# Improper integrals

1. May 8, 2013

### Denisse

How can I get the improper integral
$\int_0^{\infty}\frac{1}{x(1+x^2)}\,dx$

First thing I tried was separating the integral like this

$\int_0^{1}\frac{1}{x(1+x^2)}\,dx + \int_1^{\infty}\frac{1}{x(1+x^2)}\,dx$

And then I tried with partial fractions but it didn't work

Last edited: May 8, 2013
2. May 8, 2013

### Simon Bridge

Hmmm... partial fractions would have been my first choice - what do you mean "didn't work"?
Have you tried a trig substitution? $x=\tan\theta$ ?

 partial fractions looks much easier than the trig sub.

Last edited: May 8, 2013
3. May 8, 2013

### Office_Shredder

Staff Emeritus
Do you think the integral converges or diverges?

4. May 8, 2013

### Denisse

I tried with x=tan u and it seems like the integral is divergent because of this:

$\int _{ 0 }^{ \infty }{ \frac { 1 }{ x\left( { x }^{ 2 }+1 \right) } dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \cos { u } }{ \sin { u } } du }$

5. May 9, 2013

### mathman

The integrand > 0 for all x, and ~ 1/x as x -> 0. Therefore it is divergent.