Results 1 to 6 of 6

Math Help - Complex Analysis

  1. #1
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411

    Complex Analysis

    I need to show that \int_{0}^{\infty}\frac{\log(x)}{(1+x^2)^2}=-\frac{\pi}{4}

    If we complexify the integrand and make a sort of pacman-contour \mu

    \mu_1 = t+ ic_0 with 0\leq t\leq R

    \mu_2 = Re^{i\pi t} with t_0\leq t \leq 2\pi-t_0

     \mu_3 = R-t-ic_0 with 0\leq t\leq R

     \mu_4 = re^{i\pi t} with 2\pi- t_0\leq t \leq t_0

    As R\to \infty, r\to 0 we let t_0,c_0\to 0
    This contour contains the 2 second order singularities p= \pm i

    And we can evaluate \int_\mu f(z)dz with the Cauchy integral formula. I was somewhat surprised when I found \frac{-\pi}{4}. It's easy to check that

     |\int_{\mu_2}f(z)dz|\to 0 and |\int_{\mu 4}f(z)dz|\to 0

    As a result we get \int_{\mu_1}f(z)dz+\int_{\mu_3}f(z)dz = -\frac{\pi}{4}

    If all steps were ok, somehow I want to conclude that \int_{\mu_1\cup\mu_3}f(z)dz tends to \int_{0}^{\infty}f(x)dx

    But I'm not sure how. Also I found working with log(z) somewhat trouble-some. I believe I need to use some nice property of log(z)

    Can anyone offer some suggestions?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Dinkydoe View Post
    I need to show that \int_{0}^{\infty}\frac{\log(x)}{(1+x^2)^2}=-\frac{\pi}{4}

    If we complexify the integrand and make a sort of pacman-contour \mu

    \mu_1 = t+ ic_0 with 0\leq t\leq R

    \mu_2 = Re^{i\pi t} with t_0\leq t \leq 2\pi-t_0

     \mu_3 = R-t-ic_0 with 0\leq t\leq R

     \mu_4 = re^{i\pi t} with 2\pi- t_0\leq t \leq t_0

    As R\to \infty, r\to 0 we let t_0,c_0\to 0
    This contour contains the 2 second order singularities p= \pm i
    I don't understand what your contour is; there may be several mistakes in your parameterization... Furthermore, what are t_0, c_0? Could you describe your contour in words, or with a sketch?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    I don't understand what your contour is; there may be several mistakes in your parameterization... Furthermore, what are ? Could you describe your contour in words, or with a sketch?
    I was afraid so. I believe it's called the key-hole contour.

    http://upload.wikimedia.org/wikipedi...ontour.svg.png

    When you see this picture, you understand why I called it a 'pacman' contour ;p

    With c_0,t_0 I meant small numbers, actually dependant on R, but it's somewhat hard to define this contour it terms of R

    I made a pretty sad mistake though. I forgot to multiply \frac{-\pi}{4} with 2\pi i.
    Last edited by Dinkydoe; May 8th 2010 at 11:29 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Thanks for the precision. Note that \log z is not defined for all z\neq 0; you must exclude a half-line, usually \mathbb{R}_-. Then the contour includes values where the function is not defined.

    However, it seems to works if you symmetrize the picture, so that Pacman "looks to the left". This way the contour avoids the negative real axis \mathbb{R}_-.

    We have -x=e^{i\pi+\log x}=e^{-i\pi+\log x}. When z converges toward -x from the upper-half plane, then \log z converges toward i\pi+\log x and when it converges from the lower-half plane, \log z converges toward -i\pi+\log x. Thus, the integral along \mu_1\cup \mu_3 is almost equal to \int_0^R \frac{2i\pi}{(1+x^2)^2}dx (real parts cancel). My computation shows that the residue at i is \frac{\pi}{8}+\frac{i}{4}, and \frac{\pi}{8}-\frac{i}{4} at -i. The conclusion is \int_0^\infty \frac{dx}{(1+x^2)^2}=\frac{\pi}{4}, which is cool, but not what we need...

    Another contour will be necessary. For instance only the upper-half of the Pacman. Like a parliamentary hemicycle. Integral on the horizontal part on the right of 0 converges to your integral, and the integral on the horizontal part on the left converges to \int_0^\infty \frac{\log x+i\pi}{(1+x^2)^2}dx. By the way you'll recover the value of the above integral by looking at the imaginary part.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Hi. I don't think Cauchy's Integral Theorem can be used on this problem since the contour contains two poles. Also, just skip that "almost on the real axis thing" with the \pm c and just go for "exactly on the real axis" in the limit. Then instead of using \log(z), use \log(z)^2 and the same contour so that when you sum the branch-cut contours, the \log(x)^2 parts cancel and you're left with:

    -4\pi i\int_0^{\infty}\frac{\log(x)}{(1+x^2)^2}dx+4\pi^2  \int_0^{\infty}\frac{1}{(1+x^2)^2}dx=2\pi i\mathop\text{Res}\limits_{z=z_n}\left\{\frac{\log  (z)^2}{(1+z^2)^2}\right\},\quad z_n=\pm i

    but keep in mind of the branch you're using: over the contour above the real axis you have \log(x) and over the contour below the real axis you have \log(x)+2\pi i and \log(-i)=3\pi i/2

    I did it quick. May need a little fixin' up but I think it's close. That's my way of sayin', you work through it and in the process, learn how to do it .
    Last edited by shawsend; May 8th 2010 at 11:24 AM. Reason: fix formulas, added some explanations
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    That's my way of sayin', you work through it and in the process, learn how to do it .
    Haha so true,

    I thank you both for your comments. I appreciate your kind help very much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 4th 2011, 05:30 AM
  2. Replies: 6
    Last Post: September 13th 2011, 07:16 AM
  3. Replies: 1
    Last Post: October 2nd 2010, 01:54 PM
  4. Replies: 12
    Last Post: June 2nd 2010, 02:30 PM
  5. Replies: 1
    Last Post: March 3rd 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum