<s>[INST] 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
 [/INST] Integrating the integral by parts thrice yields
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\sin ^{4} x}{x^{4}} d x \\
=& \int_{0}^{\infty} \frac{\left(\sin ^{4} x\right)^{(3)}}{3 \times 2 \cdot x} d x \\
=& \int_{0}^{0} \frac{4 \sin 2 x-8 \sin 4 x}{6 x} d x \\
=& \frac{4}{3}\left(-\int_{0}^{\infty} \frac{\sin 2 x}{x} d x+2 \int_{0}^{\infty} \frac{\sin 4 x}{x}\right) \\
=& \frac{2}{3}\left[-\frac{\pi}{2}+2\left(\frac{\pi}{2}\right)\right] \\
=& \frac{\pi}{3}
\end{aligned}
$$
</s>