# Calculate real integrals using complex analysis

1. Apr 6, 2014

### skrat

1. The problem statement, all variables and given/known data
Calculate real integrals using complex analysis
a) $\int_{-\infty}^{\infty}\frac{dx}{x^2+1}$
b) $\int_0^\infty \frac{sin(x)}{x}dx$

2. Relevant equations

3. 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.

2. Apr 6, 2014

### vela

Staff Emeritus
Try using the fact that the integrand is even and that sin x = Im(eix).

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