Integrating Powers of $\frac{\sin x}{x}$ using Fourier Transforms

This is a problem from a past qualifying exam:

The Fourier transform of the characteristic function $h=\chi_{[-1,1]}$ of the interval $[-1,1]$ is $$\hat h(\xi) =\sqrt{\frac{2}{\pi}} \frac{\sin \xi}{\xi}.$$ Using various properties of the Fourier transform, calculate $$\int_0^\infty \frac{\sin x}{x}dx$$ $$\int_0^\infty \big(\frac{\sin x}{x}\big)^2dx$$ $$\int_0^\infty \big(\frac{\sin x}{x}\big)^4dx.$$ Note: Here we are using the definition of the Fourier transform $$\hat f(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-ix\xi} f(x) dx.$$ To evaluate the third integral, you may make use of the formula $$ (h\ast h)(x) = \begin{cases} 2-|x| & \ |x|<2 \\ 0 & \ |x| \geq 2 \end{cases}$$.

I figured out the $\int_0^\infty \big(\frac{\sin x}{x}\big)^2dx$ portion by simply using Plancherel, i.e. $||h||_2^2 = ||\hat h||_2^2$. I am still struggling with the other cases, however. I have tried the Fourier Inversion as well as multiplication formula. Because of the hint, I would gather that in the last case you are also supposed to use the fact that $\widehat{h\ast h} = \hat h \cdot \hat h$.

Applying the inversion theorem, we may write

$$h(x) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{i\xi x}\sqrt{\frac{2}{\pi}}\frac{\sin(\xi)}{\xi} \mathrm{d}\xi$$ which leads to $$\int_{-\infty}^{\infty} \frac{1}{\pi}\frac{\sin(\xi)}{\xi} \mathrm{d}\xi = h(0) = 1. $$ This in turn implies

$$\bbox[5px,border:2px solid #CAAA00]{\int_{0}^{\infty} \frac{\sin(x)}{x} \mathrm{d}x = \frac{\pi}{2}}$$

To calculate the second integral it is sufficient to use Plancherel. You should get $\pi/2$ as well if I remember correctly. The third one can be calculated in a similar fashion: $$\int_{-\infty}^{\infty} \left(\sqrt{\frac{2}{\pi}}\frac{\sin(\xi)} {\xi}\right)^4 \mathrm{d}\xi =\|\hat{h}^2\|_{L^2}^2 = \frac{1}{2\pi} \|\widehat{h\ast h}\|_{L^2}^2 = \frac{1}{2\pi}\|h \ast h\|_{L^2}^2 = \frac{1}{2\pi}\frac{16}{3} $$ so we obtain

$$\bbox[5px,border:2px solid #CAAA00]{\int_{0}^{\infty}\frac{\sin^4(x)}{x^4} \mathrm{d}x = \frac{\pi}{3}}$$

You are correct. To find the integral $\int_{-\infty}^\infty \frac{\sin^4(x)}{x^4}\,dx=\int_{-\infty}^\infty \frac{\sin^2(x)}{x^2}\frac{\sin^2(x)}{x^2}\,dx$, we simply need to convolve the Fourier Transform of $\frac{\sin^2(x)}{x^2}$ with itself, and evaluate this at $\omega=0$.

PRIMER: FOURIER-TRANSFORM PAIRS:

We have the Fourier Transform pairs

$$\begin{align} f(x) &\leftrightarrow F(\omega)\\\\ f^2(x) &\leftrightarrow \frac{1}{\sqrt {2\pi}}F(\omega)*F(\omega)\\\\ \frac{\sin(x)}{x}&\leftrightarrow \sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\\\\ \frac{\sin^2(x)}{x^2}&\leftrightarrow \frac{1}{\sqrt{2\pi}}\left(\sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\right)*\left(\sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\right)=\sqrt{\frac\pi8}2 \text{tri}(\omega/2) \end{align}$$

where $\text{rect}(t)$ and $\text{tri}(t)$ are the Rectangle Function and Triangle Function, respectively.

Therefore, we have

$$\begin{align} \int_{-\infty}^\infty \frac{\sin^4(x)}{x^4}\,dx &=\left.\left(\left(\sqrt{\frac\pi8}2 \text{tri}(\omega/2)\right)*\left(\sqrt{\frac\pi8}2 \text{tri}(\omega/2)\right)\right)\right|_{\omega =0}\\\\ &=\int_{-2}^2 \left(\sqrt{\frac{\pi}{8}}\frac{}{}(2-|\omega'|)\right)^2\,d\omega'\\\\ &=\left(\frac{\pi}{8}\right)\,2\int_0^2 (\omega'-2)^2\,d\omega'\\\\ &=\left(\frac{\pi}{8}\right)\,2\left(\frac83\right)\\\\ &=\frac{2\pi}{3} \end{align}$$

Hence, by even symmetry we can assert that

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{\sin^4(x)}{x^4}\,dx=\frac{\pi}{3}}$$