Editing Dirichlet integral (section)

=== Complex contour integration ===
Consider <math display="block">f(z) = \frac{e^{iz}} z.</math>

As a function of the complex variable <math>z,</math> 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<ref>Appel, Walter. ''Mathematics for Physics and Physicists''. Princeton University Press, 2007, p. 226. {{ISBN|978-0-691-13102-3}}.</ref>
<math display="block">g(z) = \frac{e^{iz}}{z + i\varepsilon}.</math>

The pole has been moved to the negative imaginary axis, so <math>g(z)</math> can be integrated along the semicircle <math>\gamma</math> of radius <math>R</math> centered at <math>z = 0</math> extending in the positive imaginary direction, and closed along the real axis. One then takes the limit <math>\varepsilon \to 0.</math>

The complex integral is zero by the [[residue theorem]], as there are no poles inside the integration path <math>\gamma</math>:
<math display="block">0 = \int_\gamma g(z) \,dz = \int_{-R}^R \frac{e^{ix}}{x + i\varepsilon} \, dx + \int_0^\pi \frac{e^{i(Re^{i\theta} + \theta)}}{Re^{i\theta} + i\varepsilon} iR \, d\theta.</math>

The second term vanishes as <math>R</math> 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 number|complex]]-valued function {{mvar|f}} defined and continuously differentiable on the real line and real constants <math>a</math> and <math>b</math> with <math>a < 0 < b</math> one finds
<math display="block">\lim_{\varepsilon \to 0^+} \int_a^b \frac{f(x)}{x \pm i \varepsilon} \,dx = \mp i \pi f(0) + \mathcal{P} \int_a^b \frac{f(x)}{x} \,dx,</math>

where <math>\mathcal{P}</math> denotes the [[Cauchy principal value]]. Back to the above original calculation, one can write
<math display="block">0 = \mathcal{P} \int \frac{e^{ix}}{x} \, dx - \pi i.</math>

By taking the imaginary part on both sides and noting that the function <math>\sin(x)/x</math> is even, we get
<math display="block">\int_{-\infty}^{+\infty} \frac{\sin(x)}{x} \,dx = 2 \int_0^{+\infty} \frac{\sin(x)}{x} \,dx.</math>

Finally,
<math display="block">\lim_{\varepsilon \to 0} \int_\varepsilon^\infty \frac{\sin(x)}{x} \, dx = \int_0^\infty \frac{\sin(x)}{x} \, dx = \frac \pi 2.</math>

Alternatively, choose as the integration contour for <math>f</math> the union of upper half-plane semicircles of radii <math>\varepsilon</math> and <math>R</math> together with two segments of the real line that connect them. On one hand the contour integral is zero, independently of <math>\varepsilon</math> and <math>R;</math> on the other hand, as <math>\varepsilon \to 0</math> and <math>R \to \infty</math> the integral's imaginary part converges to <math>2 I + \Im\big(\ln 0 - \ln(\pi i)\big) = 2I - \pi</math> (here <math>\ln z</math> is any branch of logarithm on upper half-plane), leading to <math>I = \frac{\pi}{2}.</math>