# Jordan's lemma

Here we follow standard texts, such as Spiegel (1964)[1] or Levinson and Redheffer (1970). [2]

We consider integrals of the form ${\displaystyle \int _{C}f(z)e^{iaz}\;dz}$, where ${\displaystyle C=C_{1}+C_{2}}$ is a closed simple contour as shown in Figure 1. The function ${\displaystyle f(z)}$ has a finite number of poles ${\displaystyle \{z_{0},z_{1},z_{2},...,z_{n}\}}$ located inside the contour. As ${\displaystyle |z|\rightarrow \infty }$ the function ${\displaystyle |f(z)|}$ in the integrand is assumed to decay at least as fast as ${\displaystyle {\frac {1}{R}}}$, where ${\displaystyle |z|=R}$ for points on ${\displaystyle C_{2}}$.

"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 ${\displaystyle 2\pi i}$ times the sum of the residues of the integrand

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

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 ${\displaystyle C_{1}}$.

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

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

By Cauchy's theorem we may assume that ${\displaystyle C_{2}}$ is a semi-circle, and may consider the polar form of ${\displaystyle z=|z|e^{i\phi }}$, ${\displaystyle dz=i|z|e^{i\phi }d\phi }$, and we may also consider ${\displaystyle z=|z|(\cos \phi +i\sin \phi )}$ in the exponential. Substituting these into the ${\displaystyle C_{2}}$ and applying the fact that as ${\displaystyle R\rightarrow \infty }$ the function ${\displaystyle |f(z)|\leq {\frac {M}{R}}}$ yielding

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

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

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

${\displaystyle \int _{C_{2}}|f(z)e^{iaz}|\;|dz|\leq \lim _{R\rightarrow \infty }M\int _{0}^{\pi /2}e^{-aR\sin(\phi )}\;d\phi \leq \lim _{R\rightarrow \infty }M\int _{0}^{\pi /2}e^{-aR{\frac {2}{\pi }}(\phi )}\;d\phi }$
${\displaystyle \qquad \qquad \qquad \qquad \qquad =\lim _{R\rightarrow \infty }\left.{\frac {-2M}{\pi aR}}e^{-aR{\frac {2}{\pi }}(\phi )}\right|_{0}^{\pi /2}={\frac {2M}{\pi aR}}(1-e^{-aR})\rightarrow 0.}$

Here we have used the fact that on the interval ${\displaystyle [0,\pi /2]}$ we have ${\displaystyle -\sin(\phi )\leq -\phi (\pi /2)}$.

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

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

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