Dirichlet integral

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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.

[edit] 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

This integral is made much simpler by recalling Euler's forumla

eiω = cosω + isinω

Then

\Im e^{i\omega}=\sin \omega, where \Im represents the imaginary part.

Rewriting the integral gives us:

-\Im\int_0^\infty e^{-\alpha\omega}e^{i\omega}d\omega=\Im\frac{1}{-\alpha+i}=\Im\frac{-\alpha-i}{\alpha^2+1}=\frac{-1}{\alpha^2+1}

So,

\frac{df}{d\alpha}=\frac{-1}{\alpha^2+1}

Integrating both sides from 0 to \infty

\int_0^\infty\frac{df}{d\alpha}d\alpha=\int_0^\infty\frac{-1}{\alpha^2+1}d\alpha
f(\infty)-f(0)=-\arctan \infty + \arctan 0
f(0)=\frac{\pi}{2} + f(\infty)

Note that f(\infty)=\lim_{\alpha\rightarrow \infty} \int_0^\infty e^{-\alpha\omega} \frac{\sin \omega}{\omega}=0

So,

f(0)=\frac{\pi}{2}

Then

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

[edit] See also

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