TITLE: Improper integral of sin(x)/x from zero to infinity
QUESTION [11 upvotes]: I was having trouble with the following integral:
$\int_{0}^\infty \frac{\sin(x)}{x}dx$. My question is, how does one go about evaluating this, since its existence seems fairly intuitive, while its solution, at least to me, does not seem particularly obvious.

REPLY [29 votes]: Let $I(s)$ be given by 
$$I(s)=\int_0^\infty \frac{e^{-sx}\sin(x)}{x}\,dx \tag1$$
for $s\ge 0$.  Note that $\lim_{s\to \infty}I(s)=0$ and that $I(0)=\int_0^\infty \frac{\sin(x)}{x}\,dx$ is the integral of interest.

Differentiating $I(s)$ as given by $(1)$ (this is justified by uniform convergence of the "differentiated" integral for $s\ge \delta>0$) reveals
$$\begin{align}
I'(s)&=-\int_0^\infty e^{-sx}\sin(x)\,dx\\\\
&=-\frac{1}{1+s^2} \tag 2
\end{align}$$
Integrating $(2)$, we find that $I(s)=\pi/2-\arctan(s)$ whence setting $s=0$ yields the coveted result
$$\int_0^\infty \frac{\sin(x)}{x}\,dx=\frac\pi2$$