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

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Discussion Overview

The discussion revolves around evaluating the improper integral ## \int_0^{\infty}\frac{1}{x(1+x^2)}\,dx ##. Participants explore various methods for solving the integral, including separation of the integral, partial fractions, and trigonometric substitution. The focus is on convergence or divergence of the integral.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests separating the integral into two parts: ## \int_0^{1}\frac{1}{x(1+x^2)}\,dx + \int_1^{\infty}\frac{1}{x(1+x^2)}\,dx##.
  • Another participant questions the effectiveness of partial fractions and proposes a trigonometric substitution, specifically ##x=\tan\theta##.
  • A participant expresses concern about the convergence of the integral, prompting a discussion on whether it diverges or converges.
  • One participant attempts the substitution ##x=\tan u## and concludes that the integral appears to be divergent, citing the behavior of the integrand as ##x \to 0##.
  • Another participant reiterates that the integrand is positive for all x and behaves like ##1/x## as ##x \to 0##, leading to a divergence conclusion.

Areas of Agreement / Disagreement

Participants express differing views on the convergence of the integral, with some suggesting it diverges based on the behavior of the integrand, while others have not reached a consensus on the method of evaluation.

Contextual Notes

There are unresolved aspects regarding the effectiveness of the proposed methods and the conditions under which the integral may converge or diverge. The discussion includes assumptions about the behavior of the integrand near critical points.

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
 
Last edited:
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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.
 
Last edited:
Do you think the integral converges or diverges?
 
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 } ##
 
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.
 

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