Calculate $\sum_\limits{n=-\infty}^{\infty} \left(\frac{\sin(w_0n)}{\pi n}\right)^4$ where $w_0<\frac{\pi}{2}$ By Parseval's theorem, this sum will be equal to the power in the Fourier transform of $$\left(\frac{\sin(w_0n)}{\pi n}\right)^2$$
$\left(\dfrac{\sin(w_0n)}{\pi n}\right)^2$ can be written as $x[n]\cdot x[n]$, where $x[n]= \dfrac{\sin(w_0n)}{\pi n}$   So we can compute the Fourier transform as the convolution:
$$\frac{1}{2\pi} \int_{-\pi}^{\pi} X(t)X(w-t)\mathrm dt$$
I'm stuck on finding this convolution. I know that $X(t)$ is $1$ when $-w_0\leq t\leq w_0$. I don't know when the product $X(t)X(w-t)$ will be $1$ and when $0$. Can someone help me?
 A: Find a function whose Fourier coefficients are a square root of your summand.  Then evaluate the integral of the square of that function.
To wit, let
$$f(x) = \begin{cases} \frac{w_0}{2 \pi} \left (2-\left | \frac{x}{w_0} \right | \right )  & |x|<2 w_0 \\0&|x|>2 w_0 \end{cases}$$
Then, if 
$$f(x) = \sum_{k=-\infty}^{\infty} c_k e^{i k x}$$
then for $0 \lt w_0 \lt \pi/2$,
$$c_k = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: f(x) e^{i k x} = \frac{\sin^2{w_0 k}}{\pi^2 k^2}$$
By Parseval's Theorem:
$$\sum_{k=-\infty}^{\infty} \frac{\sin^4{w_0 k}}{\pi^4 k^4} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: |f(x)|^2 = \frac{2 w_0^3}{3 \pi^3} $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Lets study the behavior of
$\ds{\,\,\,
\bracks{\sin\pars{w_{0}z} \over \pi z}^{4}_{\,z\ \in\ \mathbb{C}}}$ as $\ds{\,\,y \equiv \Im\pars{z} \to \pm\infty}$.
Namely,
\begin{align}
&\expo{-2\pi\verts{y}}\,
\bracks{\sin\pars{w_{0}z} \over \pi z}_{\,x\ =\ \Re\pars{z}}^{4}
\\[5mm] 
\stackrel{\mrm{as}\ y\ \to\ \pm\infty}{\sim}\,\,\,&
\expo{-2\pi\verts{y}}\,\,
{\expo{\pm 4\ic x}\expo{4\verts{w0}\verts{y}} \over 16\pi^{4}y^{4}}
\\[5mm] = &\
{\expo{\pm 4\ic x} \over 16\pi^{4}}\,\,
y^{-4}\,\exp\pars{\vphantom{\huge A}-4\verts{y}
\bracks{\color{red}{\vphantom{\LARGE A}{\pi \over 2} - \verts{w_{0}}}}}
\\[5mm]
\stackrel{\mrm{as}\ y\ \to\ \pm\infty}{\to} &
\,\,\,\,\,{\large 0}\quad
\mbox{whenever}\quad \color{red}{\verts{w_{0}} <
{\pi \over 2}}.
\end{align}
It enforces the validity of the Abel-Plana Formula I'm using in the following evaluation. Additional details are given in the above cited link.

Then,
\begin{align}
&\bbox[5px,#ffd]{\sum_{n = -\infty}^{\infty}\ \bracks{{\sin\pars{w_{0}\,n} \over \pi n}}^{4}} =
{w_{0}^{4} \over \pi^{4}}\sum_{n = -\infty}^{\infty}\ \on{sinc}^{4}\pars{w_{0}\,n}
\\[5mm] = &\
-\,{w_{0}^{4} \over \pi^{4}} + {2w_{0}^{4} \over \pi^{4}}\sum_{n = 0}^{\infty}\ \on{sinc}^{4}\pars{w_{0}\,n}
\\[5mm] = &\
-\,{w_{0}^{4} \over \pi^{4}} +
{2w_{0}^{4} \over \pi^{4}}\bracks{\int_{0}^{\infty}\ \on{sinc}^{4}\pars{w_{0}\,n}\,\dd n +
{1 \over 2}\,\on{sinc}\pars{0}}
\\[5mm] = &\
{2w_{0}^{3} \over \pi^{4}}\
\underbrace{\int_{0}^{\infty}\ {\sin^{4}\pars{n} \over n^{4}}\,\dd n}_{\ds{\pi \over 3}} =
\bbx{{2 \over 3\pi^{3}}\,w_{0}^{3}}
\end{align}
