Question

Using the method of residues, show that:


Integral( 0 to infinity )(dx/1+x^2)=pie/2

Answer 1

$$\intop_0^\infin\frac{1}{1+x^2}dx=\frac{\pi}{2}$$

The integrand is even, so the integral can be replaced with

$$\intop_0^\infin\frac{1}{1+x^2}dx=\frac{1}{2}\intop_{-\infin}^\infin\frac{1}{1+x^2}dx$$

We use the formula:

$$\oint_Cf(z)dz=2\pi i\sum Res{f(z)}$$

Find the singular point(s):

$$f(z)=\frac{1}{1+z^2}=\frac{1}{(z-i)(z+i)}$$

$z=i$ is a singular point located in the upper complex half-plane.

$Res_{z=i}f(z)=\lim_{z\rightarrow i}f(z)(z-i)=\lim_{z\rightarrow i}\frac{1}{(z+i)(z-i)}(z-i)=\lim_{z\rightarrow i}\frac{1}{z+i}=\frac{1}{2i}$

$$\intop_{-\infin}^\infin\frac{1}{1+x^2}dx=2\pi i\cdot\frac{1}{2i}=\pi$$

We get

$$\intop_0^\infin\frac{1}{1+x^2}dx=\frac{1}{2}\intop_{-\infin}^\infin\frac{1}{1+x^2}dx=\frac{π}{2}$$