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

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.