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. The lemma is named after the French mathematician Camille Jordan.

Statement[]

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

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

of positive radius R lying in the upper half-plane, centered at the origin. If the function f is of the form

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

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

${\displaystyle \left|\int _{C_{R}}f(z)\,dz\right|\leq {\frac {\pi }{a}}M_{R}\quad {\text{where}}\quad M_{R}:=\max _{\theta \in [0,\pi ]}\left|g\left(Re^{i\theta }\right)\right|.}$

with equality when g vanishes everywhere, in which case both sides are identically zero. 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 C_R for all large R and

${\displaystyle \lim _{R\to \infty }M_{R}=0}$

(*)

then by Jordan's lemma

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

• 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 C_R.

Application of Jordan's lemma[]

Jordan's lemma yields a simple way to calculate the integral along the real axis of functions f(z) = e^{i 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, …, z_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:

${\displaystyle \int _{C_{2}}f(z)\,dz=\int _{-R}^{R}f(x)\,dx\,.}$

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

${\displaystyle \oint _{C}f(z)\,dz=2\pi i\sum _{k=1}^{n}\operatorname {Res} (f,z_{k})\,,}$

where Res(f, z_k) denotes the residue of f at the singularity z_k. 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

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

Example[]

The function

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

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

${\displaystyle M_{R}=\max _{\theta \in [0,\pi ]}{\frac {1}{|1+R^{2}e^{2i\theta }|}}={\frac {1}{R^{2}-1}}\,,}$

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

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

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

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

so that

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

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,

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

Now the inequality

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

yields

${\displaystyle I_{R}:={\biggl |}\int _{C_{R}}f(z)\,dz{\biggr |}\leq R\int _{0}^{\pi }{\bigl |}g(Re^{i\theta })\,e^{aR(i\cos \theta -\sin \theta )}\,ie^{i\theta }{\bigr |}\,d\theta =R\int _{0}^{\pi }{\bigl |}g(Re^{i\theta }){\bigr |}\,e^{-aR\sin \theta }\,d\theta \,.}$

Using M_R as defined in (*) and the symmetry sin θ = sin(π − θ), we obtain

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

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

${\displaystyle \sin \theta \geq {\frac {2\theta }{\pi }}\quad }$

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

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

See also[]

• Estimation lemma

References[]

• Brown, James W.; Churchill, Ruel V. (2004). Complex Variables and Applications (7th ed.). New York: McGraw Hill. pp. 262–265. ISBN 0-07-287252-7.