$\sum_{n=-\infty}^{\infty}\frac{1}{(u +n)^2}=\frac{\pi^2}{(\sin \pi u)^2}$

Question

I've already see a proof by Marko Riedel which I list it follows:

The standard way to treat these sums is to integrate $$ f(z) = \frac{1}{(z+\alpha)^2} \pi \cot(\pi z)$$ along a contour consisting of a circle of radius $R$ and with $R$ going to infinity and hence being larger than $\alpha$, where the circle does not pass through the poles on the real axis. Now along the semicircle in the upper half plane we have $$|f(z)| \le \frac{1}{(R-|\alpha|)^2}\pi \left|\frac{e^{i\pi R\exp(i\theta)} + e^{-i\pi R\exp(i\theta)}} {e^{i\pi R\exp(i\theta)} - e^{-i\pi R\exp(i\theta)}}\right|= \frac{1}{(R-|\alpha|)^2} \pi \left|\frac{e^{2i\pi R\exp(i\theta)}+1}{e^{2i\pi R\exp(i\theta)}-1}\right| < \frac{1}{(R-|\alpha|)^2} \pi \frac{1+e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}} {1-e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}}$$ This last term is clearly $O(1/R^2)$ as $R$ goes to infinity as the quotient of the two exponentials goes to one since $\exp(-R)$ vanishes and there is no singularity when $\theta = 0$ or $\theta = \pi$ as $R\cos\theta = \pm R$, which is not an integer by the assumption that the circle avoids the poles and hence cannot be one.

My question:
It is clear that for each $z=Re^{i \theta}$ , $f(z)=O(1/R^2)$ ，but can this guarantee that the last term on RHS is uniform bounded ? Since $\theta$ varies on a compact interval $[0, \pi]$ , I want to show that for each $ \theta \in [0, \pi]$ , and a fixed positive number $M \gt 1 $ there exist an open ball contains $\theta$ and a fixed $R_{\theta}\gt 0$ , for every $R \ge R_{\theta}$ and $x \in $ the open ball $$\frac{1+e^{-2\pi R\sin(x)}e^{2i\pi R\cos(x)}} {1-e^{-2\pi R\sin(x)}e^{2i\pi R\cos(x)}} \le M$$ For $\theta \neq 0,\pi$ , it is easy to find the desired $R_{\theta}$ and the open ball , but if $\theta = 0$ or $ \theta = \pi$ , for every open ball containing $\theta$ , it behave erratically near $\theta$ , I have no idea how to deal with this .

Answer 1

With the quoted proof being unsatisfactory we try again. With the goal of evaluating

$$\sum_{n=-\infty}^\infty \frac{1}{(u+n)^2}$$

where $u$ is not an integer we study the function

$$f(z) = \frac{1}{(u+z)^2} \pi\cot(\pi z).$$

which has the property that with $S$ being our sum,

$$S = \sum_n \mathrm{Res}_{z=n} f(z) = \sum_n \frac{1}{(u+n)^2}.$$

We examine what happens when we integrate $f(z)$ along the rectangle $$\Gamma =\pm (N+1/2) \pm i N$$ with $N$ a large positive integer.

There are no poles on this contour and seeing that $\frac{1}{(u+z)^2}\in\Theta(1/N^2)$ on the contour, the integral $$\int_\Gamma f(z) dz$$ goes to zero as $N$ goes to infinity. This will be shown below.

This implies that the sum of the residues at the poles of $f(z)$ is zero, giving

$$S + \mathrm{Res}_{z=-u} \frac{1}{(u+z)^2} \pi\cot(\pi z) = 0.$$

The residue at the double pole at $z=u$ is given by

$$\left.(\pi \cot(\pi z))'\right|_{z=-u} = \left. - \frac{\pi^2}{\sin(\pi z)^2} \right|_{z=-u}$$

so that we have

$$\bbox[5px,border:2px solid #00A000]{ \sum_n \frac{1}{(u+n)^2} = \frac{\pi^2}{\sin(\pi u)^2}.}$$

To convince yourself that the integral really does vanish consider the two lines $\Gamma_1$ which is $N+1/2\pm iN$ (right vertical) and $\Gamma_2$ which is $\pm N+1/2+iN$ (top horizontal). With $N$ going to infinity we may suppose that $N\gt 1$ and also $N\gt \max(|\Re(u)|, |\Im(u)|).$

We parameterize $\Gamma_1$ with $z=N+1/2+it$ so that $$\left|\int_{\Gamma_1} f(z) dz \right| = \left|\int_{-N}^N f(N+1/2+it) i dt\right|.$$

The norm of the fractional term attains its maximum at $t=0$ when we cross the real axis where $|u+z|$ is minimized, giving an upper bound on the norm which is

$$\frac{1}{(\Re(u)+N+1/2)^2} = \frac{1}{N^2} \frac{1}{(1+(\Re(u)+1/2)/N)^2}.$$

For the norm of the trigonometric term we get $$|\pi\cot(\pi (N+1/2) + \pi it)| =\pi\left|\frac{e^{i\pi (N+1/2) - \pi t}+e^{-i\pi (N+1/2) + \pi t}} {e^{i\pi (N+1/2) - \pi t}-e^{-i\pi (N+1/2) + \pi t}}\right| \\ = \pi\left|\frac{i(-1)^N e^{- \pi t} - i(-1)^N e^{\pi t}} {i(-1)^N e^{- \pi t} + i(-1)^N e^{\pi t}}\right| = \pi|\tanh(\pi t)|.$$

Observe that with $t$ real $\pi \tanh(\pi t)$ has no poles and its norm is bounded by $\pi$. Therefore the norm of the integral along $\Gamma_1$ is bounded by

$$2N \times \frac{\pi}{(\Re(u)+N+1/2)^2} \in 2N \times \Theta(1/N^2) = \Theta(1/N)$$

and the integral vanishes as $N$ goes to infinity as claimed.

For $\Gamma_2$ we parameterize with $z = t + i N$ so that $$\left|\int_{\Gamma_2} f(z) dz \right| = \left|\int_{-(N+1/2)}^{N+1/2} f(t+iN) dt\right|.$$

The norm of the fractional term is minimized when we cross the imaginary axis, giving an upper bound on the norm which is

$$\frac{1}{(\Im(u)+N)^2} = \frac{1}{N^2} \frac{1}{(1+\Im(u)/N)^2}.$$

For the norm of the trigonometric term we get $$|\pi\cot(\pi t + \pi i N)| = \pi\left|\frac{e^{i\pi t - \pi N} + e^{-i\pi t + \pi N}} {e^{i\pi t - \pi N} - e^{-i\pi t + \pi N}}\right| \le \pi\left|\frac{e^{\pi N}+e^{-\pi N}}{e^{\pi N}-e^{-\pi N}}\right| =\pi|\coth(\pi N)|.$$

There aren't any poles here either and this term is bounded above by $\pi\coth(\pi)$ because $N>1.$ This gives the following bound on the norm of the integral along $\Gamma_2:$

$$(2N+1) \times \frac{\pi\coth(\pi)}{(\Im(u)+N)^2} \in (2N+1) \times \Theta(1/N^2) = \Theta(1/N)$$

and this integral also vanishes as $N$ goes to infinity as claimed.

The other two line segments can be bounded by the same technique.