Evaluate $\int_0^\infty\frac {\sin^4(x)} {x^4} \operatorname dx$ How to evaluate the definite integral:
$$
\int \limits_0^\infty\frac {\sin^4(x)} {x^4} \operatorname dx
$$
Also provide the reference to various theorems to be to used to evaluate it!
Thank you.
 A: Recalling the result

$$ \int_{0}^{\infty} G(u)f(u) du = \int_{0}^{\infty} g(u)F(u) du, $$

where $F(u)$ and $G(u)$ are the Laplace transform of $f$ and $g$. Now, applying this to our problem gives

$$ \int_0^\infty\frac {\sin^4(x)} {x^4} dx = 4\int_{0}^{\infty} {\frac {x^2}{ \left( {x}^{2}+4 \right)  \left( {x}^{2}+16 \right) }
}=\frac{\pi}{3}\,.$$

Note: 
1) We used the following Laplace transforms

$$ \mathcal{L} (\sin(x)^4) =  {\frac {24}{s \left( {s}^{2}+4 \right)  \left( {s}^{2}+16 \right) }
},$$
$$ \mathcal{L} ( \frac{x^3}{6} )= \frac{1}{s^4}. $$

2) Use the partial fraction to evaluate the integral

$$ {\frac {x^2}{ \left( {x}^{2}+4 \right)  \left( {x}^{2}+16 \right) }
}= \frac{4}{3}\, \frac{1}{ {x}^{2}+16 } - \frac{1}{3}\, \frac{1}{ {x}^{2}+4 }.$$

A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\int_{0}^{\infty}{\sin^4\pars{x} \over x^4}\,\dd x:\ {\large ?}}$.
Let's
$\ds{{\cal I}\pars{\mu} \equiv \half\int_{-\infty}^{\infty}
{\sin^4\pars{\mu x} \over x^4}\,\dd x\ \mbox{with}\ \mu > 0}$ and such that
$\ds{\int_{0}^{\infty}{\sin^4\pars{x} \over x^4}\,\dd x = {\cal I}\pars{1}}$. Also, $\ds{{\cal I}\pars{0^{+}} = 0}$.

\begin{align}
{\cal I}'\pars{\mu}&=2\int_{-\infty}^{\infty}{\sin^{3}\pars{\mu x}\cos\pars{\mu x}
\over x^3}\,\dd x
=\int_{-\infty}^{\infty}{\sin^{2}\pars{\mu x}\sin\pars{2\mu x} \over x^{3}}\,\dd x
\\[3mm]&=\half\int_{-\infty}^{\infty}
{\sin\pars{2\mu x} - \sin\pars{2\mu x}\cos\pars{2\mu x} \over x^{3}}\,\dd x
={1 \over 4}\int_{-\infty}^{\infty}{2\sin\pars{2\mu x} - \sin\pars{4\mu x} \over x^{3}}
\,\dd x\\[3mm]&\mbox{with}\ {\cal I}'\pars{0^{+}} = 0
\end{align}

\begin{align}
{\cal I}''\pars{\mu}&=\int_{-\infty}^{\infty}
{\cos\pars{2\mu x} - \cos\pars{4\mu x} \over x^{2}}\,\dd x\,,
\qquad{\cal I}''\pars{0^{+}} = 0
\end{align}

\begin{align}
{\cal I}'''\pars{\mu}&=\int_{-\infty}^{\infty}
{-2\sin\pars{2\mu x} + 4\sin\pars{4\mu x} \over x}\,\dd x\
=-2\int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\dd x +
4\int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\dd x
\\[3mm]&=2\pi\quad\mbox{where we used the well know result}\
\int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\dd x = \pi 
\end{align}

Then,
$$
{\cal I}''\pars{\mu} = 2\pi\mu\,,\quad{\cal I}'\pars{\mu} = \pi\mu^{2}\quad
\mbox{and}\quad{\cal I}\pars{\mu} = {1 \over 3}\,\pi\mu^{3}
$$
$$\color{#00f}{\large%
\int_{0}^{\infty}{\sin^{4}\pars{x} \over x^{4}}\,\dd x = {1 \over 3}\,\pi}
$$
A: We can use a version of Parseval's equality:
$$\int_{-\infty}^{\infty} dx \, f(x)^2 = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k)^2$$
where
$$F(k) = \int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x}$$
Now, we may use the well-known FT:
$$f(x) = \frac{\sin^2{x}}{x^2} \implies F(k) = \pi \left (1-\frac{|k|}{2} \right ) $$
(This result may be easily derived using, e.g., the convolution theorem on the even more basic FT of $\sin{x}/x$.)
Thus the integral is $1/2$ of
$$\frac1{2 \pi} \pi^2 \int_{-2}^2 dk \, \left (1-\frac{|k|}{2} \right )^2  = \pi \int_0^2 \left ( 1-k+\frac14 k^2\right ) = \frac{2 \pi}{3}$$
or $\pi/3$.
A: NOTE: This is a sketch of what a solution may look like, but it is what it is and nothing more - a sketch.
I would do it 'a la residue'.
First, note that:
$$\sin^4{x} = \Re \frac{e^{4it}-4e^{2it}+3}{8}$$
Then note that if we choose a contour integral over the semicircle in the complex plane - that is on $\Gamma_{R}$ as $R \rightarrow \infty$ - we are bounded by $\frac{1}{R^3} \rightarrow 0$ as $R \rightarrow \infty$. We note that the only pole is at $z=0$ and so this problem reduces to $2\cdot \int_{0}^{\infty} \frac{\sin^4(x)}{x^4} = -\pi \cdot i \cdot Res_{z=0} \frac{\sin^4(z)}{z^4}$. Now the third derivative is $\frac{-64ie^{4it} + 32ie^{2it}}{8}$. This gives a residue of $1/3! \cdot \frac{-32i}{8} = -2i/3$. 
Hence, we conclude that $\int_{0}^{\infty} \frac{\sin^4(x)}{x^4} dx = 1/2\cdot \pi i \cdot -2i/3 = \pi/3$.
