Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$

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 $$\begin{eqnarray} \int_0^\infty \frac{\sin^4(x)}{x^4} \mathrm{d}x &=& \frac{2}{3} \int_0^\infty \frac{\sin^2(2x)}{x^2} \mathrm{d} x - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=& \frac{4}{3} \int_0^\infty \frac{\sin^2(y)}{y^2} \mathrm{d} y - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=& \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x = \frac{\pi}{3} \end{eqnarray} $$

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

$$\int_{-\infty}^{\infty} dx \: \frac{\sin^4{x}}{x^4} = \frac{1}{2 \pi} \int_{-2 }^{2 } dk \: \pi^2 \left ( 1 - \frac{|k|}{2} \right )^2 = \frac{\pi}{2} \frac{4}{3} = \frac{2 \pi}{3}$$

$$\therefore \: \int_{0}^{\infty} dx \: \frac{\sin^4{x}}{x^4} = \frac{\pi}{3}$$

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.