Here we follow standard texts, such as Spiegel (1964)^{[1]} or Levinson and Redheffer (1970). ^{[2]}
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 Fourierlike 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
$\left\int _{C_{2}}f(z)e^{iaz}\;dz\right\leq \int _{C_{2}}\leftf(z)e^{iaz}\right\;dz$.
By Cauchy's theorem we may assume that $C_{2}$ is a semicircle, and may consider the polar form of $z=ze^{i\phi }$,
$dz=ize^{i\phi }d\phi$, 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
$\int _{C_{2}}\leftf(z)e^{iaz}\right\;dz=\int _{0}^{\pi }\leftf(z)e^{iaz(\cos \phi \sin \phi )}\rightize^{i\phi }d\phi \leq \lim _{R\rightarrow \infty }{\frac {M}{R}}\int _{0}^{\pi }e^{aR\sin(\phi )}\;Rd\phi ,$
where $R$ is the radius of the semicircle $C_{2}.$ 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
$\int _{C_{2}}f(z)e^{iaz}\;dz\leq \lim _{R\rightarrow \infty }M\int _{0}^{\pi /2}e^{aR\sin(\phi )}\;d\phi \leq \lim _{R\rightarrow \infty }M\int _{0}^{\pi /2}e^{aR{\frac {2}{\pi }}(\phi )}\;d\phi$
$\qquad \qquad \qquad \qquad \qquad =\lim _{R\rightarrow \infty }\left.{\frac {2M}{\pi aR}}e^{aR{\frac {2}{\pi }}(\phi )}\right_{0}^{\pi /2}={\frac {2M}{\pi aR}}(1e^{aR})\rightarrow 0.$
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
 ↑ Spiegel, Murray R. "Theory and problems of complex variables, with an introduction to Conformal Mapping and its applications." Schaum's outline series (1964).
 ↑ Levinson, Norman, and Raymond M. Redheffer. "Complex variables." (1970), HoldenDay, New York.
External links
find literature about Jordan's lemma







