Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$ I need to show that 
$$
\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}
$$
I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from.  Using a change of variable $ x \mapsto 2x $ :
$$
\int_0^\infty \frac{\sin^2(2x)}{x^2}dx = \pi
$$
Now using the identity $\sin^2(2x) = 4\sin^2x - 4\sin^4x $, we obtain
$$
\int_0^\infty \frac{\sin^2x - \sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\frac{\pi}{2} - \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
But I am now at a loss as to how to make $x^4$ appear at the denominator.  Any ideas appreciated.
Important: I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and use the change of variable and identity mentioned above
 A: You mentioned that you used complex analysis to evaluate $\int_{0}^{\infty} \frac{\sin^{2}(x)}{x^{2}} \, \mathrm dx$.
We can also use complex analysis to evaluate $\int_{0}^{\infty} \frac{\sin^{4}(x) }{x^{4}} \, \mathrm dx$.
Using the trigonometric identity $ \displaystyle \sin^{4} x = \frac{1}{8} \Big(\cos 4x - 4 \cos 2x + 3 \Big)$, we get
$$ \begin{align} \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \, \mathrm dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{4} x}{x^4} \, \mathrm dx \\ &= \frac{1}{16} \int_{-\infty}^{\infty} \Re \ \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \, \mathrm  dx \\ &= \frac{1}{16}\int_{-\infty}^{\infty} \Re \ \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, \mathrm dx \\ &= \frac{1}{16} \, \Re \, \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, \mathrm dx. \end{align}$$
So let's integrate the function $$f(z) = \frac{e^{4iz}-4e^{2iz}+3+4iz}{z^{4}}$$ around a contour the consists of the real axis from $-R$ to $R$, $R>0$, and the upper half of the circle $|z|=R$. To avoid the simple pole at the origin, the contour needs to be indented at the origin.
Letting the radius of the indentation go to zero and $R \to \infty$, we get
$$ \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, \mathrm dx- i \pi \ \text{Res}[f(z),0] = 0,$$
where
$$ \begin{align} \operatorname{Res}[f(z),0] &= \lim_{z \to 0} \frac{e^{4iz}-4e^{2iz}+3+4iz}{z^{3}} \\ &= \lim_{z \to 0} \frac{-64ie^{4iz}+32ie^{2iz} }{6} \\ &= - \frac{16i}{3}. \end{align}$$
Therefore,
$$ \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \, dx = \frac{1}{16} \left(\frac{16 \pi}{3} \right)=\frac{\pi}{3} .$$

Technically, it wasn't necessary to add $4ix$ to the numerator.
For reasons explained here, the Cauchy principal value of $ \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \, \mathrm dx $ exists even though $\frac{e^{4iz}-4e^{2iz}+3}{z^{4}}$ has a pole of order $3$ at the origin.
A: HINT: Use the relation 
$$\int_0^\infty  \left(\frac{\sin x}{x}\right)^n \mathrm{d}x
= \frac{\pi}{2^n (n-1)!}  
    \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}$$
You may find a proof here A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}{\sin^{4}\pars{x} \over x^{4}}\,\dd x = {\pi \over 3}:\
     {\large ?}}$

$\large\tt \mbox{METHOD}\ 0:$ Let's
  $\ds{{\cal F}\pars{\mu} \equiv
     \int_{0}^{\infty}{\sin^{4}\pars{\mu x} \over x^{4}}\,\dd x}$ such that
  $\ds{\int_{0}^{\infty}{\sin^{4}\pars{\mu x} \over x^{4}}\,\dd x ={\cal F}\pars{1}}$.
  \begin{align}
\color{#c00000}{{\cal F}'\pars{\mu}}
&=\int_{0}^{\infty}{4\sin^{3}\pars{\mu x}\cos\pars{\mu x} \over x^{3}}\,\dd x
=\int_{0}^{\infty}{\bracks{1 - \cos\pars{2\mu x}}\sin\pars{2\mu x}
                           \over x^{3}}\,\dd x
\\[3mm]&=\half\int_{0}^{\infty}{2\sin\pars{2\mu x} - \sin\pars{4\mu x}
                           \over x^{3}}\,\dd x
\quad\mbox{with}\ {\cal F}\pars{0} = 0\\[5mm]&\mbox{}
\end{align}
  \begin{align}
\color{#c00000}{{\cal F}''\pars{\mu}}&=
2\int_{0}^{\infty}{\cos\pars{2\mu x}
- \cos\pars{4\mu x} \over x^{2}}\,\dd x
\quad\mbox{with}\ {\cal F}'\pars{0} = 0\\[5mm]&\mbox{}
\end{align}
  \begin{align}
\color{#c00000}{{\cal F}'''\pars{\mu}}&=
2\int_{0}^{\infty}{-2\sin\pars{2\mu x}
+ 4\sin\pars{4\mu x} \over x}\,\dd x
=4\sgn\pars{\mu}\
\overbrace{\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x}
^{\ds{=\ {\pi \over 2}\,. \mbox{See below}}} = 2\pi\sgn\pars{\mu}
\\[3mm]&\mbox{with}\ {\cal F}''\pars{0} = 0
\end{align}
  $$
\mbox{With}\ \mu > 0\,,\ {\cal F}'''\pars{\mu} = 2\pi\ \imp\ {\cal F}''\pars{\mu}
=2\pi\mu\ \imp\ {\cal F}'\pars{\mu}=\pi\mu^{2}\ \imp\
{\cal F}\pars{\mu} = {\pi \over 3}\,\mu^{3}
$$
  $$
\color{#00f}{\large\int_{0}^{\infty}{\sin^{4}\pars{x} \over x^{4}}\,\dd x
={\cal F}\pars{1} = {\pi \over 3}\large}
$$
  $-----------------------------------------$
  Also
  \begin{align}
\color{#c00000}{\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x}&=
\half\int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\dd x
=\half\int_{-\infty}^{\infty}\
\overbrace{\bracks{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k}}
^{\ds{=\ {\sin\pars{x} \over x}}}\ \dd x
\\[3mm]&={\pi \over 2}\int_{-1}^{1}\
\overbrace{\bracks{\int_{-\infty}^{\infty}\expo{\ic kx}\,{\dd x \over 2\pi}}}
^{\ds{=\ \delta\pars{k}}}\ \dd k
={\pi \over 2}\int_{-1}^{1}\delta\pars{k}\,\dd k =\color{#c00000}{\pi \over 2}
\end{align}
  $\ds{\delta\pars{k}}$ is the
  Dirac Delta Function.

A: You are likely expected to integrate by parts (twice)
$$ \begin{eqnarray}
   \int \frac{\sin^4(x)}{x^4} \mathrm{d}x &=&  -\frac{1}{3} \frac{\sin^4(x)}{x^3} + \frac{4}{3} \int \frac{\cos(x) \sin^3(x) }{x^3} \mathrm{d} x
\\ 
  &=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \frac{3 \cos^2(x) \sin^2(x) - \sin^4(x)}{x^2} \mathrm{d} x
\\
  &=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \left(\frac{\sin^2(2x)}{x^2}  - \frac{\sin^2(x)}{x^2} \right) \mathrm{d}x
\end{eqnarray}
$$
where the last equality used 
$$\begin{eqnarray} 
3 \cos^2(x) \sin^2(x) - \sin^4(x) &=& 3 \cos^2(x) \sin^2(x) - \sin^2(x) (1-\cos^2(x)) \\ &=& \left(2 \sin(x) \cos(x) \right)^2 - \sin^2(x) = \sin^2(2x) - \sin^2(x)
\end{eqnarray}
$$
Now
$$\begin{eqnarray} 
   \int_0^\infty \frac{\sin^4(x)}{x^4} \mathrm{d}x &=& \frac{2}{3} \int_0^\infty \frac{\sin^2(2x)}{x^2} \mathrm{d} x - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=& \frac{4}{3} \int_0^\infty \frac{\sin^2(y)}{y^2} \mathrm{d} y - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=&
  \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x = \frac{\pi}{3}
\end{eqnarray}
$$
A: Hint: use Parseval/Plancherel theorem on $(\sin{x}/x)^2$.
That is, the FT of $(\sin{x}/x)^2$ is 
$$\int_{-\infty}^{\infty} dx \: \frac{\sin^2{x}}{x^2} e^{i k x} = \begin{cases} \\\pi \left (1 - \frac{|k|}{2} \right ) & |k| \le 2  \\ 0 & |k| > 2  \end{cases}$$
Plancherel/Parseval says that
$$\int_{-\infty}^{\infty} dx \: \frac{\sin^4{x}}{x^4} = \frac{1}{2 \pi} \int_{-2 }^{2 } dk \: \pi^2 \left ( 1 - \frac{|k|}{2} \right )^2 = \frac{\pi}{2} \frac{4}{3} = \frac{2 \pi}{3}$$
$$\therefore \: \int_{0}^{\infty} dx \: \frac{\sin^4{x}}{x^4} = \frac{\pi}{3}$$
A: Integrating the integral by parts thrice yields
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\sin ^{4} x}{x^{4}} d x \\
=& \int_{0}^{\infty} \frac{\left(\sin ^{4} x\right)^{(3)}}{3 \times 2 \cdot x} d x \\
=& \int_{0}^{0} \frac{4 \sin 2 x-8 \sin 4 x}{6 x} d x \\
=& \frac{4}{3}\left(-\int_{0}^{\infty} \frac{\sin 2 x}{x} d x+2 \int_{0}^{\infty} \frac{\sin 4 x}{x}\right) \\
=& \frac{2}{3}\left[-\frac{\pi}{2}+2\left(\frac{\pi}{2}\right)\right] \\
=& \frac{\pi}{3}
\end{aligned}
$$
