Today I have encounter a series:

$$\sum_{n=-\infty}^{+\infty}\frac{1}{(u+n)^2}=\frac{\pi^2}{(\sin \pi u)^2}$$ where $u \not \in \Bbb{Z}$ . I have known a method to compute it (by residue formula): $$\int_{|z|=N+1/2}\frac{\pi \cot \pi z}{(u+z)^2}\text{dz}=-\frac{\pi^2}{\sin^2 \pi u}+\sum_{k=-N}^{N}\frac{1}{(u+k)^2}$$

where $-u$ is a pole of order 2, and $n \in \Bbb{Z}$ and $|n| \leq N$ are poles of order 1. Because $$\left|\int \frac{\pi \cot \pi z}{(u+z)^2}\text{dz}\right|\leq \int_0^{2\pi}\pi\left|1+\frac{2}{e^{i2\pi z}-1}\right|\frac{1}{|((N+1/2)e^{i\varphi}+u)^2|}\text{d}\varphi \rightarrow 0$$ as $N \rightarrow +\infty$.

Does there exist some other ways to compute it?

  • 2
    $\begingroup$ It is a direct application of Fourier transform and Poisson summation formula. See here for a related problem. $\endgroup$ Commented Feb 26, 2013 at 16:21

2 Answers 2


Consider the Weierstrass form of the Gamma function:

$$\frac{1}{\Gamma (x)}=xe^{\gamma x}\prod_{n\geq 1}\left(1+\frac{x}{n}\right)e^{-\frac{x}{n}}$$ Then:

$$\ln\Gamma (x)=-\ln x-\gamma x+\sum_{n\geq 1}\left[\frac{x}{n}-\ln \left(1+\frac{x}{n}\right)\right]$$

$$\frac{d}{dx}\ln\Gamma(x)=-\frac{1}{x}-\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+x}\right)$$

$$\begin{align*}\frac{d^2}{dx^2}\ln\Gamma (x)&=\frac{1}{x^2}+\sum_{n\geq 1}\frac{1}{(n+x)^2}\\[7pt]&=\sum_{n\geq 0}\frac{1}{(n+x)^2}\quad(1)\end{align*}$$

Note that by letting $x\to 1-x$ and $n\to-n-1$ we also have:

$$\frac{d^2}{dx^2}\ln\Gamma(1-x)=\sum_{n\leq -1}\frac{1}{(n+x)^2}\quad(2)$$

Combine $(1)$ and $(2):$

$$ \begin{align*}\sum_{\mathbb{Z}}\frac{1}{(n+x)^2}&=\frac{d^2}{dx^2}\ln\Gamma(1-x)+\frac{d^2}{dx^2}\ln\Gamma(x)\\&=\frac{d^2}{dx^2}\ln \Gamma(x)\Gamma(1-x)\\[7pt]&=\frac{d^2}{dx^2}\ln\pi\csc \pi x\quad(x\not\in\mathbb{Z})\\[7pt]&=\pi^2\csc^2\pi x\end{align*}$$

  • $\begingroup$ It is a very nice proof and I like it. This identity has lots of applications. $\endgroup$
    – xpaul
    Commented Apr 25, 2013 at 20:48

Let $f(z) =\displaystyle \sum_{n\in \mathbb{Z}} \frac{1}{(n+z)^2}$ and $g(z) = \dfrac{\pi^2}{\sin^2(\pi z)}.$ Both functions are $1$ periodic, so it suffices to understand their behavior in the strip $\mathfrak{R}(z)\in [0,1).$ It is straightforward to verify that both these functions have a singularity only at the origin and are bounded on the strip excluding any neighborhood of the origin.

Now look near $z=0.$ We have $f(z) =\displaystyle \frac{1}{z^2} + \sum_{n\in\mathbb{Z}\setminus\{0\}} \frac{1}{(z+n)^2}.$ By the power series expansion of sine and geometric series, we have $$ g(z) = \frac{\pi^2}{\pi^2 z^2 - (\pi^4/3)z^4 + \mathcal{O}(z^6) } = \frac{1}{z^2} \left( \frac{1}{1- (\pi^4/3) z^2 + \mathcal{O}(z^4) } \right) = \frac{1}{z^2} \left( 1 + (\pi^2/3) z^2 + \mathcal{O}(z^4) \right) = \frac{1}{z^2} +\pi^2/3 + \mathcal{O}(z^2).$$

So in fact $f(z)-g(z)$ is bounded and analytic on the strip, and hence the whole complex plane. By Liouville's theorem, $f(z)-g(z) = f(0)-g(0)=0$ so $f(z) = g(z).$

  • $\begingroup$ Yes, this is the classical proof from Ahlfors. $\endgroup$ Commented May 29, 2014 at 7:46

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