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

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

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