Jordan's lemma

Template:Short description 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 Template:Math, defined on a semicircular contour

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

of positive radius Template:Math lying in the upper half-plane, centered at the origin. If the function Template:Math is of the form

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

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

$|\underset{C_R}{\int }f(z)dz|\le \frac{\pi }{a}{M}_R\text{where}{M}_R:=\underset{\theta \in [0,\pi ]}{\max }\left|g\left(R{e}^{i\theta }\right)\right|.$

with equality when Template:Math vanishes everywhere, in which case both sides are identically zero. An analogous statement for a semicircular contour in the lower half-plane holds when Template:Math.

Remarks

• If Template:Math is continuous on the semicircular contour Template:Math for all large Template:Math and

then by Jordan's lemma $$\underset{R\to \mathrm{\infty }}{\lim }\underset{C_R}{\int }f(z)dz=0.$$

• For the case Template:Math, 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 Template:Math.

Application of Jordan's lemma

The path Template:Math is the concatenation of the paths Template:Math and Template:Math.

Jordan's lemma yields a simple way to calculate the integral along the real axis of functions Template:Math holomorphic on the upper half-plane and continuous on the closed upper half-plane, except possibly at a finite number of non-real points Template:Math, Template:Math, …, Template:Math. Consider the closed contour Template:Math, which is the concatenation of the paths Template:Math and Template:Math shown in the picture. By definition,

${\oint }_{C}f(z)dz=\underset{C_1}{\int }f(z)dz+\underset{C_2}{\int }f(z)dz.$

Since on Template:Math the variable Template:Math is real, the second integral is real:

$\underset{C_2}{\int }f(z)dz=\underset{-R}{\overset{R}{\int }}f(x)dx.$

The left-hand side may be computed using the residue theorem to get, for all Template:Math larger than the maximum of Template:Math, Template:Math, …, Template:Math,

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

where Template:Math denotes the residue of Template:Math at the singularity Template:Math. Hence, if Template:Math satisfies condition (Template:EquationNote), then taking the limit as Template:Math tends to infinity, the contour integral over Template:Math vanishes by Jordan's lemma and we get the value of the improper integral

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)dx=2\pi i\sum _{k=1}^n\mathrm{Res}(f,z_k).$

Example

The function

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

satisfies the condition of Jordan's lemma with Template:Math for all Template:Math with Template:Math. Note that, for Template:Math,

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

hence (Template:EquationNote) holds. Since the only singularity of Template:Math in the upper half plane is at Template:Math, the above application yields

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

Since Template:Math is a simple pole of Template:Math and Template:Math, we obtain

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

so that

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\cos x}{1+x^2}dx=\mathrm{Re}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{ix}}{1+x^2}dx=\frac{\pi }{e}.$

This result exemplifies the way some integrals difficult to compute with classical methods are easily evaluated with the help of complex analysis.

This example shows that Jordan's lemma can be used instead of a much simpler estimation lemma. Indeed, estimation lemma suffices to calculate $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{ix}}{1+x^2}dx$, as well as $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\cos x}{1+x^2}dx$, Jordan's lemma here is unnecessary.

Proof of Jordan's lemma

By definition of the complex line integral,

$\underset{C_R}{\int }f(z)dz=\underset{0}{\overset{\pi }{\int }}g(R{e}^{i\theta })e^{iaR(\cos \theta +i\sin \theta )}iR{e}^{i\theta }d\theta =R\underset{0}{\overset{\pi }{\int }}g(R{e}^{i\theta })e^{aR(i\cos \theta -\sin \theta )}i{e}^{i\theta }d\theta .$

Now the inequality

$|\underset{a}{\overset{b}{\int }}f(x)dx|\le \underset{a}{\overset{b}{\int }}|f(x)|dx$

yields

$I_R:=|\underset{C_R}{\int }f(z)dz|\le R\underset{0}{\overset{\pi }{\int }}|g(R{e}^{i\theta })e^{aR(i\cos \theta -\sin \theta )}i{e}^{i\theta }|d\theta =R\underset{0}{\overset{\pi }{\int }}|g(R{e}^{i\theta })|e^{-aR\sin \theta }d\theta .$

Using Template:Math as defined in (Template:EquationNote) and the symmetry Template:Math, we obtain

$I_R\le R{M}_R\underset{0}{\overset{\pi }{\int }}{e}^{-aR\sin \theta }d\theta =2R{M}_R\underset{0}{\overset{\pi /2}{\int }}{e}^{-aR\sin \theta }d\theta .$

Since the graph of Template:Math is concave on the interval Template:Math, the graph of Template:Math lies above the straight line connecting its endpoints, hence

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

for all Template:Math, which further implies

$I_R\le 2R{M}_R\underset{0}{\overset{\pi /2}{\int }}{e}^{-2aR\theta /\pi }d\theta =\frac{\pi }{a}(1-e^{-aR})M_R\le \frac{\pi }{a}{M}_R.$

See also

• Estimation lemma

References

• Template:Cite book