Integration by Laplace Transform
Laplace Transform is usually used to “decompose” a function into its constituents, or to solve differential equations. However, it can also be used to solve integrals. Below, I will show how it can be used to prove:
\[\int_0^{\infty} \frac{\sin(x)}{x} \, dx=\frac{\pi}{2}\]Transforming the problem into a Laplace Transformation Problem
Observe that for any \(g(t)\),
\[\mathcal{L}\{g(t)\}=F(s)=\int_0^{\infty} e^{-s t} g(t) d t.\]Taking \(s=0\), we obtain:
\[\int_0^{\infty} g(t) d t\]Now, we simply substitute \(g(t)=\frac{\sin(t)}{t}\) and solve the Laplace Transformation.
Solving the Laplace Transformation
Recall:
\[\mathcal{L}\{\sin t\}=\frac{1}{s^2+1} .\]Next, use the standard result:
\[\mathcal{L}\left[\frac{f(t)}{t}\right]=\int_s^{\infty} F(s) d s\]Where \(F(s)\) is the Laplace Transformation of \(f(t)\).
Combining the results, we arrive at:
\[\begin{align} \mathcal{L}\left[\frac{\sin(t)}{t}\right]&=\int_s^{\infty} \frac{1}{s^2+1} d s \\ &=\frac{\pi}{2}-\arctan(s) \end{align}\]Taking \(s=0\) we obtain \(\frac{\pi}{2}\) as expected.
Therefore,
\[\int_0^{\infty} \frac{\sin(x)}{x} \, dx=\frac{\pi}{2}\]As required!
Credits
The post is inspired by this video.
Exercises for the reader
Using the same technique, calculate
\[\int_0^{\infty} \frac{1-\cos t}{t e^t} d t\]The answer is presented by blackpenredpen:
