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, because the Dirichlet integral is infinite in the sense of Lebesgue integration. It is, however, finite in the sense of the improper Riemann integral or the generalized Riemann or Henstock–Kurzweil integral. This can be seen by using Dirichlet's test for improper integrals.
It is a good illustration of special techniques for evaluating definite integrals. The sine integral, an antiderivative of the sinc function, is not an elementary function. However the improper definite integral can be determined in several ways: 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.
In what follows, one needs the result which is the Laplace transform of the function (see the section 'Differentiating under the integral sign' for a derivation) as well as a version of Abel's theorem (a consequence of the final value theorem for the Laplace transform).
Therefore,
Double integration
Evaluating the Dirichlet integral using the Laplace transform is equivalent to calculating the same double definite integral by changing the order of integration, namely,
Differentiation under the integral sign (Feynman's trick) First rewrite the integral as a function of the additional variable namely, the Laplace transform of So let
In order to evaluate the Dirichlet integral, we need to determine The continuity of can be justified by applying the dominated convergence theorem after integration by parts. 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 the sine function in terms of complex exponentials:
Therefore,
Integrating with respect to gives
where is a constant of integration to be determined. Since using the principal value. This means that for
Finally, by continuity at we have as before.
Complex contour 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 to the negative imaginary axis, so can be integrated along the semicircle of radius centered at extending in the positive imaginary direction, and closed along 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 (here is any branch of logarithm on upper half-plane), leading to
Dirichlet kernel
Consider the well-known formula for 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:
(The form of the Riemann-Lebesgue Lemma used here is proven in the article cited.)
We would like to compute:
However, we must justify switching the real limit in to the integral limit in which will follow from showing that the limit does exist.
Using integration by parts, we have: