The question at hand is to use Parseval's theorem to solve the following integral:
$$\int_{-\infty}^{\infty} sinc^4 (kt) dt$$
I understand Parseval's theorem to be:
$$E_g = \int_{-\infty}^{\infty} g^2(t) = \int_{-\infty}^{\infty} |G(f)|^2 df $$
I began by doing the obvious and removing the squared such that:
$$g^2(t) = Sinc^4 (kt)$$
$$g(t) = Sinc^2 (kt)$$
Following the table of Fourier transforms in my book, I see that
$B*sinc^2({\pi}Bt)$ has the transform $\Delta(\frac{f}{2B})$. However, I'm stuck at this becuase I'm not sure how I can integrate the $\Delta$. I also feel as though I am overthinking this problem – any assistance would be greatly appreciated! Thank you so much in advanced!
Best Answer
Using Parseval's Theorem, we have
$$\begin{align} \int_{-\infty}^\infty \text{sinc}^4(kt)\,dt&=\frac{1}{k}\int_{-\infty}^\infty \text{sinc}^4(t)\,dt\\\\ &=\frac{1}{2\pi k}\int_{-\infty}^\infty \left|\mathscr{F}\left(\text{sinc}^2\right)(\omega)\right|^2\,d\omega \end{align}$$
where
$$\begin{align} \mathscr{F}\left(\text{sinc}^2\right)(\omega)&=\int_{-\infty}^\infty \text{sinc}^2(t)e^{i\omega t}\,dt\\\\ &=\frac{\pi }{4}\left(|\omega -2|-2|\omega|+|\omega+2|\right) \end{align}$$
Therefore,
$$\begin{align} \int_{-\infty}^\infty \text{sinc}^4(kt)\,dt&=\frac{1}{2\pi k} \int_{-\infty}^\infty \left(\frac{\pi }{4}\left(|\omega -2|-2|\omega|+|\omega+2|\right)\right)^2 \,d\omega\\\\ &=\frac{\pi}{16 k}\int_0^2 \left(4-2\omega \right)^2 \,d\omega\\\\ &=\frac{2\pi}{3k} \end{align}$$