Compute $\int_{0}^{\infty} \frac{\sin^4x}{x^4}dx$ Compute
$$
\int_{0}^{\infty} \frac{\sin^4x}{x^4}dx\;\;.
$$
Well, I'm not sure how to approach this. I thought about using Plancheral, and using $f(x) = \frac{\sin^2x}{x^2}$, and finding the fourier transform of $f(x)$. Any other better approach?
 A: I think you sould use integration by parts to do it more easily in the following way:
$$   \int \frac{\sin^4(x)}{x^4}dx =  -\frac{1}{3} \frac{\sin^4(x)}{x^3} + \frac{4}{3} \int \frac{\cos(x) \sin^3(x) }{x^3} dx $$
$$ =
  -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \frac{3 \cos^2(x) \sin^2(x) - \sin^4(x)}{x^2} dx$$
$$= -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \left(\frac{\sin^2(2x)}{x^2}  - \frac{\sin^2(x)}{x^2} \right) dx$$
Last line I have used $\sin^4(x) = \sin^2(x)(1-cos^2(x)) $ and simplify it.
In general we know that $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis.
Also put $2x = y$ and find the integral foe the other part.
A: I would use Parseval, which states that the integral of the square of the a function is related to the integral of the square of the FT of that function.  In this case, the function is $(\sin{x}/x)^2$, which has FT equal to $\pi (1-|k|/2)$ for $|k| \le 2$ and $0$ otherwise.  The integral here is equal to
$$\frac1{2 \pi} \int_0^2 dk \, \pi^2  \left (1-\frac{k}{2} \right )^2 = \frac{\pi}{3}$$
A: Using Plancherel:
$$ \begin{aligned} \int_{0}^{+\infty} \frac{ \sin^4 (x)}{x^4} \ \mathrm dx &= \frac{1}{2} \int_{-\infty}^{+\infty} \left (\frac{ \sin^2 (x)}{x^2} \right )^2 \ \mathrm dx \\&= \frac{\pi}{2} \int_{-\infty}^{+\infty} | f(\omega) |^2 \ \mathrm d\omega        \end{aligned} $$
Where:
$$ f(\omega) := \begin{cases} 1 - |\omega|, & |\omega| < 1 \\ 0, & \text{otherwise} \end{cases} $$
Therefore:
$$ \int_{0}^{+\infty} \frac{ \sin^4 (x)}{x^4} \ \mathrm dx = \frac{\pi}{2} \int_{-1}^{1} (1-|\omega|)^2 \ \mathrm d\omega = \frac{\pi}{3} $$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[15px,#ffd]{\int_{0}^{\infty}{\sin^{4}\pars{x} \over x^{4}}\,\dd x} =
{1 \over 2}\,\lim_{N \to \infty}\int_{-N\pi}^{N\pi}
{\sin^{4}\pars{x} \over x^{4}}\,\dd x
\\[5mm] = &\
{1 \over 2}\,\lim_{N \to \infty}\bracks{%
\int_{-N\pi}^{-\pars{N - 1}\pi}{\sin^{4}\pars{x} \over x^{4}}\,\dd x + \cdots +
\int_{\pars{N - 1}\pi}^{N\pi}{\sin^{4}\pars{x} \over x^{4}}\,\dd x}
\\[5mm] = &\
{1 \over 2}\int_{0}^{\pi}
\sin^{4}\pars{x}\sum_{k = -\infty}^{\infty}{1 \over \pars{x + k\pi}^{4}}\,\dd x
\\[5mm] = &\
{1 \over 2}\int_{0}^{\pi}
\sin^{4}\pars{x}
\,{2\cot^{2}\pars{x}\csc^{2}\pars{x} + \csc^{4}\pars{x} \over 3}\,\dd x
\\[5mm] = &\
{1 \over 3}\int_{0}^{\pi}\cos^{2}\pars{x}\,\dd x +
{1 \over 6}\int_{0}^{\pi}\dd x = {1 \over 6}\,\pi + {1 \over 6}\,\pi =
\bbx{\ds{\pi \over 3}} \\ &
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[15px,#ffd]{\int_{0}^{\infty}{\sin^{4}\pars{x} \over x^{4}}\,\dd x}
\\ = &\
\int_{0}^{\infty}\ \overbrace{\cos\pars{4x} - 4\cos\pars{2x} + 3 \over 8}^{\ds{\sin^{4}\pars{x}}}\
\overbrace{{1 \over 6}\int_{0}^{\infty}t^{3}\expo{-xt}\dd t}
^{\ds{1 \over x^{4}}}\ \dd x
\\[5mm] = &\
{1 \over 48}\int_{0}^{\infty}t^{3}\Re\int_{0}^{\infty}
\bracks{\expo{-\pars{t - 4\ic}x} -
4\expo{-\pars{t - 2\ic}x} + 3\expo{-tx}}\dd x\,\dd t
\\[5mm] = &\
{1 \over 48}\int_{0}^{\infty}
\bracks{{t^{4} \over t^{2} + 16} -
{4t^{4} \over t^{2} + 4} + 3t^{2}}\dd x\,\dd t
\\[1cm] = &\
{1 \over 48}\int_{0}^{\infty}
\left[\pars{t^{4} + 16t^{2}} - 16\pars{t^{2} + 16} + 256 \over t^{2} + 16\right.
\\[2mm] & \left.\phantom{{1 \over 48}\int_{0}^{\infty}\,\,\,\,\,\,\,}-
{4\pars{t^{4} + 4t^{2}} - 16\pars{t^{2} + 4} + 64\over t^{2} + 4} + 3t^{2}\right]\dd x\,\dd t
\\[1cm] = &\ 
{1 \over 48}\int_{0}^{\infty}
\pars{{256 \over t^{2} + 16} - {64 \over t^{2} + 4}}\dd t
\\[5mm] = &\
{1 \over 48}\,{1 \over 16}\,256 \times 4\int_{0}^{\infty}{\dd t \over t^{2} + 1} -
{1 \over 48}\,{1 \over 4}\,64 \times 2\int_{0}^{\infty}
{\dd t \over t^{2} + 1}
\\[5mm] = &\
\pars{{4 \over 3} - {2 \over 3}}\,{\pi \over 2} =
\bbx{\large{\pi \over 3}}
\end{align}
