Evaluating Integrals with the Laplace Transform

The Laplace transform is usually taught as a tool for solving linear differential equations, but it can also evaluate integrals. Here we use it to prove the classic result

\[\int_0^{\infty} \frac{\sin x}{x} \, dx=\frac{\pi}{2}.\]

Turning the integral into a Laplace transform

For a function \(g(t)\), the Laplace transform is

\[\mathcal{L}\{g\}(s)=G(s)=\int_0^{\infty} e^{-s t}\, g(t)\, d t .\]

Setting \(s=0\) formally gives \(\int_0^{\infty} g(t) \, d t\). So we take \(g(t)=\frac{\sin t}{t}\), compute its Laplace transform, and let \(s \to 0^+\).

Computing the transform

Recall

\[\mathcal{L}\{\sin t\}(s)=\frac{1}{s^2+1},\]

and the standard “division by \(t\)” rule: if \(F(s)\) is the Laplace transform of \(f(t)\), then

\[\mathcal{L}\left\{\frac{f(t)}{t}\right\}(s)=\int_s^{\infty} F(u)\, d u .\]

Combining the two,

\[\mathcal{L}\left\{\frac{\sin t}{t}\right\}(s)=\int_s^{\infty} \frac{d u}{u^2+1} =\frac{\pi}{2}-\arctan s .\]

Letting \(s \to 0^+\) gives \(\frac{\pi}{2}\). Therefore

\[\int_0^{\infty} \frac{\sin x}{x} \, dx=\frac{\pi}{2}.\]

(The integral converges only conditionally, so the limit \(s \to 0^+\) needs some care. Abel’s theorem for Laplace transforms justifies it.)

Exercise

Using the same technique, show that

\[\int_0^{\infty} \frac{1-\cos t}{t\, e^t}\, d t = \frac{\ln 2}{2}.\]

A worked solution by blackpenredpen:

Credits

This post is inspired by this video.

Changes on 25 October 2026

What was wrong. In the key formula \(\int_s^\infty F(s)\,ds\) the integration variable had the same name as the lower limit. The limit \(s \to 0^+\) at the end was also taken without comment.

What changed. The integration variable is now \(u\), the limit is explained, and the exercise has its answer, \(\tfrac12 \ln 2\). Previously titled “Integration by Laplace Transform”.