integration - Improper integral of sin(x)/x from zero to infinity

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.

Answer

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

$$\int_0^\infty \frac{\sin(x)}{x}\,dx=\frac\pi2$$