Calculate $\sum_\limits{n=-\infty}^{\infty} \left(\frac{\sin(w_0n)}{\pi n}\right)^4$ where $w_0<\frac{\pi}{2}$

Question

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?

Answer 1

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:

$$\sum_{k=-\infty}^{\infty} \frac{\sin^4{w_0 k}}{\pi^4 k^4} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: |f(x)|^2 = \frac{2 w_0^3}{3 \pi^3} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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.

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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}