# Jordan's lemma

We consider integrals of the form $\int _{C}f(z)e^{iaz}\;dz$ , where $C=C_{1}+C_{2}$ is a closed simple contour as shown in Figure 1. The function $f(z)$ has a finite number of poles located inside of the contour. As $|z|\rightarrow \infty$ $|f(z)|$ is assumed to decay at least as fast as ${\frac {1}{|z|}}$ . By the Residue Theorem the value of this integral may be written as

$[itex]2\pi i\sum {\mbox{Res}}f(z)e^{iaz}=\int _{C_{1}}f(z)e^{iaz}\;dz+\int _{C_{2}}f(z)e^{iaz}\;dz.$ Typically, the application of such a contour integral is to find the value of the integral along the real axis, which is the integral along $C_{1}$ .

The next step is to evaluate the integral over the contour $C_{2}$ and show that this vanishes. We consider the following estimate

$|\int _{C_{2}}f(z)e^{iaz}\;dz|\leq \int _{C_{2}}|f(z)e^{iaz}|\;|dz|$ .

By Cauchy's theorem we may assume that $C_{2}$ is a semi-circle, and may consider the polar form of $z=|z|e^{i\phi }$ , $dz=i|z|e^{i\phi }d\phi$ , and we may also consider $z=|z|(\cos \phi +i\sin \phi )$ in the exponential. Substituting these into the $C_{2}$ integral yields

$\int _{C_{2}}|f(z)e^{iaz}|\;|dz|=\int _{0}^{\pi }|f(z)e^{ia|z|(\cos \phi -\sin \phi )}||i|z|e^{i\phi }d\phi |\leq \lim _{R\rightarrow \infty }{\frac {M}{R}}\int _{0}^{\pi }e^{-aR\sin(\phi )}\;Rd\phi ,$ where $R$ is the radius of the semi-circle $C_{2}.$ for this integral to decay the ${\mbox{Re}}a>0.$ Here, the Maximum Modulus theorem has been applied, and we have

We may further proceed with the estimate by recognizing that integral is twice the same integral, with upper integration limit of ${\frac {\pi }{2}}$ and the extra factor of 2 is absorbed into the $M$ yielding

$\int _{C_{2}}|f(z)e^{iaz}|\;|dz|\leq \lim _{R\rightarrow \infty }M\int _{0}^{\pi /2}e^{-aR\sin(\phi )}\;d\phi \leq \lim _{R\rightarrow \infty }M\int _{0}^{\pi /2}e^{-aR{\frac {2}{\pi }}(\phi )}\;d\phi$ $\qquad \qquad \qquad \qquad \qquad =\lim _{R\rightarrow \infty }\left.{\frac {-2M}{\pi aR}}e^{-aR{\frac {2}{\pi }}(\phi )}\right|_{0}^{\pi /2}={\frac {2M}{\pi aR}}(1-e^{-aR})\rightarrow 0.$ Here we have used the fact that on the interval $[0,\pi /2]$ we have $-\sin(\phi )\leq -\phi (\pi /2)$ .