Jordan's Lemma

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

In the computation of Fourier transform-like integrals as contour integrals, we often encounter the issue of the contribution of a semicircular contour that has infinite radius.

Statement of Jordan's lemma

Given a complex valued function $f(z)$ such that $|f(z)|=o(1)$ as $|z|\rightarrow \infty$ . This means that $|f(z)|\rightarrow 0$ as $|z|$ becomes large. We require also that $f(z)e^{iaz}$ be analytic at infinity in some half-plane of $z$ we have the result that the integral over a semicircular contour in that half-plane vanishes

$\int _{C_{1}}f(z)e^{iaz}\;dz=0$ .

The quantity $a$ is real valued, and finite.

Proof of Jordan's lemma "Figure 2: the estimate of $-\sin(\phi )$ for Jordan's lemma"

Following the hypothesis of the lemma we consider the following contour integral

$\int _{C_{1}}f(z)e^{iaz}\;dz$ .

If we let $R=|z|$ and note by hypothesis that $|f(z)|=o(1)$ on $C_{1}$ , we have

$\left|\int _{C_{1}}f(z)e^{iaz}\;dz\right|=\left|\int _{C_{1}}f(z)e^{iaR(\cos(\phi )+i\sin(\phi )}\;dz\right|\leq \int _{C_{1}}\left|f(z)e^{iaR(\cos(\phi )+i\sin(\phi ))}\right|\;|dz|$ $\leq o(1)R\int _{0}^{\pi }e^{-aR\sin(\phi )}\;d\phi =o(1)\;2R\int _{0}^{\pi /2}e^{-aR\sin(\phi )}\;d\phi$ ,

where the last equality follows because the integrand is symmetric about the point $\phi =\pi /2$ .

The function $-\sin(\phi )\leq -2/\pi \phi$ on the interval $[0,\pi /2]$ allowing the further estimate (see Figure 2)

$o(1)2R\int _{0}^{\pi /2}e^{-aR\sin(\phi )}\;d\phi \leq o(1)2R\int _{0}^{\pi /2}e^{-aR2/\pi \phi }\;d\phi$ .

The last integral may be solved by elementary means to yield

$o(1)2R\int _{0}^{\pi /2}e^{-aR2/\pi \phi }\;d\phi =\left.{\frac {o(1)}{a\pi }}e^{-aR2/\pi \phi }\right|_{0}^{\pi /2}=o(1)(1-e^{-aR})\rightarrow 0$ as $R\rightarrow \infty$ .

Though we have depicted the contour of integration in the upper half-plane of $z$ , the result is actually general. The value of the parameter $a$ takes on the appropriate sign as to make the exponential always decaying. So, a similar result exists for closure of the contour in the lower half-plane of $z$ , for Fourier-like integrals. Furthermore, for Laplace-like integrals we may consider closure in the right or left half-planes of the integration variable and obtain a similar result.