31 Jul 2025

Contour Integration for Definite Integrals

Residue calculations for a real rational integral and a trigonometric integral on the unit circle.

bsc semester-ii mathematical-physics contour-integration definite-integrals

Contour choice is controlled by decay and singularities. If $f(z)=P(z)/Q(z)$ and $\deg Q\geq\deg P+2$, then on a semicircle of radius $R$ the integrand is $O(R^{-2})$ or smaller, while the arc length is $\pi R$; hence the arc contribution tends to zero. For $e^{ibz}$ with $b>0$, the upper half-plane is preferred because

\[\lvert e^{ib(x+iy)}\rvert=e^{-by}\leq1.\]

If a pole lies on the real axis, the contour must be indented and the result interpreted as a Cauchy principal value; an ordinary improper integral cannot be assumed to exist.

A rational integral on the real line

Let $a>0$ and consider

\[I=\int_{-\infty}^{\infty}\frac{dx}{x^2+a^2}.\]

Use

\[f(z)=\frac1{z^2+a^2} =\frac1{(z-ia)(z+ia)}\]

and close the real segment $[-R,R]$ by the upper semicircle. Only the pole $z=ia$ lies inside.

The upper semicircle encloses \(z=ia\) but not \(z=-ia\). Counterclockwise orientation fixes the positive sign in the residue theorem.

The residue is

\[\operatorname{Res}(f,ia) =\lim_{z\to ia}\frac{z-ia}{(z-ia)(z+ia)} =\frac1{2ia}.\]

Thus

\[\int_{-R}^{R}\frac{dx}{x^2+a^2} +\int_{\Gamma_R}\frac{dz}{z^2+a^2} =2\pi i\frac1{2ia} =\frac{\pi}{a}.\]

On the arc $\lvert z\rvert=R>a$,

\[\lvert z^2+a^2\rvert\ge \left\lvert\lvert z\rvert^2-a^2\right\rvert=R^2-a^2.\]

Since the arc length is $\pi R$,

\[\left\lvert \int_{\Gamma_R}\frac{dz}{z^2+a^2} \right\rvert \le\frac{\pi R}{R^2-a^2} \longrightarrow0.\]

Taking $R\to\infty$ gives

\[\boxed{ \int_{-\infty}^{\infty}\frac{dx}{x^2+a^2} =\frac{\pi}{a}}.\]

If $x$ and $a$ have units of length, both sides have units of inverse length.

A trigonometric integral

For real $a>\lvert b\rvert>0$, evaluate

\[J=\int_0^{2\pi}\frac{d\theta}{a+b\cos\theta}.\]

Set $z=e^{i\theta}$ on the unit circle. Then

\[d\theta=\frac{dz}{iz}, \qquad \cos\theta=\frac12\left(z+\frac1z\right).\]

Therefore

\[\begin{aligned} J &=\oint_{\lvert z\rvert=1} \frac{1}{a+\dfrac b2(z+z^{-1})}\frac{dz}{iz}\\ &=\frac2i\oint_{\lvert z\rvert=1} \frac{dz}{bz^2+2az+b}. \end{aligned}\]

The poles are the roots

\[z_\pm=\frac{-a\pm\sqrt{a^2-b^2}}{b}.\]

Their product is $z_+z_-=1$. Because $a>\lvert b\rvert>0$, $z_+$ has magnitude less than one and $z_-$ has magnitude greater than one. Only $z_+$ is enclosed.

The derivative of the quadratic denominator is $2bz+2a$, so

\[\operatorname{Res} \left(\frac1{bz^2+2az+b},z_+\right) =\frac1{2(bz_++a)}.\]

Using $bz_++a=\sqrt{a^2-b^2}$,

\[\begin{aligned} J &=\frac2i(2\pi i) \frac1{2\sqrt{a^2-b^2}}\\ &=\boxed{\frac{2\pi}{\sqrt{a^2-b^2}}}. \end{aligned}\]

Both definite-integral results are independently reduced from their residues in the Unit III Maxima worksheet.

Solved Problems

1. Evaluate

   \[I=\int_{-\infty}^{\infty}\frac{dx}{(x^2+1)^2}\]

   by residues.

   Close the contour in the upper half-plane. The only enclosed singularity is the double pole $z=i$. Write

   \[f(z)=\frac1{(z-i)^2(z+i)^2}.\]

   Its residue is

   \[\begin{aligned} \operatorname{Res}(f,i) &=\left.\frac{d}{dz}\frac1{(z+i)^2}\right\rvert_{z=i}\\ &=\left.-\frac2{(z+i)^3}\right\rvert_{z=i} =-\frac2{(2i)^3} =\frac1{4i}. \end{aligned}\]

   On $\lvert z\rvert=R>1$,

   \[\lvert(z^2+1)^2\rvert \geq(R^2-1)^2,\]

   so the arc magnitude is at most $\pi R/(R^2-1)^2\to0$. Therefore

   \[I=2\pi i\left(\frac1{4i}\right) =\boxed{\frac{\pi}{2}}.\]

   The integrand is positive and behaves as $x^{-4}$, consistent with a finite positive result.

2. For $a>0$ and $b>0$, evaluate

   \[I(a,b)=\int_{-\infty}^{\infty} \frac{\cos(bx)}{x^2+a^2}\,dx.\]

   Integrate

   \[f(z)=\frac{e^{ibz}}{z^2+a^2}\]

   over the upper semicircle. On its arc,

   \[\lvert e^{ibz}\rvert=e^{-b\,\operatorname{Im}z}\leq1,\]

   and

   \[\left\lvert\int_{\Gamma_R}f(z)\,dz\right\rvert \leq\frac{\pi R}{R^2-a^2}\longrightarrow0.\]

   The enclosed pole is $z=ia$, with residue

   \[\operatorname{Res}(f,ia) =\frac{e^{ib(ia)}}{2ia} =\frac{e^{-ab}}{2ia}.\]

   Hence

   \[\int_{-\infty}^{\infty} \frac{e^{ibx}}{x^2+a^2}\,dx =2\pi i\frac{e^{-ab}}{2ia} =\frac{\pi}{a}e^{-ab}.\]

   The sine part is odd and integrates to zero, so taking the real part gives

   \[\boxed{I(a,b)=\frac{\pi}{a}e^{-ab}}.\]

   At $b\to0^+$ this reduces to $\pi/a$, the rational integral derived above in the article.

Descriptive Questions

1. Explain how decay, pole locations, and exponential factors determine the choice and orientation of a contour.
2. Derive the semicircle estimate required for rational functions whose denominator degree exceeds the numerator degree by at least two.
3. Derive the substitution $z=e^{i\theta}$ for trigonometric integrals and explain how poles inside the unit circle are selected.
4. Explain why poles on the real axis require indentation and a Cauchy principal-value prescription.

Numerical Problems

1. Evaluate $\int_0^\infty dx/(x^4+1)$.

   Answer: $\pi/(2\sqrt2)$.

2. Evaluate $\int_0^\infty dx/(1+x^6)$.

   Answer: $\pi/3$.

3. Evaluate $\int_{-\infty}^{\infty}x\sin(2x)\,dx/(x^2+1)$.

   Answer: $\pi e^{-2}$.

4. Evaluate $\int_0^{2\pi}\cos\theta\,d\theta/(5+4\cos\theta)$.

   Answer: $-\pi/3$.

5. Evaluate $\operatorname{PV}\int_{-\infty}^{\infty}dx/(x^2-1)$.

   Answer: $0$.

All added pole residues, arc reductions, Fourier-type values, and principal-value checks are verified in the Unit III Maxima worksheet; every printed residual is zero.

References

1. Contour integration — Wikipedia
2. MIT OpenCourseWare 18.04, lecture notes: Topic 9, Definite Integrals Using the Residue Theorem
3. James Ward Brown and Ruel V. Churchill, Complex Variables and Applications, 9th ed., Chapter 7.