Jordan's lemma
/ zhor-DAHN /
The plain ML bound says the integral over a semicircular arc is at most (arc length) times (max of the integrand). For an oscillatory integrand like e^(i a z) f(z) with f decaying only like 1/R, this crude bound gives roughly (pi R) times (1/R) = pi, which does not go to zero — the method would seem to stall. Jordan's lemma is the sharper estimate that rescues exactly this case by taking the oscillating exponential seriously.
The statement: let f be continuous on the upper semicircle C_R (radius R) and suppose the maximum of |f(z)| on C_R, call it M_R, tends to 0 as R grows. Then for any positive constant a, the integral over C_R of e^(i a z) f(z) dz tends to 0 as R goes to infinity. The mechanism is that e^(i a z), with z = R e^(i theta) in the upper half-plane, equals e^(i a R cos theta) times e^(-a R sin theta); the second factor decays exponentially wherever sin theta is positive, which is most of the upper arc. The key inequality (Jordan's inequality, that sin theta is at least 2 theta / pi on [0, pi/2]) tames the small region near the endpoints, and the whole arc integral is bounded by pi M_R / a, which goes to 0 because M_R does.
This is the indispensable tool for Fourier-type integrals, the integral from minus infinity to infinity of f(x) e^(i a x) dx, where you want a real integral of something times cos(a x) or sin(a x). It lets you close the contour upward (for a positive) knowing the arc vanishes even though f decays too slowly for the plain semicircle argument. The honest point worth stressing: the arc vanishing is not free — without the exponential's directional decay, the integral over the big arc need not die, and Jordan's lemma is precisely the certificate that, for e^(i a z), it does.
To evaluate the integral over the real line of (cos x) / (x^2 + 1) dx, write it as the real part of the integral of e^(i x) / (x^2 + 1) dx. The factor f(z) = 1/(z^2+1) decays like 1/R^2, so Jordan's lemma kills the upper arc; the upper pole z = i gives the value pi / e.
Jordan's lemma lets the arc vanish for e^(i a z) f(z) even when f decays slowly.
The lemma needs a positive (for closing upward) and is for e^(i a z), not cos(a z) or sin(a z) directly — those grow in one half-plane. Always convert cos and sin into the real or imaginary part of an exponential before applying it.