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Improper Integral

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]\int_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}dy[/itex]


    2. Relevant equations
    The answer is [itex]\pi\log(x+1)[/itex].


    3. The attempt at a solution
    I have attempted many different substitutions like [itex]y=\tan\theta[/itex]. I have also tried breaking up the log but nothing definitive comes out. Any help would be appreciated. I would prefer to not use any complex analysis.
     
    Last edited: Apr 17, 2013
  2. jcsd
  3. Apr 17, 2013 #2

    Mark44

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    Which is the variable of integration? Without either dx or dy, I can't tell.
     
  4. Apr 17, 2013 #3
    Sorry. It is now fixed. We are integrating with respect to y.
     
  5. Apr 18, 2013 #4
    I am beginning to think that there is a way to show that the integral is equal to [itex]\pi\log(x+1)[/itex] without directly computing the integral. Any thoughts?
     
  6. Apr 18, 2013 #5

    SammyS

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    Perhaps, ...

    consider that [itex]\displaystyle \ \frac{d}{dx}\pi\log(x+1)=\frac{\pi}{x+1}\ .[/itex]
     
  7. Apr 18, 2013 #6

    Zondrina

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    Yes, this.

    Expressing it as an expansion can also work.
     
  8. Apr 18, 2013 #7
    Hmm. That looks very interesting. Thanks guys. I will play with that.
     
  9. Apr 18, 2013 #8
    I am confused. How do I use the fact that [itex]\frac{d}{dx}\pi\log(x+1)=\frac{\pi}{x+1}[/itex]? Shouldn't we be considering the changes in [itex]y[/itex] and not [itex]x[/itex]?
     
  10. Apr 18, 2013 #9

    Zondrina

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    Are you familiar with the expansion for ##\frac{1}{1-x}##?

    Use that to find an expansion for ##\frac{\pi}{1-(-x)}##

    What do you get?
     
  11. Apr 18, 2013 #10
    I got [itex]\pi\sum\limits_{n=0}^{\infty}(-1)^{n}x^{n}[/itex] but I don't see how I can apply this to solving the integral.
     
  12. Apr 19, 2013 #11

    haruspex

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    Let the original integral be F(x). What does dF/dx look like? Can you solve the integral wrt y in that?
     
  13. Apr 19, 2013 #12
    [itex]F(x)=\int\limits_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}dy[/itex]
    so
    [itex]\frac{d}{dx}F(x)=\frac{d}{dx}\int\limits_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}dy=\int\limits_{0}^{\infty}\frac{2xy^{2}}{(y^{2}+1)(x^{2}y^{2}+1)}dy=2x\int\limits_{0}^{\infty}\frac{y^{2}}{(y^{2}+1)(x^{2}y^{2}+1)}dy=2x\int\limits_{0}^{\infty}\left[\frac{Ay+B}{y^{2}+1}+\frac{Cy+D}{x^{2}y^{2}+1}\right][/itex]
    where [itex]A,C=0[/itex] and [itex]B=-\frac{1}{1-x^{2}},D=\frac{1}{1-x^{2}}[/itex]. Then
    [itex]2x\int\limits_{0}^{\infty}\left[\frac{Ay+B}{y^{2}+1}+\frac{Cy+D}{x^{2}y^{2}+1}\right]=\frac{2x}{1-x^{2}}\int\limits_{0}^{\infty}\frac{1}{x^{2}y^{2}+1}dy-\frac{2x}{1-x^{2}}\int\limits_{0}^{\infty}\frac{1}{y^{2}+1}dy=\frac{2x}{1-x^{2}}\left(\frac{\pi}{2x}\right)-\frac{2x}{1-x^{2}}\left(\frac{\pi}{2}\right)=\frac{\pi-x\pi}{1-x^{2}}=\pi\left(\frac{1-x}{1-x^{2}}\right)=\frac{\pi}{x+1}[/itex].

    This shows that
    [itex]\frac{d}{dx}F(x)=\frac{d}{dx}\int\limits_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}dy=\frac{d}{dx}\pi\log(x+1)[/itex]
    but I don't see how it shows that
    [itex]\int\limits_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}dy=\pi\log(x+1)[/itex].
    When we integrate them both in terms of [itex]x[/itex], would we not have a constant on both sides that could be different?
     
  14. Apr 19, 2013 #13

    haruspex

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    Yes, so the final step is to show that the constant is zero. You have f(x) = g(x) + c for all x. How might you determine c? Any ideas?
     
  15. Apr 19, 2013 #14
    Hmm... Let me sit on this for a while. I will post again once I have an idea! Thanks!
     
  16. Apr 19, 2013 #15
    Ok. Here's an idea. Took longer to think of than it should of..
    Since
    [itex]F(x)=\int\limits_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}=\pi\log(x+1)+C[/itex]
    we can let [itex]x=0[/itex] such that
    [itex]F(0)=\int\limits_{0}^{\infty}\frac{\log(1)}{y^{2}+1}=\pi\log(1)+C=0[/itex].
    Hence it must be that [itex]C=0[/itex].
     
  17. Apr 20, 2013 #16

    haruspex

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    That'll do it.
     
  18. Apr 20, 2013 #17
    Thanks a ton!
     
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