Dirichlet integral

Summary

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.[1][2] This can be seen by using Dirichlet's test for improper integrals. The value of the integral (in the Riemann or Henstock sense) 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.[3]

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 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 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[4]

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 f 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

Let

be the Dirichlet kernel.[5]

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 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.[6]

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.

See also

References

  1. ^ Bartle, Robert G. (10 June 1996). "Return to the Riemann Integral" (PDF). The American Mathematical Monthly. 103 (8): 625–632. doi:10.2307/2974874. JSTOR 2974874.
  2. ^ Bartle, Robert G.; Sherbert, Donald R. (2011). "Chapter 10: The Generalized Riemann Integral". Introduction to Real Analysis. John Wiley & Sons. pp. 311. ISBN 978-0-471-43331-6.
  3. ^ Zill, Dennis G.; Wright, Warren S. (2013). "Chapter 7: The Laplace Transform". Differential Equations with Boundary-Value Problems. Cengage Learning. pp. 274-5. ISBN 978-1-111-82706-9.
  4. ^ Appel, Walter. Mathematics for Physics and Physicists. Princeton University Press, 2007, p. 226. ISBN 978-0-691-13102-3.
  5. ^ Chen, Guo (26 June 2009). A Treatment of the Dirichlet Integral Via the Methods of Real Analysis (PDF) (Report).
  6. ^ R.C. Daileda. Improper Integrals (PDF) (Report).

External links