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Improper integrals

  1. May 8, 2013 #1
    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. jcsd
  3. May 8, 2013 #2

    Simon Bridge

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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: May 8, 2013
  4. May 8, 2013 #3

    Office_Shredder

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    Do you think the integral converges or diverges?
     
  5. May 8, 2013 #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 } ##
     
  6. May 9, 2013 #5

    mathman

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