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Math Help - improper integral #4

  1. #1
    Super Member Random Variable's Avatar
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    improper integral #4

    Evaluate  \int_{0}^{\infty} \frac{\ln x}{(x+a)(x+b)} \ dx \  \ a,b>0 \ (a \ne b)

    Unlike the last integral, try not to use contour integration.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: improper integral #4

    Quote Originally Posted by Random Variable View Post
    Evaluate  \int_{0}^{\infty} \frac{\ln x}{(x+a)(x+b)} \ dx \  \ a,b>0 \ (a \ne b)

    Unlike the last integral, try not to use contour integration.
    Spoiler:


    We first make the observation that


    \displaystyle \frac{1}{(x+a)(x+b)}=\frac{1}{b-a}\int_a^b\frac{dy}{(x+y)^2} \text{ }dy



    And thus, if I(a,b) is the given integral we have that


    \displaystyle I(a,b)=\frac{1}{b-a}\int_0^\infty \int_a^b \frac{\log(x)}{(x+y)^2}\text{ }dy\,dx


    Now, since


    \displaystyle \int_{\mathbb{R}\times[a,b]}\left|\frac{\log(x)}{(x+y)^2}\right|\text{ }d(x,y)<\infty


    (which is easily verified via estimation) we may apply Fubini's theorem to conclude that


    \displaystyle I(a,b)=\frac{1}{b-a}\int_a^b\int_0^\infty \frac{\log(x)}{(x+y)^2}\text{ }dx\,dy


    But, it is easy to see that


    \displaystyle \int\frac{\log(x)}{(x+y)^2}\text{ }dx=\frac{x\log(x)}{y(y+x)}-\frac{\log(y+x)}{y}


    So that


    \displaystyle \int_0^{\infty}\frac{\log(x)}{(x+y)^2}\text{ }dx=\frac{\log(y)}{y}


    and so


    \begin{aligned}I(a,b) &=\frac{1}{b-a}\int_a^b\int_0^\infty\frac{\log(x)}{(x+y)^2} \text{ }dx\\ &=\frac{1}{b-a}\int_a^b\frac{\log(y)}{y}\;dy\\ &=\frac{\log^2(b)-\log^2(a)}{2(b-a)}\end{aligned}

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  3. #3
    Super Member Random Variable's Avatar
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    Re: improper integral #4

    There is an easier approach.

    Make the substitution  u = \frac{ab}{x} and see what happens.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: improper integral #4

    Quote Originally Posted by Random Variable View Post
    There is an easier approach.

    Make the substitution  u = \frac{ab}{x} and see what happens.
    Yes, that may work. But it wasn't what was most natural to me. I evaluated the integral on Wolfram Alpha for a couple of specific values and noticed that it was always of the form F(b)-F(a) where b>a that just screamed use a double integral to me.
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  5. #5
    Super Member Random Variable's Avatar
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    Re: improper integral #4

    Quote Originally Posted by Drexel28 View Post
    Yes, that may work. But it wasn't what was most natural to me. I evaluated the integral on Wolfram Alpha for a couple of specific values and noticed that it was always of the form F(b)-F(a) where b>a that just screamed use a double integral to me.
    The substitution is not something I came up with. My automatic approach would be to find the Cauchy principal value using contour integration and then drop the P.V. label because the integral is convergent.

    But if you make that substitution you get

     \int^{\infty}_{0} \frac{ \ln x}{(x+a)(x+b)} \ dx = \ln (ab) \int^{\infty}_{0} \frac{du}{(u+a)(u+b)} - \int^{\infty}_{0} \frac{ \ln u}{(u+a)(u+b)} \ du


    so  \int^{\infty}_{0} \frac{\ln x}{(x+a)(x+b)} \ dx = \frac{\ln (ab)}{2} \int^{\infty}_{0} \frac{du}{(x+a)(x+b)} \ dx


    EDIT:  = \frac{\ln a + \ln b}{2(a-b)} \int^{\infty}_{0} \Big(\frac{1}{x+b} - \frac{1}{x+a} \Big) \ dx = \frac{\ln a + \ln b}{2(a-b)} \ (\ln a - \ln b)


     = \frac{\ln^{2} a - \ln^{2} b}{2(a-b)}
    Last edited by Random Variable; June 14th 2011 at 07:43 PM.
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  6. #6
    Super Member Random Variable's Avatar
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    Re: improper integral #4

    The same substitution can be used to evaluate  \int^{b}_{a} \frac{\ln x}{(x+a)(x+b)} \ dx .
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