Evaluating $\sum_{k=1}^{\infty}\left(\frac{\sin(tk)}{k}\right)^2$ Using Poisson Summation Formula, how do you evaluate the following infinite sum $\sum_{k=1}^{\infty}\left(\frac{\sin(tk)}{k}\right)^2$? 
The Poisson Summation Formula states that: $\sum_{k=-\infty}^{\infty}f(2\pi k)=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}\hat{f}(k)$.
I tried playing around with the idea that if $f(x)=\mathbf{1}_{[-1,1]}$ then $\hat{f}(\xi)=2\frac{\sin(i\xi)}{i\xi}$. How should I go ahead from here to compute the summation? 
 A: Compute the Fourier Transform
$$
\begin{align}
&\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,e^{-2\pi ix\xi}\,\mathrm{d}x\\
&=\int_{-\infty-i}^{\infty-i}\frac{e^{2iz}-2+e^{-2iz}}{-4z^2}\,e^{-2\pi iz\xi}\,\mathrm{d}z\tag1\\
&=\int_{-\infty-i}^{\infty-i}\frac{e^{2iz(1-\pi\xi)}-2e^{-2\pi iz\xi}+e^{-2iz(1+\pi\xi)}}{-4z^2}\,\mathrm{d}z\tag2\\[6pt]
&=\pi(1-\pi\xi)[\pi\xi\le1]+2\pi(\pi\xi)[\pi\xi\lt0]-\pi(1+\pi\xi)[\pi\xi\lt-1]\tag3\\[12pt]
&=\pi(1-\pi\xi)[0\le\pi\xi\le1]+\pi(1+\pi\xi)[-1\le\pi\xi\le0]\tag4\\[12pt]
&=\pi(1-\pi|\xi|)\big[\,\pi|\xi|\le1\,\big]\tag5
\end{align}
$$
Explanation:
$(1)$: write $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\phantom{\text{(1):}}$ shift the contour since there are no singularities
$(2)$: combine exponents
$(3)$: use the contour $[-R-i,R-i]\cup Re^{i\pi[0,1]}-i$
$\phantom{\text{(3):}}$ for exponentials with a positive coefficient of $iz$
$\phantom{\text{(3):}}$ use the contour $[-R-i,R-i]\cup Re^{-i\pi[0,1]}-i$
$\phantom{\text{(3):}}$ for exponentials with a negative coefficient of $iz$
$\phantom{\text{(3):}}$ we need only count the residues from
$\phantom{\text{(3):}}$ the exponentials with a positive coefficient of $iz$
$(4)$: simplify
$(5)$: simplify
For $t\gt0$, substitute $x\mapsto x/t$ and apply $(5)$:
$$
\begin{align}
\int_{-\infty}^\infty\frac{\sin^2(tx)}{x^2}\,e^{-2\pi ix\xi}\,\mathrm{d}x
&=t\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,e^{-2\pi ix\xi/t}\,\mathrm{d}x\tag6\\[6pt]
&=\pi(t-\pi|\xi|)\big[\,\pi|\xi|\le t\,\big]\tag7
\end{align}
$$

Apply Poisson Summation
$$
\begin{align}
t^2+2\sum_{k=1}^\infty\frac{\sin^2(tk)}{k^2}
&=\sum_{k\in\mathbb{Z}}\frac{\sin^2(tk)}{k^2}\tag8\\[6pt]
&=\sum_{k\in\mathbb{Z}}\pi(t-\pi|k|)\big[\,\pi|k|\le t\,\big]\tag9\\
&=\pi t+2\sum_{k=1}^{\lfloor t/\pi\rfloor}\pi(t-\pi k)\tag{10}\\[9pt]
&=\pi t+\left(2\pi t-\pi^2\right)\lfloor t/\pi\rfloor-\pi^2\lfloor t/\pi\rfloor^2\tag{11}
\end{align}
$$
Explanation:
$\phantom{1}(8)$: make a sum over $\mathbb{Z}$
$\phantom{1}(9)$: Poisson summation
$(10)$: make a sum over $\mathbb{N}$
$(11)$: sum in $k$
Solve for the sum:
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty\frac{\sin^2(tk)}{k^2}=\frac12\left((\pi-t)t+\left(2\pi t-\pi^2\right)\lfloor t/\pi\rfloor-\pi^2\lfloor t/\pi\rfloor^2\right)}\tag{12}
$$

A Dilogarithmic Identity
As Claude Leibovici shows using $\sin(tk)=\frac{e^{itk}-e^{-itk}}{2i}$,
$$\newcommand{\Li}{\operatorname{Li}}
\sum_{k=1}^\infty\frac{\sin^2(tk)}{k^2}=\frac{\pi^2}{12}-\frac14\left(\Li_2\left(e^{2it}\right)+\Li_2\left(e^{-2it}\right)\right)\tag{13}
$$
which gives a nice identity:
$$
\hspace{-18pt}\bbox[5px,border:2px solid #C0A000]{\Li_2\left(e^{it}\right)+\Li_2\left(e^{-it}\right)=\frac{\pi^2}3-\frac12\left((2\pi-t)t+4\!\left(\pi t-\pi^2\right)\left\lfloor\frac{t}{2\pi}\right\rfloor-4\pi^2\left\lfloor\frac{t}{2\pi}\right\rfloor^2\right)}\tag{14}
$$
A: Without Poisson summation formula
$$S=\sum_{k=1}^{\infty}\left(\frac{\sin(tk)}{k}\right)^2=\sum_{k=1}^{\infty}\frac{\sin^2(tk)}{k^2}=\frac 12\sum_{k=1}^{\infty}\frac{1-\cos(2tk)}{k^2}$$
make the cosine exponentials to get
$$S=\frac{\pi^2}{12}-\frac{1}{4} \left(\text{Li}_2\left(e^{-2 i t}\right)+\text{Li}_2\left(e^{2 i   t}\right)\right)$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[15px,#ffe]{\sum_{k = 1}^{\infty}
\bracks{\sin\pars{tk} \over k}^{2} =
\sum_{k = 1}^{\infty}
\bracks{\sin\pars{\verts{t}k} \over k}^{2}}:\
{\Large ?}}$.

With $\ds{z \equiv x + y\ic \in \mathbb{C}}$ and $\ds{x,y \in \mathbb{R}}$, I'll consider
$\ds{\bracks{\sin\pars{\verts{t}z}/z}^{\, 2}}/
\expo{2\pi\verts{y}}$ as $\ds{y \to \pm\infty}$. Namely,
\begin{align}
{\bracks{\sin\pars{\verts{t}z}/z}^{\,2} \over \expo{2\pi\verts{y}}}
& \,\,\,\stackrel{\mrm{as}\ y\ \pm\infty}{\sim}\,\,\,
{1 \over 4}\expo{2\ic\verts{t}x}\,
{\exp\pars{-2\pars{\pi - \verts{t}}\verts{y}} \over y^{2}}
\\[5mm] & \stackrel{\mrm{as}\ y\ \to\ \pm\infty}{\to}\,\,\, 0 \quad
\mbox{as long as}\quad\color{red}{\verts{t} < \pi}
\end{align}
which sets a $\ds{\underline{condition}}$ for the validity of the
Abel-Plana Formula. Also, see
this link.
Then,
\begin{align}
&\bbox[10px,#ffe]{\left.\sum_{k = 1}^{\infty}
\bracks{\sin\pars{tk} \over k}^{2}
\right\vert_{\ 0\ <\ \verts{t}\ <\ \pi}}
\\[5mm] = &\
t^{2}\sum_{k = 1}^{\infty}\mrm{sinc}^{2}\pars{\verts{t}k} =
-t^{2} + t^{2}\sum_{k = 0}^{\infty}\mrm{sinc}^{2}\pars{\verts{t}k}
\\[5mm] = &\
-t^{2} + t^{2}
\braces{\int_{0}^{\infty}\mrm{sinc}^{2}\pars{\verts{t}k}
\,\dd k +
\bracks{{1 \over 2}\,\mrm{sinc}^{2}\pars{\verts{t}\xi}}
_{\ \xi\ =\ 0}}
\label{1}\tag{1}
\\[5mm] = &\
-t^{2} + t^{2}\pars{{1 \over \verts{t}}\,{\pi \over 2} +
{1 \over 2}} =
{1 \over 2}\verts{t}\pars{\pi - \verts{t}}  
\end{align}
In line (\ref{1}), I used the
Abel-Plana.

In general,
\begin{align}
&\bbox[10px,#ffe]{\left.\bracks{%
\sum_{k = 1}^{\infty}{\sin\pars{tk} \over k}}^{2}
\,\right\vert_{\ t\ \in\ \mathbb{R}}}
\\[5mm] = &\
\left\{\begin{array}{lcl}
\ds{0} & \mbox{if} &
\ds{\verts{t}}\ \mbox{is a multiple of}\ \pi
\\[4mm]
\ds{{1 \over 2}\,\pi^{2}\braces{\verts{t} \over \pi}
\pars{1 - \braces{\verts{t} \over \pi}}} &&
\mbox{otherwise}
\end{array}\right.
\end{align}
