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Complex integral

  1. Apr 10, 2008 #1
    Hi,

    I have a problem with the following complex integral.

    Integral from 0 to+infinty sin²(a.lnz)/(x - 1)² = pi.a coth(2.pi.a) - 1/2 a>0

    I tried different contours and methods,but without result.

    Can you help me to find out the complex contour integration.

    Thanks
     
  2. jcsd
  3. Apr 10, 2008 #2
    Do you mean

    [tex]
    \int_0^\infty{}\frac{\sin^2{(a \log z)}}{(z-1)^2}
    [/tex]
    with z in the denominator instead of x. What contour did you try?
     
  4. Apr 10, 2008 #3

    tiny-tim

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    Welcome to PF!

    Hi Belgium 12! Welcome to PF! :smile:

    (do you mean (z - 1)² ?)

    Hint: sin²θ = [1 - cos2θ]/2; and cos(2a ln(z)) = … ? :smile:
     
  5. Apr 10, 2008 #4
    Yes that's the integral.

    I tried a contour in upper half plane with an identitation at 1.

    Then key-hole contour with the branch 0<phi<2.pi.

    The a rectangular contour after the subtitution z= exp u

    tanks
     
  6. Apr 10, 2008 #5

    This would be

    [tex]
    \frac{1}{2}\left[e^{2ia\log z}+e^{-2ia\log z}\right]=\frac{1}{2}\left[z^{2ia}+z^{-2ia}\right]
    [/tex]

    How does this help?
     
  7. Apr 10, 2008 #6
    Hi,

    I try it out and come later with a possible answer
     
  8. Apr 11, 2008 #7
    Hi,

    I have no result with cos(2alnz)=1/2(z exp2ia + z exp -2ia)

    you have then sin²=(1-cos 2 phi)/2

    thus 1/2{1- 1/2(zexp 2ia+zexp -2ia)}/ (x-1)²

    We already have a second order pole on the real axis (x-1)²/2

    I used the key-hole contour.

    thanks
     
  9. Apr 11, 2008 #8
    hi,
    the result of the integral pi.a coth(2pia)-1/2 is correct.I saw it in a book "integral tables"

    The integral with just the answer comes from a book in french "Recueil de problemes

    sur la theorie des fonctions analytiques" by Evgrafov MIR publishers Moscow.
     
  10. Apr 12, 2008 #9
    Hi,
    tiny tim

    Maybe I should try sin²(a.lnz)/(x-1)²=(exp 2ialnz -2 +exp2ialnz).-1/4/(x-1)²

    ans with the term -2/(x-1) partial integration in the PV sence.

    int from 1+e to infinty +int from 0 to 1-e.
    Then the two other integrals 1)zexp2ia/(x-1)² and zexp -2ia/(x-1)² and the adding.

    what's your opinion???

    thanks
     
  11. Apr 12, 2008 #10

    tiny-tim

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    Hi Belgium 12! :smile:

    Yes, that's what I was thinking of originally … it gives you something like:
    [tex]\frac{\left[z^{2ia}+z^{-2ia}\,-\,2\right]}{4 (z\,-\,1)^2}\,.[/tex]​
    Unfortunately, as Pere Callahan asks, does that actually get us anywhere?

    I'm very rusty on contour integrals.

    I suspect it has something to do with a quadrant integral, going down the positive real axis, up the positive imaginary axis, and then round in a huge 90º arc. But that still needs some way of (literally!) getting round the pole at z = 1. :confused:

    Sorry … I'm stuck. :redface:
     
  12. Apr 16, 2008 #11
    Hi,
    Tiny Tim
    Callehan
    I found this in a bib,in a french book .Integral with double pole on the real axis.
    It's a bit difficult for me for writing the math.expressions.I don't have latex or an other program.

    Sorry for that
     

    Attached Files:

    Last edited: Apr 16, 2008
  13. Apr 17, 2008 #12
    Hi Belgium,

    I have to admit your file is too long for me to read right now ... and the math expression don't speed up the reading. Why don't you try learning latex (it's nothing about the keyboard, just some markup language, similar to HTML) and post your specific question about the method (if there are any)?
     
  14. Apr 18, 2008 #13
    Hi,
    Pere callahan

    About your answer ysterday evening about the integral.I would say its not a specifc

    method.Ist just a complex contour integration,but you don't see it many books on

    complex integration.

    The method is just they insert a convergence factor.

    sin²(a.lnz)/(x-1)²=-1/4(z.exp 2ia -2 +z.exp -2ia)/(x-1)² they insert the factor let us say

    z.expb

    or -1/4(z.exp(2ia+b) -2.z.exp(b) -z.exp(-2ia+b))/(x-1)².

    then integrate the three individual integrals take the sum en let b->0

    that was it

    Belgium 12
     
  15. Apr 18, 2008 #14
    an then employ the formula

    {int from o to 1-r xexp.a/(x-1)² + int from 1+r to inf. xexp.a/(x-1)² - 2/r}=-pi.a.cotg(pi.a)

    to the three terms a=2ia and a=-2ia
     
  16. Apr 18, 2008 #15

    tiny-tim

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  17. Apr 19, 2008 #16
    yes that's is.

    Belgium 12
     
  18. Apr 19, 2008 #17

    HallsofIvy

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    I've merged the threads.
     
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