Use parseval's identity to prove $\int_0^\infty {\sin^4t \over t^2} dt = {\pi \over 4}$ I know how to calculate
$$\int_0^\infty {\sin^4t \over t^4} dt$$
by taking the function $f(x)=1-|x|$ and it will be $\pi \over 3$.
But here the denominator is not raised to power $4$ but $2$.
How should I approach it?

$$\begin{align}
I&=-\frac{\sin^4(t)}{t}|_0^\infty+\int_0^\infty \frac{4\sin^3(t)\cos(t)}{t}~dt\\
\\
&=\int_0^\infty \frac{\left(1-\cos(2t)\right)\cdot\sin(2t)}{t}~dt\\
\\
&=\int_0^\infty \frac{\sin(2t)-\frac{1}2\sin(4t)}{t}~dt\\
\\
&=\frac{\pi}2-\frac{\pi}4=\frac{\pi}{4}\end{align}$$
