Integral through Fourier Transform and Parseval's Identity $$
\int_{-\infty}^{\infty}{\rm sinc}^{4}\left(\pi t\right)\,{\rm d}t\,.
$$
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..
 A: 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 
$$
\int_{\mathbb R} \operatorname{sinc}^4(2 \pi a \xi)\,d\xi
 = \frac{1}{16a^4}\cdot \frac{16a^3}{3} = \frac{1}{3a}
$$
A: $\newcommand{\+}{^{\dagger}}%
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$\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}

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



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