I'm trying to show that $\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(n-\alpha)^2}=\pi^2\csc \pi \alpha \cot \pi \alpha$ for $0<\alpha<1$. The method that I decided to use is contour (square) integrals and the residue theorem.


I noticed that $f(z)=\frac{1}{z^2\sin\pi(z+\alpha)}$ has poles at $z=0, k-\alpha, k\in \mathbb{Z}$ with residues $$\mbox{res}(f(z);0)=-\pi \cot (\pi \alpha) \csc( \pi \alpha) \\ \mbox{res}(f(z);k-\alpha)=\frac{(-1)^k}{\pi(k-\alpha)^2}$$ which one can obtain by simply calculating. So then I considered a path $\Gamma_N$ defined by a square with vertices at $N(1, i), N(1, -1), N(-1, i), N(-1, -i)$, $N \in \mathbb{N}$. So if we let $a_k$ be the list of poles of $f$ inside the region formed by $\Gamma_N$, by the residue theorem, we get:

$$\begin{align*} \oint_{\Gamma_N}f(z)dz &= 2\pi i\sum_{a_i} \mbox{res}(f(z);a_k) \\ &= 2\pi i \left[ -\pi \cot (\pi \alpha) \csc( \pi \alpha)+\sum_{n=-N+1}^N \frac{(-1)^n}{\pi(n-\alpha)^2}\right] \\ &\to 2\pi i \left[ -\pi \cot (\pi \alpha) \csc( \pi \alpha)+\sum_{-\infty}^{\infty} \frac{(-1)^n}{\pi(n-\alpha)^2}\right] \end{align*}$$

Now, supposing $\oint_{\Gamma_N}f(z)dz \to 0$, we get $$0=-\pi \cot (\pi \alpha) \csc( \pi \alpha)+\sum_{-\infty}^{\infty} \frac{(-1)^n}{\pi(n-\alpha)^2}$$ and then we're done.

But I cannot seem to prove that $\oint_{\Gamma_N}f(z)dz \to 0$. I'm pretty sure the Estimation Lemma is used here, but I am not sure how to go about it. Is this even true? Are there any other ways of proving it?

  • $\begingroup$ Doing this rigorously with residues is tricky. It's easier to prove that starting from known expansions, especially 4.22.4 in the standard tables. $\endgroup$ – Professor Vector Sep 29 '17 at 4:11

Hint: You can use an analogous function $$f(z) = \frac{\csc(\pi z)}{(z-a)^2}$$ integrate around the square $\Gamma_N$ with vertices $\pm N \pm Ni$, where $N$ is a half integer. Regarding your question of vanishing of the integral, prove the following:

$\csc(z)$ is uniformly bounded on $\Gamma_N$

Since the denominator is of $O(1/z^2)$, this shows the integral tends to 0.

The following result is also handy at residue summation, they're not difficult to prove either.

$\csc(z), \cot(z)$ are uniformly bounded on $\Gamma_N$ with $N$ half integer.

$\tan(z), \sec(z)$ are uniformly bounded on $\Gamma_N$ with $N$ integer.

  • $\begingroup$ Cheers got it ${}{}$ $\endgroup$ – Jihoon Kang Sep 29 '17 at 7:45

In $(3)$ from this answer, it is shown that $$ \sum_{k\in\mathbb{Z}}\frac{(-1)^k}{k+z}=\pi\csc(\pi z)\tag{1} $$ Substituting $z\mapsto-z$ in $(1)$ and taking the derivative yields $$ \sum_{k\in\mathbb{Z}}\frac{(-1)^k}{(k-z)^2}=\pi^2\csc(\pi z)\cot(\pi z)\tag{2} $$


The OP had asked, "Are there any other ways of proving it?"

We begin by writing the Fourier series,

$$\cos(\alpha x)=a_0/2+\sum_{n=1}^\infty a_n\cos(nx) \tag1$$

for $x\in [-\pi/\pi]$. The Fourier coefficients in $(1)$ are given by

$$\begin{align} a_n&=\frac{2}{\pi}\int_0^\pi \cos(\alpha x)\cos(nx)\,dx\\\\ &=\frac1\pi (-1)^n \sin(\pi \alpha)\left(\frac{1}{\alpha +n}+\frac{1}{\alpha -n}\right)\tag2 \end{align}$$

Substituting $2$ into $1$, dividing by $\sin(\pi y)$, and setting $x=0$ n $(2)$ reveals

$$\begin{align} \pi \csc(\pi \alpha)&=\frac1\alpha +\sum_{n=1}^\infty (-1)^n\left(\frac{1}{\alpha -n}+\frac{1}{\alpha +n}\right)\\\\ &=\sum_{n=-\infty}^\infty \frac{(-1)^n}{\alpha-n}\tag3 \end{align}$$

Now differentiating with respect to $\alpha$, enforcing the substitution $\alpha\to \alpha/\pi$, and dividing by $\pi^2$ yields the coveted result

$$\csc(\alpha)\cot(\alpha)=\sum_{n=-\infty}^\infty \frac{(-1)^n}{(n-\alpha)^2}$$

  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Jan 30 at 5:03

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