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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:

- Fourier Transform - Laplace Equation on infinite strip - weird solution involving series
- Dirac Delta and Exponential integral
- Fourier Cosine Transform (Parseval Identity) for definite integral
- Fourier Transform of $\exp{(A\sin(x))}$
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$$E_g = \int_{-\infty}^{\infty} g^2(t) = \int_{-\infty}^{\infty} |G(f)|^2 df $$

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!

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- Distribution theory and differential equations.

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:

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

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