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?
 A: Let $f(x) := \begin{cases}-\frac14 & \text{if $0 \le x \le 2$}\\ \frac14 & \text{if $-2 \le x < 0$}\\ 0 & \text{otherwise}\end{cases}.$
\begin{align}
F\left(\omega\right) &= \int_{-\infty}^{\infty}f(x)e^{-i\omega x}\mathrm d x\\
&= -\frac14\int_{0}^2 e^{-i\omega x}\mathrm d x + \frac14\int_{-2}^0 e^{-i\omega x}\mathrm d x\\
&= \frac1{4i\omega}\left(e^{2i\omega} - 1 - 1 + e^{-2i\omega}\right)\\
&= \frac1{i\omega}\left(\frac{e^{i\omega} - e^{-i\omega}}{2i}\right)^2\\
&= i\times\frac{1}{\omega}\sin^2\left(\omega\right)
\end{align}
By parceval identity:
\begin{align}
\int_{0}^\infty \frac1{\omega^2}\sin^4\left(\omega\right)\mathrm d \omega
&= \frac1{2}\int_{-\infty}^\infty\left|F(\omega)\right|^2 \mathrm d \omega\\
&= \frac1{2} \times 2\pi \int_{-\infty}^{\infty}\left|f(x)\right|^2\mathrm d x\\ &= \pi \int_{-2}^{2}\frac1{16}\mathrm d x = \frac\pi{16} \times 4 = \frac\pi 4
\end{align}
A: $$\begin{align}
I&=-\frac{\sin^4(t)}{t}|_0^\infty+\int_0^\infty \frac{4\sin^3(t)\cos(t)}{t}~dt\\
\\
&=\int_0^\infty \frac{\left(1-\cos(2t)\right)\cdot\sin(2t)}{t}~dt\\
\\
&=\int_0^\infty \frac{\sin(2t)-\frac{1}2\sin(4t)}{t}~dt\\
\\
&=\frac{\pi}2-\frac{\pi}4=\frac{\pi}{4}\end{align}$$
