Integral through Fourier Transform and Parseval's Identity

Question

$$ \int_{-\infty}^{\infty}{\rm sinc}^{4}\left(\pi t\right)\,{\rm d}t\,. $$

Can you help me evaluate this integral with the help of Fourier Transform and Parseval Identity. I could not see how it is implemented. Thank you..

Answer 1

There are two correct answers to this question, depending on how you understand sinc. My guess is that your convention is $\operatorname{sinc}x = \frac{\sin x}{x}$.

I don't know your FT convention, but I will use $\hat f(\xi)=\int f(x)e^{-2\pi i \xi x}\,dx$. Then $$\hat \chi_{[-a,a]}(\xi)= \int_{-a}^a e^{-2\pi i \xi x}\,dx = \frac{e^{2\pi i a\xi}-e^{-2 \pi i a\xi}}{2\pi i \xi} = 2a\operatorname{sinc}(2 \pi a \xi)$$ To square the righthand side, convolve $\chi_{[-a,a]}$ with itself. This convolution is $f(x) = (2a-|x|)^+$. Thus, $\hat f(\xi) = 4a^2 \operatorname{sinc}^2(2 \pi a \xi)$. Since $$ \int_{\mathbb R} f^2 = 2\int_0^{2a} (2a-x)^2\,dx = \frac{16 a^3}{3} $$ Parseval's identity implies $$ \int_{\mathbb R} \operatorname{sinc}^4(2 \pi a \xi)\,d\xi = \frac{1}{16a^4}\cdot \frac{16a^3}{3} = \frac{1}{3a} $$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{-\infty}^{\infty}{\sin^{4}\pars{\pi t} \over \pars{\pi t}^{4}}\,\dd t ={1 \over \pi}\,{\cal F}\pars{1}\quad\mbox{where}\quad{\cal F}\pars{\mu} \equiv \int_{-\infty}^{\infty}{\sin^{4}\pars{\mu t} \over t^{4}}\,\dd t\,,\quad{\large \mu > 0}}$

\begin{align} &{\cal F}'\pars{\mu}= \int_{-\infty}^{\infty}{4\sin^{3}\pars{\mu t}\cos\pars{\mu t} \over t^{3}}\,\dd t = \int_{-\infty}^{\infty}{2\sin^{2}\pars{\mu t}\sin\pars{2\mu t} \over t^{3}}\,\dd t \\[3mm]&= \int_{-\infty}^{\infty}{\bracks{1 - \cos\pars{2\mu t}}\sin\pars{2\mu t} \over t^{3}}\,\dd t = \int_{-\infty}^{\infty}{\sin\pars{2\mu t} - \sin\pars{2\mu t}\cos\pars{2\mu t} \over t^{3}}\,\dd t \\[3mm]&= \half\int_{-\infty}^{\infty}{2\sin\pars{2\mu t} - \sin\pars{4\mu t} \over t^{3}}\,\dd t \end{align} ${\large\tt\mbox{Notice that}\ {\cal F}'\pars{0} = {\cal F}''\pars{0} = 0}$

\begin{align} &{\cal F}'''\pars{\mu}= \half\int_{-\infty}^{\infty}{-8t^{2}\sin\pars{2\mu t} + 16t^{2}\sin\pars{4\mu t} \over t^{3}}\,\dd t = \int_{-\infty}^{\infty}{8\sin\pars{4\mu t} - 4\sin\pars{2\mu t} \over t}\,\dd t \\[3mm]&= 8\int_{-\infty}^{\infty}{\sin\pars{4\mu t} \over t}\,\dd t - 4\int_{-\infty}^{\infty}{\sin\pars{2\mu t} \over t}\,\dd t = 4\int_{-\infty}^{\infty}{\sin\pars{t} \over t} = 4\int_{-\infty}^{\infty}\pars{\half\int_{-1}^{1}\expo{\ic kt}\,\dd k}\,\dd t \\[3mm]&= 4\pi\int_{-1}^{1}\dd k\int_{-\infty}^{\infty}\expo{\ic k t}\,{\dd t \over 2\pi} = 4\pi\int_{-1}^{1}\delta\pars{k}\,\dd k = 4\pi \end{align}

$$ {\cal F}''\pars{\mu} = 4\pi\mu\,,\quad {\cal F}'\pars{\mu} = 2\pi\mu^{2}\,, \quad{\cal F}\pars{\mu} = {2\pi \over 3}\,\mu^{3} $$

$$\color{#0000ff}{\large% \int_{-\infty}^{\infty}{\sin^{4}\pars{\pi t} \over \pars{\pi t}^{4}}\,\dd t = {2 \over 3}} $$