Results 1 to 6 of 6

Thread: improper integral #4

  1. June 13th 2011, 08:57 PM #1
    Random Variable
    Random Variable is offline
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    2

    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.
    Follow Math Help Forum on Facebook and Google+
  2. June 14th 2011, 02:38 PM #2
    Drexel28
    Drexel28 is offline
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    College Park, Maryland
    Posts
    4,542
    Thanks
    11

    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}

    Follow Math Help Forum on Facebook and Google+
  3. June 14th 2011, 04:30 PM #3
    Random Variable
    Random Variable is offline
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    2

    Re: improper integral #4

    There is an easier approach.

    Make the substitution u = \frac{ab}{x} and see what happens.
    Follow Math Help Forum on Facebook and Google+
  4. June 14th 2011, 04:58 PM #4
    Drexel28
    Drexel28 is offline
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    College Park, Maryland
    Posts
    4,542
    Thanks
    11

    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.
    Follow Math Help Forum on Facebook and Google+
  5. June 14th 2011, 05:48 PM #5
    Random Variable
    Random Variable is offline
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    2

    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 06:43 PM.
    Follow Math Help Forum on Facebook and Google+
  6. June 14th 2011, 07:41 PM #6
    Random Variable
    Random Variable is offline
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    2

    Re: improper integral #4

    The same substitution can be used to evaluate \int^{b}_{a} \frac{\ln x}{(x+a)(x+b)} \ dx .
    Follow Math Help Forum on Facebook and Google+
« integral | Induction: an inequality »

Similar Math Help Forum Discussions

  1. Improper integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 19th 2010, 01:19 PM
  2. Improper Integral help?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 15th 2010, 04:03 PM
  3. Improper Integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 27th 2010, 11:58 AM
  4. Improper integral 2
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 26th 2008, 10:54 AM
  5. Improper integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 25th 2008, 11:14 PM

Search Tags


/mathhelpforum @mathhelpforum