Infinite Series $\sum\limits_{n=-\infty}^{\infty}\frac{1}{(x+n\pi)^m}$

Question

How can we find a closed form for the following infinite series for any $m\in\mathbb N$? $$\sum_{n=-\infty}^{\infty}\frac{1}{(x+n\pi)^m}$$

Answer 1

To obtain convergence in the specific case $m=1$ let's rewrite your series $$f_m(x):=\sum_{n=-\infty}^{\infty}\frac{1}{(x+n\,\pi)^m}$$

as : \begin{align} f(x)&:=\frac 1x+\sum_{n=1}^\infty\frac{1}{x+n\,\pi}+\frac{1}{x-n\,\pi}\\ &=\frac 1x+\sum_{n=1}^\infty\frac{2x}{x^2-(n\,\pi)^2}\\ \\ &=\cot(x)\\ \end{align} This was deduced by Euler (proof using Fourier series for $\,x:=\pi\,z$).

It is straightforward to obtain : $\;\displaystyle f_{m+1}(x)=\frac {(-1)^{m}}{m!}\left(\frac d{dx}\right)^mf(x)\;$ so that \begin{align} f_{m+1}(x)&=\frac {(-1)^m\;\cot^{(m)}(x)}{m!} \\ \end{align} The first results for $\;m!\cdot f_{m+1}(x)\,$ are in the following table (supposing $\,t:=\tan(x)$) :

\begin{array} {l|c} m!\cdot f_{m+1}&(-1)^m\,\cot^{(m)}(x)\\ \hline 0!\cdot f_1&\frac 1{t}\\ 1!\cdot f_2&\frac {1+1t^2}{t^2}\\ 2!\cdot f_3&\frac {2+2t^2}{t^3}\\ 3!\cdot f_4&\frac {6+8t^2+2t^4}{t^4}\\ 4!\cdot f_5&\frac {24+40t^2+16t^4}{t^5}\\ \end{array} Since $\;\left(\tan(x)^n\right)'=+n\,\tan(x)^{n-1}+n\,\tan(x)^{n+1}\;$
while $\;\left(\cot(x)^n\right)'=-n\,\cot(x)^{n-1}-n\,\cot(x)^{n+1}\;$
the triangle of coefficients obtained at the numerator will be the same than the triangle for the derivatives of the $\tan$ function (OEIS A101343 with more references from Knuth like the paper 'Computation of tangent, Euler, and Bernoulli numbers', Foata and others).

This allows to propose the general formula :

$$f_{m+1}(x)=\frac {\sum_{k=0}^{\lfloor (m+1)/2\rfloor}\;c_{m,k}\;\tan(x)^{2k}}{m!\;\tan(x)^{m+1}}=\frac 1{m!}\sum_{k=0}^{\lfloor (m+1)/2\rfloor}\;c_{m,k}\;\cot(x)^{m+1-2k}$$ with $\;\;\displaystyle c_{m,k}:=(m-2(k-1))\;c_{m-1,k-1}+(m-2k)\;c_{m-1,k}\;$ and $\;c_0,k=\delta_k^0$

An 'explicit formula' was proposed at OEIS : $\;c_{m,k}=\operatorname{tr}_{n,k}+\operatorname{tr}_{n,k-1}\;$ with $\operatorname{tr}$ defined by : $$\operatorname{tr}_{n,i}=\sum_{j=j_0}^{2i} \binom{j+n-2\,i-1}{n-2\,i-1}\,(j+n-2\,i)!\;2^{2\,i-j}(-1)^{j-i}\;\operatorname{Stirling}_2(n,j+n-2\,i)$$

(with a Stirling number of the second kind at the right and $j_0=1$ if $2\,i>n$ and $j_0=0$ else)