Math 2.0

statement lemma

Jordan's lemma

Statement

Let a>0a > 0 and R>0R > 0 be real numbers. Let CRC_R be the semicircle {Reiθ:θ∈[0,π]}\{ R e^{i\theta} : \theta \in [0, \pi] \} in the upper half-plane, with the parametrization θ↦Reiθ\theta \mapsto R e^{i\theta}. Let gg be a continuous complex function on CRC_R, and let f(z)=eiazg(z)f(z) = e^{iaz} g(z) for z∈CRz \in C_R, where ewe^w is the complex exponential. Let MR=max⁡θ∈[0,π]∣g(Reiθ)∣M_R = \max_{\theta \in [0, \pi]} |g(R e^{i\theta})|. Then the contour integral of ff along CRC_R satisfies ∣∫CRf(z) dz∣≤πaMR.\left| \int_{C_R} f(z) \, dz \right| \le \frac{\pi}{a} M_R.

Remarks

If MR→0M_R \to 0 as R→∞R \to \infty, then ∫CRf(z) dz→0\int_{C_R} f(z) \, dz \to 0 as R→∞R \to \infty. The bound does not contain the length πR\pi R of CRC_R, so it is stronger than the estimation lemma for this type of integrand. A similar result holds for the semicircle in the lower half-plane when a<0a < 0. The lemma is a main step in the evaluation of real integrals such as ∫−∞∞eiaxx2+1 dx\int_{-\infty}^{\infty} \frac{e^{iax}}{x^2 + 1} \, dx.

Explanations

L1 human

On the semicircle, ∣eiaz∣=e−aRsin⁡θ|e^{iaz}| = e^{-aR \sin \theta}, and this number is very small except near the two ends of the semicircle. So the factor eiaze^{iaz} kills most of the long path. This is why the bound stays small even when the radius RR is large.

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  1. Complex numbers object
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Sources

  1. Web https://en.wikipedia.org/wiki/Jordan%27s_lemma, Statement