Jordan's lemma

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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.


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

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

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

f(z) = e^{i a z} 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| \le \frac{\pi}{a} M_R \quad \text{where} \quad M_R := \max_{\theta \in [0,\pi]} \left| g \left(R e^{i \theta}\right) \right| .

where equal sign is when g vanishes everywhere. An analogous statement for a semicircular contour in the lower half-plane holds when a < 0.


  • 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.
  • 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[edit]

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) = 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)\,.


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,


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[edit]

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)}\,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\,.

Now the inequality

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


 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\,.

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

\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\,.

See also[edit]


  • 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.