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.
$\endgroup$ 21 Answer
$\begingroup$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
$$\int_0^\infty \frac{\sin(x)}{x}\,dx=\frac\pi2$$
$\endgroup$ 6
