Jordan's lemma

Here we follow standard texts, such as Spiegel (1964) or Levinson and Redheffer (1970).

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 $\{z_{0},z_{1},z_{2},...,z_{n}\}$ located inside the contour. As $|z|\rightarrow \infty$ the function $|f(z)|$ in the integrand is assumed to decay at least as fast as ${\frac {1}{R}}$, where $|z|=R$ for points on $C_{2}$.

"Figure 1: the contour C for a Fourier-like integration (upward closure only)"

By the Residue Theorem the value of this integral may be written as $2\pi i$ times the sum of the residues of the integrand

$2\pi i\sum _{k=0}^{n}{\mbox{Res}}(f(z)e^{iaz};z=z_{k})=\int _{C_{1}}f(z)e^{iaz}\;dz+\int _{C_{2}}f(z)e^{iaz}\;dz.$

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

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

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 $C_{2}$ and applying the fact that as $R\rightarrow \infty$ the function $|f(z)|\leq {\frac {M}{R}}$ yielding

$\int _{C_{2}}\left|f(z)e^{iaz}\right|\;|dz|=\int _{0}^{\pi }\left|f(z)e^{ia|z|(\cos \phi -\sin \phi )}\right||i|z|e^{i\phi }d\phi |\leq \lim _{R\rightarrow \infty }{\frac {M}{R}}\int _{0}^{\pi }e^{-aR\sin(\phi )}\;Rd\phi ,$

where $R$ is the radius of the semi-circle $C_{2}.$ For the integral to exist, the integrand must decay sufficiently rapidly. Thus, the ${\mbox{Re}}\;a>0.$Here, the Maximum Modulus theorem has been applied to generate the estimate of $|f(z)|\leq M.$

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 $-\sin(\phi )\leq -\phi (\pi /2)$.

The case of closure in the lower half plane of $z$ follows the same way, with the exception that the ${\mbox{Re}}\;a<0$.

Thus, the integral over $C_{1}$ is given by the expression

$2\pi i\sum _{k=0}^{n}{\mbox{Res}}(f(z)e^{iaz};z=z_{k})=\int _{C_{1}}f(z)e^{iaz}\;dz.$

References

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