Dirichlet integral


In mathematics, there are several integrals known as the Dirichlet integral, after the German mathematician Peter Gustav Lejeune Dirichlet, one of which is the improper integral of the sinc function over the positive real line:
This integral is not absolutely convergent, meaning is not Lebesgue-integrable, and so the Dirichlet integral is undefined in the sense of Lebesgue integration. It is, however, defined in the sense of the improper Riemann integral or the generalized Riemann or Henstock–Kurzweil integral. The value of the integral can be derived using various ways, including the Laplace transform, double integration, differentiating under the integral sign, contour integration, and the Dirichlet kernel.

Evaluation

Laplace transform

Let be a function defined whenever. Then its Laplace transform is given by
if the integral exists.
A property of the Laplace transform useful for evaluating improper integrals is
provided exists.
One may use this property to evaluate the Dirichet integral as follows:
because is the Laplace transform of the function.

Double integration

Evaluating the Dirichlet integral using the Laplace transform is equivalent to attempting to evaluate the same double definite integral in two different ways, by reversal of the order of integration, namely:

Differentiation under the integral sign (Feynman's trick)

First rewrite the integral as a function of the additional variable. Let
In order to evaluate the Dirichlet integral, we need to determine.
Differentiate with respect to and apply the Leibniz rule for differentiating under the integral sign to obtain
Now, using Euler's formula one can express a sinusoid in terms of complex exponential functions. We thus have
Therefore,
Integrating with respect to gives
where is a constant of integration to be determined. Since using the principal value. This means
Finally, for, we have, as before.

Complex integration

The same result can be obtained by complex integration. Consider
As a function of the complex variable, it has a simple pole at the origin, which prevents the application of Jordan's lemma, whose other hypotheses are satisfied.
Define then a new function
The pole has been moved away from the real axis, so can be integrated along the semicircle of radius centered at and closed on the real axis. One then takes the limit.
The complex integral is zero by the residue theorem, as there are no poles inside the integration path
The second term vanishes as goes to infinity. As for the first integral, one can use one version of the Sokhotski–Plemelj theorem for integrals over the real line: for a complex-valued function defined and continuously differentiable on the real line and real constants and with one finds
where denotes the Cauchy principal value. Back to the above original calculation, one can write
By taking the imaginary part on both sides and noting that the function is even, we get
Finally,
Alternatively, choose as the integration contour for the union of upper half-plane semicircles of radii and together with two segments of the real line that connect them. On one hand the contour integral is zero, independently of and ; on the other hand, as and the integral's imaginary part converges to , leading to.

Dirichlet kernel

Let
be the Dirichlet kernel.
It immediately follows that
Define
Clearly, is continuous when, to see its continuity at 0 apply L'Hopital's Rule:
Hence, fulfills the requirements of the Riemann-Lebesgue Lemma. This means
Choose limits and. We would like to say that
In order to do so, however, we must justify switching the real limit in to the integral limit in. This is in fact justified if we can show the limit does exist, which we do now.
Using integration by parts, we have:
Now, as and the term on the left converges with no problem. See the list of limits of trigonometric functions. We now show that is absolutely integrable, which implies that the limit exists.
First, we seek to bound the integral near the origin. Using the Taylor-series expansion of the cosine about zero,
Therefore,
Splitting the integral into pieces, we have
for some constant. This shows that the integral is absolutely integrable, which implies the original integral exists, and switching from to was in fact justified, and the proof is complete.