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,
Evaluating the Dirichlet integral using the Laplace transform is equivalent to calculating the same double definite integral by changing the order of integration, namely,
The change of order is justified by the fact that for all , the integral is absolutely convergent.
Differentiation under the integral sign (Feynman's trick)
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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
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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 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 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
Consider the well-known formula for 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 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:
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.