Jordan's lemma

Jordan’s lemma is a refinement of the ML estimate for integrands containing an oscillatory factor $e^{iaz}$ with $a>0$, in the upper half-plane.

Statement

Let $C_R$ be the upper semicircle $|z|=R$, $\mathrm{Im}(z)\ge 0$. Let $f$ be continuous on $C_R$ for all large $R$, with $|f(z)|\to 0$ as $|z|\to \infty$ in the upper half-plane (uniformly in $\arg z\in [0,\pi ]$). Then for any $a>0$,

${\int }_{C_R}f(z)e^{iaz}dz\to 0\text{as }R\to \infty .$

Why this is finer than ML

On $C_R$, $z=R{e}^{i\theta }$, so $|e^{iaz}|=e^{-aR\sin \theta }$. For $\theta \in (0,\pi )$, $\sin \theta >0$, so $|e^{iaz}|$ decays exponentially in $R$ on the strict interior of the upper semicircle. Near $\theta =0,\pi$ (the endpoints of the semicircle on the real axis), $\sin \theta \to 0$ and the decay is weak.

The standard ML estimate $|f|\le M$ on the contour, length $\pi R$, gives a bound $M\pi R$ — this can blow up if $M$ doesn’t decay fast enough. But if $f$ also has $|e^{iaz}|$ multiplying it, the exponential decay in the interior dominates and the integral still goes to zero — even when $|f(z)|\cdot R\nrightarrow 0$.

The careful estimate uses the inequality $\sin \theta \ge 2\theta /\pi$ for $\theta \in [0,\pi /2]$ to bound the exponential decay, then splits the semicircle into pieces near and away from the real axis. Result: as long as $f\to 0$ uniformly at infinity (no specific rate required), the integral vanishes.

Why we need it

ML alone wouldn’t kill the semicircle contribution for ${\int }_{-\infty }^{\infty }f(x)\cos (ax)dx$ when $f$ decays only as $1/x$ — the magnitude $\pi R\cdot (M/R)=\pi M$ doesn’t go to zero. With Jordan’s lemma and the $e^{iaz}$ factor (replacing $\cos (ax)$ with $e^{iax}$ and taking real parts), the exponential decay in the upper half-plane provides the needed control.

Standard application

To compute ${\int }_{-\infty }^{\infty }f(x)\cos (ax)dx$ or $\int f(x)\sin (ax)dx$ with $f$ rational and $a>0$:

1. Replace $\cos (ax)$ or $\sin (ax)$ with $e^{iax}$, intending to take real or imaginary part of the result.
2. Close the contour with the upper semicircle (since $|e^{iaz}|$ decays for $\mathrm{Im}(z)>0$).
3. The semicircle contribution vanishes by Jordan’s lemma.
4. The closed-contour integral equals $2\pi i\sum \mathrm{Res}$ in the upper half-plane.
5. Take real or imaginary part to get the original integral.

Example. ${\int }_{-\infty }^{\infty }\cos x/(x^2+1)dx$.

Consider $\int e^{iz}/(z^2+1)dz$. Upper-half-plane pole: $z=i$. Residue at $z=i$: $e^{i\cdot i}/(2i)=e^{-1}/(2i)$.

Contour integral = $2\pi i\cdot e^{-1}/(2i)=\pi /e$.

By Jordan’s lemma the semicircle goes to zero, so the real-axis integral equals $\pi /e$. Take real part of $\int e^{ix}/(x^2+1)=\pi /e$. So $\int \cos x/(x^2+1)dx=\pi /e$.

What goes wrong with $a<0$

If $a<0$, you’d need to close in the lower half-plane instead (where $|e^{iaz}|$ decays for $\mathrm{Im}(z)<0$). The choice of which half-plane to close in depends on the sign of $a$.

If you tried to close in the wrong half-plane, the exponential would grow on the semicircle and the contribution wouldn’t vanish — the trick wouldn’t work.

In context

Jordan’s lemma is essential to the technique of computing real integrals involving sin / cos via complex contour integration. Without it, the semicircle would dominate and the contour trick wouldn’t apply. With it, the residue theorem handles a wide class of integrals that are intractable by purely real methods.