# 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

$C_{R}=\{Re^{i\theta }\mid \theta \in [0,\pi ]\}$ of positive radius R lying in the upper half-plane, centered at the origin. If the function f is of the form

$f(z)=e^{iaz}g(z),\quad z\in C_{R},$ with a positive parameter a, then Jordan's lemma states the following upper bound for the contour integral:

$\left|\int _{C_{R}}f(z)\,dz\right|\leq {\frac {\pi }{a}}M_{R}\quad {\text{where}}\quad M_{R}:=\max _{\theta \in [0,\pi ]}\left|g\left(Re^{i\theta }\right)\right|.$ with equality when g vanishes everywhere. An analogous statement for a semicircular contour in the lower half-plane holds when a < 0.

### Remarks

• If f is continuous on the semicircular contour CR for all large R and

$\lim _{R\to \infty }M_{R}=0$ (*)

then by Jordan's lemma
$\lim _{R\to \infty }\int _{C_{R}}f(z)\,dz=0.$ • For the case a = 0, see the estimation lemma.
• 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

Jordan's lemma yields a simple way to calculate the integral along the real axis of functions f(z) = ei a z g(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,

$M_{R}=\max _{\theta \in [0,\pi ]}{\frac {1}{|1+R^{2}e^{2i\theta }|}}={\frac {1}{R^{2}-1}}\,,$ 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)(zi), 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 the way some integrals difficult to compute with classical methods are easily evaluated with the help of complex analysis.

## Proof of Jordan's lemma

By definition of the complex line integral,

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

${\biggl |}\int _{a}^{b}f(x)\,dx{\biggr |}\leq \int _{a}^{b}\left|f(x)\right|\,dx$ yields

$I_{R}:={\biggl |}\int _{C_{R}}f(z)\,dz{\biggr |}\leq R\int _{0}^{\pi }{\bigl |}g(Re^{i\theta })\,e^{aR(i\cos \theta -\sin \theta )}\,ie^{i\theta }{\bigr |}\,d\theta =R\int _{0}^{\pi }{\bigl |}g(Re^{i\theta }){\bigr |}\,e^{-aR\sin \theta }\,d\theta \,.$ Using MR as defined in (*) and the symmetry sin θ = sin(πθ), we obtain

$I_{R}\leq RM_{R}\int _{0}^{\pi }e^{-aR\sin \theta }\,d\theta =2RM_{R}\int _{0}^{\pi /2}e^{-aR\sin \theta }\,d\theta \,.$ 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

$\sin \theta \geq {\frac {2\theta }{\pi }}\quad$ for all θ ∈ [0, π ⁄ 2], which further implies

$I_{R}\leq 2RM_{R}\int _{0}^{\pi /2}e^{-2aR\theta /\pi }\,d\theta ={\frac {\pi }{a}}(1-e^{-aR})M_{R}\leq {\frac {\pi }{a}}M_{R}\,.$ ## See also

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