Showing that $\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(n-\alpha)^2}=\pi^2\csc \pi \alpha \cot \pi \alpha$ 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?
 A: 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
$$
\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{(k-z)^2}=\pi^2\csc(\pi z)\cot(\pi z)\tag{2}
$$
A: 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.

A: 
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
$$\csc(\alpha)\cot(\alpha)=\sum_{n=-\infty}^\infty \frac{(-1)^n}{(n-\alpha)^2}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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$\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}
