# 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!

#### Solutions Collecting From Web of "Using Parseval's theorem to solve an integral"

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}

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.