Jordan's Lemma

Complex Analysis · Axiom Academy

When you close a Fourier integral with a big arc, this is what makes the arc vanish — even when the function barely decays. 1. The Exponential Does the Work Walk along the arc . Split the exponent into real and imaginary parts: the imaginary part only spins the phase, while the real part sets the size. On the upper arc that real part is negative, so e^ iaz decays exponentially the higher you climb. 2. The Inequality That Tames the Integral To turn the decay into a usable bound we need a clean handle on . The trouble is that is curved. Jordan's trick: on , the curve never dips below the straight chord joining its endpoints. on — the chord sits under the arc because is concave there. A bigger exponent that we subtract gives a smaller exponential: . The right-hand bound is a plain exponential in — its integral over is elementary. The arc's two halves match, so — we only need the bound on the first quarter-turn. Replacing by its chord gives . The whole bound now shrinks like 1/R — fast enough to win. 3. Putting It Together: the Arc Vanishes Parameterize the arc, pull the magnitude through the integral, bound |f| by its maximum M_R on C_R , and apply Jordan's inequality. Everything telescopes into a single clean estimate that goes to zero: The hypothesis: we only assume as . The factor R from the arc length is exactly cancelled by the 1/R from the integral — what survives is , which dies because M_R does.

This is the written version of the interactive lesson above. See the full Complex Analysis course.