# Jordan's lemma

In complex analysis, Jordan's lemma is a result frequently used in conjunction with the residue theorem to evaluate contour integrals and improper integrals. It is named after the French mathematician Camille Jordan.

## Statement

Consider a complex-valued, continuous function f, defined on a semicircular contour

$C_R=\{z : z=R e^{i \theta}, \theta\in [0,\pi]\}$

of radius R > 0 lying in the upper half-plane, centred at the origin. If the function f is of the form

$f(z)=e^{iaz} g(z)\,,\quad z\in C_R,$

with a parameter a > 0, then Jordan's lemma states the following upper bound for the contour integral:

$\biggl|\int_{C_R} f(z)\, dz\biggr| \le \frac\pi{a}\max_{\theta\in [0,\pi]} \bigl|g \bigl(R e^{i \theta}\bigr)\bigr|\,.$

where equal sign is when g(z) is identically zero. An analogous statement for a semicircular contour in the lower half-plane holds when a < 0.

### Remarks

• If f is defined and continuous on the semicircular contour CR for all large R and
$M_R:=\max_{\theta\in [0,\pi]} \bigl|g \bigl(R e^{i \theta}\bigr)\bigr| \to 0\quad \mbox{as } R \to \infty\,,\qquad(*)$
then by Jordan's lemma
$\lim_{R \to \infty} \int_{C_R} f(z)\, dz = 0.$
• Compared to the estimation lemma, the upper bound in Jordan's lemma does not explicitly depend on the length of the contour CR.

## Application of Jordan's lemma

The path C is the concatenation of the paths C1 and C2.

Jordan's lemma yields a simple way to calculate the integral along the real axis of functions f (z) = eiazg(z) holomorphic on the upper half-plane and continuous on the closed upper half-plane, except possibly at a finite number of non-real points z1, z2, ..., zn. Consider the closed contour C, which is the concatenation of the paths C1 and C2 shown in the picture. By definition,

$\oint_{C} f(z)\, dz = \int_{C_1}f(z)\,dz + \int_{C_2} f(z)\,dz\,.$

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

$\int_{C_2} f(z)\,dz = \int_{-R}^{R} f(x)\,dx\,.$

The left-hand side may be computed using the residue theorem to get, for all R larger than the maximum of |z1|, |z2|, ..., |zn|,

$\oint_{C} f(z)\, dz = 2\pi i \sum_{k=1}^n \operatorname{Res}(f, z_k)\,,$

where Res(f, zk) denotes the residue of f at the singularity zk. Hence, if f satisfies condition (*), then taking the limit as R  tends to infinity, the contour integral over C1 vanishes by Jordan's lemma and we get the value of the improper integral

$\int_{-\infty}^{\infty} f(x)\,dx = 2\pi i \sum_{k=1}^n \operatorname{Res}(f, z_k)\,.$

## Example

The function

$f(z)=\frac{e^{iz}}{1+z^2},\qquad z\in{\mathbb C}\setminus\{i,-i\},$

satisfies the condition of Jordan's lemma with a = 1 for all R > 0 with R ≠ 1. Note that, for R > 1,

$M_R=\max_{\theta\in[0,\pi]}\frac1{|1+R^2e^{2i\theta}|}=\frac1{R^2-1}\,,$

hence (*) holds. Since the only singularity of f in the upper half plane is at z = i, the above application yields

$\int_{-\infty}^\infty \frac{e^{ix}}{1+x^2}\,dx=2\pi i\,\operatorname{Res}(f,i)\,.$

Since z = i is a simple pole of f and 1 + z2 = (z + i)(z - i), we obtain

$\operatorname{Res}(f,i)=\lim_{z\to i}(z-i)f(z) =\lim_{z\to i}\frac{e^{iz}}{z+i}=\frac{e^{-1}}{2i}$

so that

$\int_{-\infty}^\infty \frac{\cos x}{1+x^2}\,dx=\operatorname{Re}\int_{-\infty}^\infty \frac{e^{ix}}{1+x^2}\,dx=\frac{\pi}{e}\,.$

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

## Proof of Jordan's lemma

By definition of the complex line integral,

\begin{align} \int_{C_R} f(z)\, dz &=\int_0^\pi g(Re^{i\theta})\,e^{iaR(\cos\theta+i \sin\theta)}\,i Re^{i\theta}\,d\theta\\ &=R\int_0^\pi g(Re^{i\theta})\,e^{aR(i\cos\theta-\sin\theta)}\,ie^{i\theta}\,d\theta\,. \end{align}

Now the inequality

$\biggl|\int_a^b f(x)\,dx\biggr|\le\int_a^b |f(x)|\,dx$

yields

\begin{align} I_R:=\biggl|\int_{C_R} f(z)\, dz\biggr| &\le R\int_0^\pi\bigl|g(Re^{i\theta})\,e^{aR(i\cos\theta-\sin\theta)}\,ie^{i\theta} \bigr|\,d\theta\\ &=R\int_0^\pi \bigl|g(Re^{i\theta})\bigr|\,e^{-aR\sin\theta}\,d\theta\,. \end{align}

Using MR as defined in (*) and the symmetry sin θ = sin(πθ), we obtain

$I_R \le RM_R\int_0^\pi e^{-aR\sin\theta}\,d\theta = 2RM_R\int_0^{\pi/2} e^{-aR\sin\theta}\,d\theta\,.$

Since the graph of sin θ is concave on the interval θ ∈ [0,π /2], the graph of sin θ lies above the straight line connecting its endpoints, hence

$\sin\theta\ge \frac{2\theta}{\pi}\quad$

for all θ ∈ [0,π /2], which further implies

$I_R \le 2RM_R \int_0^{\pi/2} e^{-2aR\theta/\pi}\,d\theta =\frac{\pi}{a} (1-e^{-a R}) M_R\le\frac\pi{a}M_R\,.$