Formula for the series $f(x):=\sum\limits_{n=1}^\infty\frac{x}{x^2+n^2}$ Good evening! Recently I had to study the properties of a function $$f(x):=\sum\limits_{n=1}^\infty\displaystyle\frac{x}{x^2+n^2}.$$ I found its supremum and proved that $\lim\limits_{x\to+\infty}f(x)=\displaystyle\frac{\pi}{2}$. But as far as I remember such sums can be calculated explicitly. Maybe someone knows a quick hint to this? And in general: a lot of sums can be calculated explicitly by reducing to some Taylor and Laurent series, special functions, etc., and the methods are usually special for every particular problem. Is there any book, where a complete list of known examples can be found?

Let me supplement another (simple) method for this particular problem( not the general summation technique the OP is looking for). It simply uses the Poisson's summation formula,
\begin{equation}
\sum_{n= -\infty}^{\infty} f(n) = \sum_{m = -\infty}^{\infty} \tilde{f}( 2\pi m )
\end{equation}
Set $g(x) = e^{-\epsilon |x| + ix \theta } $, then the Fourier transform of $g(x)$ is
\begin{equation}
\tilde{g}(k) =\int_{-\infty}^{\infty} e^{-\epsilon |x| + ix \theta } e^{ik x} dx = \frac{2 \epsilon}{ \epsilon^2 + (k - \theta)^2 }
\end{equation}
so apply Poisson's summation formula to $g(x)$,
\begin{equation}
\sum_{n = -\infty}^{\infty} g(n ) = \frac{\sinh \epsilon}{\cosh \epsilon - \cos \theta} = \sum_{n = -\infty}^{\infty}  \frac{2 \epsilon}{ \epsilon^2 + (2\pi n - \theta )^2 }
\end{equation}
For the present case, set $\theta = 0$, $\epsilon = 2\pi x$,
\begin{equation}
RHS = \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{x^2 + n^2} + \frac{1}{\pi x }= LHS = \coth( \pi x )
\end{equation}
hence,
\begin{equation}
\sum_{n=1}^{\infty} \frac{x}{x^2 + n^2} = \frac{\pi}{2}\coth( \pi x) - \frac{1}{2x }
\end{equation}
Edit: the original motivation to add the $\theta$ term is prove that following identity,
\begin{equation}
\frac{1}{2\pi }\frac{1- r^2}{1 -2r \cos \theta + r^2} = - \frac{1}{\pi} \sum_{-\infty}^{\infty} \frac{\ln r}{ (\ln r)^2 + ( \theta + 2\pi n )^2 }
\end{equation}
namely, the Poisson kernel is equal to the sum of Lorentzian(this is how I know this sum has closed form). For the present case, one can of course set $\theta = 0$ in the first place to make it "complex-analysis-free".
