# 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

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

# References

1. Spiegel, Murray R. "Theory and problems of complex variables, with an introduction to Conformal Mapping and its applications." Schaum's outline series (1964).
2. Levinson, Norman, and Raymond M. Redheffer. "Complex variables." (1970), Holden-Day, New York.