Using Parseval's theorem to solve an integral 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!
 A: 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, 
$$\begin{align}
\int_{-\infty}^\infty \text{sinc}^4(kt)\,dt&=\frac{1}{2\pi k} \int_{-\infty}^\infty \left(\frac{\pi }{4}\left(|\omega -2|-2|\omega|+|\omega+2|\right)\right)^2 \,d\omega\\\\
&=\frac{\pi}{16 k}\int_0^2 \left(4-2\omega \right)^2 \,d\omega\\\\
&=\frac{2\pi}{3k}
\end{align}$$
A: Another approach, maybe easier.
$$ I(k)=\int_\mathbb{R}\text{sinc}(kt)^4\,dt = \frac{1}{k}\int_{-\infty}^{+\infty}\frac{\sin(x)^4}{x^4}\,dx \tag{1}$$
but $\sin(x)^4 = \frac{3}{8}-\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$ by De Moivre's formula, so by applying integration by parts three times:
$$ I(k) = \frac{1}{6k}\int_{-\infty}^{+\infty}\frac{\frac{d^3}{dx^3}\sin(x)^4}{x}\,dx = \frac{1}{6k}\int_{-\infty}^{+\infty}\frac{8\sin(4x)-4\sin(2x)}{x}\,dx\tag{2}$$
and:

$$ I(k) = \frac{(8-4)\pi}{6k} = \color{red}{\frac{2\pi}{3k}}\tag{3}$$

follows.
