Solving Improper Integral: \int_0^{\infty}\frac{1}{x(1+x^2)}

[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

 

[Simon Bridge]

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

[edit] partial fractions looks much easier than the trig sub.

 

[Office_Shredder]

Do you think the integral converges or diverges?

 

[Denisse]

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

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

 

[mathman]

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

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