Fourier-type oscillatory integrals
/ FOOR-ee-ay /
A whole family of integrals from physics and signal processing has the shape: the integral over the real line of f(x) cos(a x) dx or f(x) sin(a x) dx, where f decays at infinity but the cosine or sine oscillates forever. The oscillation makes these delicate, and contour methods evaluate them cleanly by treating the oscillation as the real or imaginary part of a complex exponential.
The procedure. First combine the trig factor into an exponential: cos(a x) = the real part of e^(i a x), and sin(a x) = the imaginary part of e^(i a x). So compute instead the integral over the real line of f(x) e^(i a x) dx, and take the real or imaginary part of the answer at the end. Now move to the complex plane: form the large semicircle contour, apply the residue theorem, and discard the arc using Jordan's lemma (which, unlike the plain ML bound, succeeds even when f decays only like 1/R because e^(i a z) decays exponentially in the upper half-plane when a is positive). The result is that the integral of f(x) e^(i a x) dx = 2 pi i times the sum of residues of f(z) e^(i a z) at the poles in the upper half-plane.
These are the integrals behind Fourier transforms, diffraction, and the response of linear systems, which is why this technique is so central in applied mathematics. The sign of a tells you which way to close: a positive closes upward, a negative closes downward (where e^(i a z) decays). And the reason you must go through e^(i a z) rather than cos(a z) directly is honest and important: cos(a z) and sin(a z) blow up exponentially in one of the half-planes, so the arc would not vanish — only the one-sided exponential decays the right way.
For the integral over the real line of (sin x) / x dx (with a = 1), this is the imaginary part of the integral of e^(i z) / z dz; the pole sits on the real axis at 0, so an indented contour plus Jordan's lemma give the famous value pi.
Convert the oscillation to e^(i a x), apply residues plus Jordan's lemma, take the real or imaginary part.
Do not naively replace cos(a x) by cos(a z) in the contour integral; cos(a z) grows in one half-plane and the arc would not vanish. Always use the one-sided exponential e^(i a z) and recover cos or sin at the very end.