Dirichlet integral.html

In mathematics, there are several integrals known as the Dirichlet integral, after the German mathematician Peter Gustav Lejeune Dirichlet.

One of those is

$\int_0^\infty \frac{\sin \omega}{\omega}\,d\omega = \frac{\pi}{2}$

This can be proven using a Fourier integral representation. It can also be evaluated quite simply using differentiation under the integral sign.

Proof Using Differentiation Under the Integral Sign

We will first rewrite the integral as a function of an arbitrary constant, $α$ and $ω$ .

Let $f(\alpha)=\int_0^\infty e^{-\alpha\omega} \frac{\sin \omega}{\omega} d\omega$

Then we need to find $f (0)$

Differentiating with respect to $α$ gives us:

$\frac{df}{d\alpha}=\frac{\partial}{\partial\alpha}\int_0^\infty e^{-\alpha\omega} \frac{\sin \omega}{\omega} d\omega$

Applying the Leibniz Integral Rule,

$\frac{\partial}{\partial\alpha}\int_0^\infty e^{-\alpha\omega} \frac{\sin \omega}{\omega} d\omega = \int_0^\infty \frac{\partial}{\partial\alpha}e^{-\alpha\omega}\frac{\sin \omega}{\omega} d\omega = -\int_0^\infty e^{-\alpha\omega} \sin \omega \,d\omega$