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Jordan's lemma

Consider a function of the form

${\displaystyle f($z)=e^{iaz}g(z)\,,\ a>0.}

Jordan's lemma states that if

${\displaystyle C_{R}=\{z:z=Re^{i\theta },\theta \in [0,\pi ]\}\,,\ R>0}$

is the semicircular path of radius R lying in the upper half plane, centred at the origin, and

${\displaystyle \quad |g\left(z\right)|\to 0{\mbox{ as }}R\to \infty {\mbox{ for all }}\theta \in [0,\pi ]\ (*)}$

then

${\displaystyle \lim _{R\to \infty }\int _{C_{R}}f($z)\,dz=0.}

A similar statement for the lower half plane exists when ${\displaystyle a<0}$.

Application of Jordan's lemma

Jordan's lemma yields a simple way to calculate the integral along the real axis of functions ${\displaystyle f($z)=e^{iaz}g(z)} holomorphic on the upper half-plane, except possibly at a finite number of non-real points a₁, a₂, ..., a_n. Consider the closed contour C, which is the concatenation of the paths C₁ and C₂ shown in the picture. By definition,

${\displaystyle \oint _{C}f($z)\,dz=\int _{C_{1}}f(z)\,dz+\int _{C_{2}}f(z)\,dz}

Since on C₂ the variable z is real, the second integral is real:

${\displaystyle \oint _{C}f($z)\,dz=\int _{C_{1}}f(z)\,dz+\int _{-R}^{R}f(x)\,dx}

Hence, if f satisfies condition (*) of Jordan's lemma, taking the limit as R tends to infinity,

${\displaystyle \oint _{C}f($z)\,dz=\int _{-\infty }^{\infty }f(x)\,dx}

Finally, the left-hand side may be computed using the residue theorem to get

${\displaystyle \int _{-\infty }^{\infty }f($x)\,dx=2\pi i\sum _{k=1}^{n}\operatorname {Res\ } {\left(f,a_{k}\right)}}

where

${\displaystyle \operatorname {Res\ } {\left(f,a_{k}\right)}}$ is the residueoff at the singularity a_k.

Example

The function

${\displaystyle f($z)={\frac {e^{iz}}{1+z^{2}}}}

satisfies the condition of Jordan's lemma. Since the only singularity of f(z) in the upper half plane is at z = i, the above yields

${\displaystyle \int _{-\infty }^{\infty }{\frac {e^{ix}}{1+x^{2}}}\,dx=2\pi i\,\operatorname {Res\ } {\left(f,i\right)}}$

An easy calculation gives

${\displaystyle \operatorname {Res\ } {\left(f,i\right)}={\frac {e^{-1}}{2i}}}$

so that

${\displaystyle \int _{-\infty }^{\infty }{\frac {\cos x}{1+x^{2}}}\,dx=\int _{-\infty }^{\infty }{\frac {e^{ix}}{1+x^{2}}}\,dx={\frac {\pi }{e}}}$

This result exemplifies how some integrals difficult to compute with classical tools are easily tackled with the help of complex analysis.

Proof

By definition,

${\displaystyle \int _{C_{R}}f($z)\,dz=\int _{0}^{\pi }g(Re^{i\theta })e^{iaR(\cos \theta +i\sin \theta )}iRe^{i\theta }\,d\theta =\int _{0}^{\pi }g(Re^{i\theta })e^{aR(i\cos \theta -\sin \theta )}iRe^{i\theta }\,d\theta }

Now the inequality

${\displaystyle \left|\int _{a}^{b}f($x)\,dx\right|<\int _{a}^{b}\left|f(x)\right|\,dx}

yields

${\displaystyle I_{R}=\left|\int _{C_{R}}f($z)\,dz\right|\leq \int _{0}^{\pi }\left|g(Re^{i\theta })e^{aR(i\cos \theta -\sin \theta )}iRe^{i\theta }\right|\,d\theta =\int _{0}^{\pi }\left|g(Re^{i\theta })\right|e^{-aR\sin \theta }\ R\,d\theta }

The inequalities

${\displaystyle \sin \theta \geq {\frac {2\theta }{\pi }}}$ for ${\displaystyle 0\leq \theta \leq {\frac {\pi }{2}}}$ and ${\displaystyle \sin \theta \geq {\frac {2(\pi -\theta )}{\pi }}}$ for ${\displaystyle {\frac {\pi }{2}}\leq \theta \leq \pi }$

imply

${\displaystyle \lim _{R\rightarrow \infty }I_{R}\leq \lim _{R\rightarrow \infty }\int _{0}^{\pi }e^{-aR\sin \theta }\ R\,d\theta \leq \lim _{R\rightarrow \infty }\left[\int _{0}^{\pi /2}e^{-2aR\theta /\pi }\ R\,d\theta \ +\ \int _{\pi /2}^{\pi }e^{-2aR(\pi -\theta )/\pi }\ R\,d\theta \right]=0}$

as can be seen by explicitly computing the integrals by parts.