Formula for the series $f(x):=\sum\limits_{n=1}^\infty\frac{x}{x^2+n^2}$

Question

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?

Answer 1

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".

Answer 2

Inspired by what JessicaK gave in her answer to a previous question (I refer to the link JessicaK gives in her comment) and using a CAS for control, what was found is $$f(x)=\sum\limits_{n=1}^\infty\displaystyle\frac{x}{x^2+n^2}=\frac{\pi x \coth (\pi x)-1}{2 x}$$ All the credit must be given to JessicaK.

Answer 3

Maybe someone knows a quick hint to this?

Take the natural logarithm of Euler's infinite product formula for the sine function, and then differentiate it. For the final touch, use Euler's formula, linking trigonometric and hyperbolic functions together.

Answer 4

Please look at complex analysis if you want powerful and elegant methods for infinite sums such as these.

Otherwise there is also this plug 'n chug formula called the (simplified) Euler-Maclaurin formula: $$\sum_{n=1}^mf(n)=\int_1^mf(x)\textrm{d}x+\frac{1}{2}(f(1)+f(m))+\int_1^mH(x)f'(x)\textrm{d}x$$ where $f$ is differentiable and $H(x):=x-\lfloor x\rfloor-\frac{1}{2}$.

With $f(n):=x/(x^2+n^2)$ we have

$$\int_1^m \frac{x}{x^2+y^2}\textrm{d}y=\arctan(m/x)-\arctan(1/x)$$ for $m\rightarrow \infty$ this becomes $\pi/2-\arctan(1/x)$. Also $\lim_{m\rightarrow \infty}f(m)=0$.

Now for example if $x=1$ then if we neglect the messy last integral in the formula then the approximation $$\sum_{n=1}^\infty \frac{x}{x^2+n^2}\approx \frac{\pi}{2}-\arctan(1/x)+\frac{x}{2x^2+2}$$is very crude. The apprximation is $1.0353981$ for $x=1$ which is quite far from the $1/2(\pi\coth(\pi)-1)\approx 1.0766740$ given in the link provided by JessicaK. However with large $x$ the approximation is serviceable. For example for $x=1000$ i get $1.570296326$ which is pretty close to $\pi/2=1.5707963267$ which is the value that you proved convergence to as $x\rightarrow \infty$.

This is how the approximation (green) looks vs. the series (blue) for $0.5\leq x\leq 10$