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

  • Thread starter Denisse
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Looks like it diverges by comparison to 1/x, too.In summary, the improper integral ## \int_0^{\infty}\frac{1}{x(1+x^2)}\,dx ## does not converge and is divergent. Attempts to solve it using partial fractions and trig substitution were unsuccessful.
  • #1
Denisse
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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
 
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  • #2
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.
 
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  • #3
Do you think the integral converges or diverges?
 
  • #4
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
Denisse said:
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.
 

What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite or the integrand is not defined at one or more points within the interval of integration.

How do you solve an improper integral?

To solve an improper integral, you first need to rewrite it as a limit of definite integrals. Then, you can use various techniques such as substitution, partial fractions, or integration by parts to evaluate the limit and find the value of the integral.

Why is the integral \int_0^{\infty}\frac{1}{x(1+x^2)} considered improper?

The integral is considered improper because the integrand has a vertical asymptote at x = 0 and the upper limit of integration, \infty, is infinite.

What is the domain of the integrand \frac{1}{x(1+x^2)}?

The domain of the integrand is all real numbers except for x = 0 and x = \infty. This is because the integrand is undefined at these points.

What is the convergence behavior of \int_0^{\infty}\frac{1}{x(1+x^2)}?

The integral converges, meaning that it has a finite value, because the integrand approaches 0 as x approaches \infty. However, it requires special techniques to evaluate due to its improper nature.

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