TITLE: Are there any ways to evaluate $\int^\infty_0\frac{\sin x}{x}dx$ without using double integral?
QUESTION [9 upvotes]: Are there any ways to evaluate $\int^\infty_0\frac{\sin x}{x}dx$ without using double integral?

I can't find any this kind of solution. Can anyone please help me? Thank you.

REPLY [3 votes]: Using the following version of the Fourier transform,
$$F(\omega)=(\mathcal{F}f)(\omega)=\int_{-\infty}^\infty e^{-i\omega t}f(t)\,dt,$$
then Parseval's theorem looks like
$$\int_{-\infty}^\infty \overline{f(t)}\,g(t)\,dt=\frac{1}{2\pi}\int_{-\infty}^\infty \overline{F(\omega)}\,G(\omega)\,d\omega.$$
Let $H(t)$ and $\delta(t)$ denote the Heaviside step function and the Dirac delta, respectively. Then
$$H(t+1)-H(t-1)\stackrel{\mathcal{F}}{\longmapsto}2\frac{\sin{\omega}}{\omega}$$
and
$$\delta(t)\stackrel{\mathcal{F}}{\longmapsto}1.$$
Now we use Parseval's theorem and get
$$\int_0^\infty \frac{\sin{x}}{x}\,dx=\frac{1}{2}\int_{-\infty}^\infty \frac{\sin{x}}{x}\,dx=\frac{1}{4}\int_{-\infty}^\infty 2\frac{\sin{\omega}}{\omega}\cdot 1\,d\omega =$$
$$=\frac{\pi}{2}\cdot\frac{1}{2\pi}\int_{-\infty}^\infty 2\frac{\sin{\omega}}{\omega}\cdot 1\,d\omega = [\text{Parseval}] = $$
$$=\frac{\pi}{2}\int_{-\infty}^\infty (H(t+1)-H(t-1))\delta(t)\,dt=\frac{\pi}{2}.$$