I want to solve the following integral: $$\int^{+\infty}_{-\infty} \frac{\sin^2 t}{t^2}dt$$
using Parseval Theorem: $$\int^{+\infty}_{-\infty} f(x)g^{*}(x)dx = \int^{+\infty}_{-\infty} F(\omega)G^{*}(\omega)d\omega$$
where $F(\omega) = \mathfrak F \{ f(x)\} $
$ G^{*}(\omega)$ is the complex conjugate of the Fourier transform of $g(x)$
I tried to do the following:
$$f(x) = \sin^2 t = \frac{1-\cos 2t}{2} \space ; \space g(x) = t^{-2}$$
However I am only able to obtain :
$$ F(\omega) = \frac{1}{2 \sqrt{2\pi}} \int^{+\infty}_{-\infty} (1-\cos 2t)e^{i\omega t}dt $$
My problem is when I am trying to obtain:
$$G(\omega) =\frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} t^{-2}e^{i\omega t}dt$$
I think the problem is the discontinuity at $t=0$, but if this is the problem I will alwyas have it since the function to integrate is divided by $t^2$
Anybody knows if the integral can be solved? or if there is a better approach to this problem?
Thanks in advance!
The idea is to use that $$ \frac{1}{\sqrt{2\pi}}\int_{-1}^{1} \frac{1}{2}e^{ix t} \, dx = \frac{1}{\sqrt{2\pi}}\frac{e^{it}-e^{-it}}{2it} = \frac{1}{\sqrt{2\pi}}\frac{\sin{t}}{t}, $$ so $(\sin{t})/t$ is the Fourier transform of the function $B(x)$ that is $\sqrt{\pi/2}$ on $[-1,1]$ and zero elsewhere. Then Parseval gives $$ \int_{-\infty}^{\infty} \left( \frac{\sin{t}}{t} \right)^2 \, dt = \int_{-1}^{1} \left( \sqrt{\frac{\pi}{2}} \right)^2 \, dx = 2 \cdot \frac{\pi}{2} = \pi. $$