\sqrt{- k} + 1 )})$$ where $\psi$ is the Polygamma Function implemented in the Mathematica way;this may be of some help.
