Dirichlet integral

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 derived from attempts to evaluate a double improper integral two different ways. It can also be derived using differentiation under the integral sign.

Evaluation

Double Improper Integral Method

Pre-knowledge of properties of Laplace transforms allows us to evaluate this Dirichlet integral succinctly in the following manner:

$\int_0^\infty\frac{\sin t}{t}\, dt=\int_{0}^{\infty}\mathcal{L}\{\sin t\}\, ds=\int_{0}^{\infty}\frac{1}{s^{2}+1}\, ds=\arctan s\bigg|_{0}^{\infty}=\frac{\pi}{2}$

This is equivalent to attempting to evaluate the same double definite integral in two different ways, by reversal of the order of integration, viz.,

$\left ( I_1=\int_0^\infty {\int _0^\infty e^{-st} \sin t\, dt}\, ds\right ) = \left ( I_2=\int_0^\infty {\int _0^\infty e^{-st} \sin t\, ds} \, dt = \int_0^\infty \sin t{\int _0^\infty e^{-st}\, ds} \, dt\right ),$

$\left ( I_1=\int_0^\infty {\frac{1}{s^2+1}}\, ds = \frac{\pi}{2}\right ) = \left ( I_2=\int_0^\infty \sin t\, \frac{1}{t} \, dt\right ) \text{, provided } s>0.$

Differentiation under the integral sign

First rewrite the integral as a function of variable $\!a$. Let

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

then we need to find $\!f(0)$.

Differentiate with respect to $\!a$ and apply the Leibniz Integral Rule to obtain:

$\frac{df}{da}=\frac{d}{da}\int_0^\infty e^{-a\omega} \frac{\sin \omega}{\omega} d\omega = \int_0^\infty \frac{\partial}{\partial a}e^{-a\omega}\frac{\sin \omega}{\omega} d\omega = -\int_0^\infty e^{-a\omega} \sin \omega \,d\omega = -\mathcal{L}\{\sin \omega\}(a).$

This integral was evaluated without proof, above, based on Laplace trasform tables; we derive it this time. It is made much simpler by recalling Euler's formula,

$\! e^{i\omega}=\cos \omega + i\sin \omega ,$

then,

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

$\therefore\frac{df}{da}=-\Im\int_0^\infty e^{-a\omega}e^{i\omega}d\omega=\Im\frac{1}{-a+i}=\Im\frac{-a-i}{a^2+1}=\frac{-1}{a^2+1} \text{, given that } a > 0 .$

Integrating with respect to $\!a$:

$f(a) = \int \frac{-da}{a^2+1} = A - \arctan a,$

where $\! A$ is a constant to be determined. As,

$f(+\infty)=0 \therefore A = \arctan (+\infty) = \frac{\pi}{2} + m\pi,$

$\therefore f(0)=\lim _{a \to 0^+} f(a) = \frac{\pi}{2} + m\pi - \arctan 0 = \frac{\pi}{2} + n\pi,$

for some integers m & n. It is easy to show that $\!n$ has to be zero, by analyzing easily observed bounds for this integral:

$0<\int _0^\infty \frac {\sin x}{x}dx < \int _0^\pi \frac {\sin x}{x}dx < \pi$

End of proof.

Extending this result further, with the introduction of another variable, first noting that $\! {\sin x}/{x}$ is an even function and therefore

$\int_0^\infty \frac{\sin x}{x}\,dx = \int_{-\infty}^0 \frac{\sin x}{x}\,dx = -\int_0^{-\infty} \frac{\sin x}{x}\,dx,$

then:

$\int_0^\infty \frac{\sin b\,\omega}{\omega}\,d\omega = \int_0^{b\,\infty} \frac{\sin b\,\omega}{b\,\omega}\,d(b\,\omega) = \int_0^{\sgn b\times\infty} \frac{\sin x}{x}\,dx = \sgn b \int_0^\infty \frac{\sin x}{x}\,dx = \frac{\pi}{2}\,\sgn b$

Complex integration

The same result can be obtained via complex integration. Let's consider

$f(z)=\frac{e^{iz}}{z}$

As a function of the complex variable z, it has a simple pole at the origin, which prevents the application of Jordan's lemma, whose other hypotheses are satisfied. We shall then define a new function^[1] g(z) as follows

$g(z)=\frac{e^{iz}}{z +i\epsilon}$

The pole has been moved away from the real axis, so g(z) can be integrated along the semicircle of radius R centered at z=0 and closed on the real axis, then the limit $\epsilon \rightarrow 0$ should be taken.

The complex integral is zero by the residue theorem, as there are no poles inside the integration path

$0 = \int_\gamma g(z) = \int_{-R}^R \frac{e^{ix}}{x +i\epsilon} dx + \int_{0}^{\pi} \frac{e^{i(Re^{i\theta} + \theta)}}{Re^{i\theta} +i\epsilon} iR d\theta$

The second therm vanishes as R goes to infinity; for arbitrarily small ε, the Sokhatsky–Weierstrass theorem applied to the first one yelds

$0= \mathrm{P.V.} \int \frac{e^{ix}}{x} dx - \pi i \int_{-\infty}^{\infty}\delta(x) e^{ix} dx$

Where P.V. indicates Cauchy Principal Value. By taking the imaginary part on both sides and noting that sinc(x) is even and by definition sinc(0) = 1, we get the desired result

$\lim_{\epsilon\rightarrow 0}\int_\epsilon^{\infty} \frac{\sin(x)}{x} = \int_0^{\infty} \frac{\sin(x)}{x} = \frac{\pi}{2}$

Notes

1. ^ Appel, Walter. Mathematics for Physics and Physicists. Princeton University Press, 2007, p. 226.

See also

• Dirichlet principle

External links

• Weisstein, Eric W., "Dirichlet Integrals" from MathWorld.