Calculate real integrals using complex analysis

[skrat]

Homework Statement

Calculate real integrals using complex analysis
a) ##\int_{-\infty}^{\infty}\frac{dx}{x^2+1}##
b) ##\int_0^\infty \frac{sin(x)}{x}dx##

Homework Equations

The Attempt at a Solution

a)
##\int_{-\infty }^{\infty }\frac{dz}{z^2+1}=\int_{-R}^{R}\frac{dx}{x^2+1}+\int _\gamma\frac{dz}{z^2+1}##

Where ##R-> \infty##.

If ##z=re^{i\varphi }## than

##\int _\gamma\frac{dz}{z^2+1}=i\int _\gamma\lim_{R->\infty }\frac{re^{i\varphi }}{r^2e^{2i\varphi }+1}d\varphi =0##

So initial equation ##\int_{-\infty }^{\infty }\frac{dz}{z^2+1}=\int_{-R}^{R}\frac{dx}{x^2+1}+\int _\gamma\frac{dz}{z^2+1}## is now ##\int_{-\infty }^{\infty }\frac{dz}{z^2+1}=\int_{-R}^{R}\frac{dx}{x^2+1}##.

There is a pole of first order in ##i##. Which gives me ##Res(f,i)=\frac{1}{2i}## and finally

##\int_{-\infty }^{\infty }\frac{dz}{z^2+1}=2\pi i\frac{1}{2i}=\pi ##

Therefore ##\int_{-\infty }^{\infty }\frac{dx}{x^2+1}=\pi ##.

b) Have no idea.

##\int_{0}^{\infty}\frac{sin(z)}{z}dz=\int_{0}^{\infty}\frac{sin(x)}{x}dx+\int_\gamma \frac{sin(z)}{z}dz##

I tried to integrate (because it is also from ##0## to ##\infty ##) ##\int_{0}^{\infty}\frac{dx}{1+x^3}## to maybe find out anything yet I could get the right result here either...

Lets get back to this: ##\int_{0}^{\infty}\frac{sin(z)}{z}dz##.

This integral is ##0##, because ##\frac{sin(z)}{z}## has a removable singularity for ##z=0##, therefore ##Res(f,z=0)=0##.

But what to do with ##\int_\gamma \frac{sin(z)}{z}dz## ... That I do not know.

 

[vela]

Try using the fact that the integrand is even and that sin x = Im(e^ix).