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It is a good illustration of special techniques for evaluating definite integrals, particularly when it is not useful to directly apply the fundamental theorem of calculus due to the lack of an elementary antiderivative for the integrand, as the sine integral, an antiderivative of the sinc function, is not an elementary function. In this case, 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.
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The Dirichlet Integral is destroyed by Feynman's Trick
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Transcription
Evaluation
Laplace transform
Let be a function defined whenever Then its Laplace transform is given by
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,
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)
First rewrite the integral as a function of the additional variable namely, the Laplace transform of So let
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.
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
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.