# Rigorously proving $\displaystyle\int^{\infty}_0 \frac{\sin{x}}{x} dx= \frac{\pi}{2}$ [duplicate]

I want to prove the famous formula: $\displaystyle\int^{\infty}_0 \frac{\sin{x}}{x} dx = \frac{\pi}{2}.$ There are many ways to do it, for example, by some Fourier analysis. But how about a simple method: integrating just $e^{-xy}\sin{x}$ by $y$ and $x$ successively? Since $|\frac{\sin{x}}{x}|$ is not in $L^1$, Fubini-Tonelli theorem cannot be applicable at first glance. Is this way rigorously justifiable?

• @JohnMa it's not quite a duplicate since the question is specifically how to justify the method in this answer. – Ben Grossmann Jul 5 '15 at 1:01

Here is one way. Let

$$I(a)=\int_0^{\infty}\frac{\sin x}{x}e^{-ax}dx$$

Taking the derivative of $I$ yields

$$I'(a)=-\int_0^{\infty}e^{-ax}\sin x\,dx=-\frac{1}{a^2+1}$$

Integrating, we find that

$$I(a)=-\arctan(a)+C$$

Noting that $\lim_{a\to \infty}I(a)=0$, we see that $C=\pi/2$. Thus,

$$I(0)=\pi/2$$

and we're done!