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

Question

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.

Progress

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?

Answer 1

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 $$ \sum_{k\in\mathbb{Z}}\frac{(-1)^k}{(k-z)^2}=\pi^2\csc(\pi z)\cot(\pi z)\tag{2} $$

Answer 2

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.

Answer 3

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

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

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[15px,#ffe]{\left.\sum_{n = -\infty}^{\infty}{\pars{-1}^{n} \over \pars{n - \alpha}^{2}} \right\vert_{\ 0\ <\ \alpha\ <\ 1} = \pi^{2}\csc\pars{\pi\alpha}\cot\pars{\pi\alpha}}:\ {\Large ?}}$.

---

\begin{align} \left.\sum_{n = -\infty}^{\infty}{\pars{-1}^{n} \over \pars{n - \alpha}^{2}} \right\vert_{\ 0\ <\ \alpha\ <\ 1} & = \sum_{n = 1}^{\infty}{\pars{-1}^{n} \over \pars{n + \alpha}^{2}} + {1 \over \alpha^{2}} + \sum_{n = 1}^{\infty}{\pars{-1}^{n} \over \pars{n - \alpha}^{2}}\label{1}\tag{1} \end{align} Then, \begin{align} \sum_{n = 1}^{\infty}{\pars{-1}^{n} \over \pars{n \pm \alpha}^{2}} & = \sum_{n = 1}^{\infty}\pars{-1}^{n}\ \overbrace{\bracks{-\int_{0}^{1}\ln\pars{x}\, x^{n \pm \alpha - 1}\,\dd x}} ^{\ds{1 \over \pars{n \pm a}^{2}}} \\[5mm] & = \int_{0}^{1}\ln\pars{x}\,x^{\pm \alpha} \sum_{n = 1}^{\infty}\pars{-x}^{n - 1}\,\dd x = \int_{0}^{1}{\ln\pars{x}\,x^{\pm \alpha} \over 1 + x}\,\dd x \\[5mm] & = \left.\partiald{}{\mu}\int_{0}^{1}{x^{\mu} \over 1 + x} \,\dd x\,\right\vert_{\ \mu\ =\ \pm\alpha} = \left.\partiald{}{\mu}\int_{0}^{1}{x^{\mu} - x^{\mu + 1} \over 1 - x^{2}} \,\dd x\,\right\vert_{\ \mu\ =\ \pm\alpha} \\[5mm] & = {1 \over 2}\partiald{}{\mu}\bracks{% \int_{0}^{1}{1 - x^{\mu/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{\mu/2 - 1/2} \over 1 - x}\,\dd x}_{\ \mu\ =\ \pm\alpha} \\[5mm] & = {1 \over 2}\partiald{}{\mu}\bracks{% \Psi\pars{{\mu \over 2} + 1} - \Psi\pars{{\mu \over 2} + {1 \over 2}}}_{\ \mu\ =\ \pm\alpha} \\[5mm] & = {1 \over 4}\bracks{% \Psi\, '\pars{{\pm\alpha \over 2} + 1} - \Psi\, '\pars{{\pm\alpha \over 2} + {1 \over 2}}}\label{2}\tag{2} \end{align}

---

Replacing (\ref{2}) in (\ref{1}): \begin{align} &\left.\sum_{n = -\infty}^{\infty}{\pars{-1}^{n} \over \pars{n - \alpha}^{2}}\right\vert_{\ 0\ <\ \alpha\ <\ 1} = {1 \over 4}\bracks{% \Psi\, '\pars{{\alpha \over 2} + 1} - \Psi\, '\pars{{\alpha \over 2} + {1 \over 2}}} + {1 \over \alpha^{2}} \\[2mm] + &\ {1 \over 4}\bracks{% \Psi\, '\pars{{-\alpha \over 2} + 1} - \Psi\, '\pars{{-\alpha \over 2} + {1 \over 2}}} \\[5mm] = &\ {1 \over \alpha^{2}} + {1 \over 4}\bracks{% \Psi\, '\pars{{\alpha \over 2} + 1} + \Psi\, '\pars{-\,{\alpha \over 2} + 1}} \\[2mm] - &\ {1 \over 4}\bracks{% \Psi\, '\pars{{\alpha \over 2} + {1 \over 2}} + \Psi\, '\pars{-\,{\alpha \over 2} + {1 \over 2}}} \\[5mm] = &\ {1 \over \alpha^{2}} + {1 \over 4}\bracks{% -\,{4 \over \alpha^{2}} + \Psi\, '\pars{{\alpha \over 2}} + \Psi\, '\pars{-\,{\alpha \over 2} + 1}} \\[2mm] - &\ {1 \over 4}\bracks{% \Psi\, '\pars{{\alpha \over 2} + {1 \over 2}} + \Psi\, '\pars{-\,{\alpha \over 2} + {1 \over 2}}} \\[5mm] = &\ {1 \over 4}\bracks{\pi^{2}\csc^{2}\pars{\pi\,{\alpha \over 2}}} - {1 \over 4}\braces{\pi^{2}\csc^{2}\pars{\pi\bracks{{1 \over 2} - {\alpha \over 2}}}} \\[5mm] = &\ \bbox[15px,#ffe,border:1px soloid navy]{\pi^{2}\csc\pars{\pi\alpha}\cot\pars{\pi\alpha}} \end{align}