1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cauchy Residue Theorem

  1. Jan 26, 2010 #1
    1. The problem statement, all variables and given/known data

    p(t) = integral[-inf,+inf] ( x/sinh(x) exp (i t x) dx)

    2. Relevant equations
    singularity @ x = n*pi*i where n = +-1, +-2, +-3,...

    Near n*pi*i one can write sinh(x) ~ (x - n*pi*i)


    3. The attempt at a solution

    I apply the cauchy residue theorem. For a positive value of t the contour of integrat
    can be closed with a semicircle in a positive complex x plane.

    Hence,

    p = (2*pi*i) * SUM ( n*pi*i) (-1)^n exp(-n*pi*t) ????

    p = pi^2/ ( 1 + cosh( pi*t) )
    ????

    According to the reference i have, the answer should have been

    p = pi / (4*cosh^2(pi*t/2) )



    I would really appreciate your help in this matter. More power.
     
  2. jcsd
  3. Jan 26, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The second answer is a lot closer to what I am getting so far, but with a pi^2. How did you change that sum into your answer?
     
  4. Jan 26, 2010 #3
    I get the series as from n=1 to infinity. I factor out pi*i from the summation.
    I have SUM (-1)^n*n exp(- pi*t*n) = -1/2 ( cosh( pi*t) +1 )^-1. I use wolframalpha to solve this.
     
  5. Jan 26, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Who am I to argue with WA? Are you really allowed to use that to solve your homework problems without figuring out how to sum the series on your own? I did it the old fashioned way by summing n(-r)^n by integrating a geometric series. That doesn't mean the reference is wrong. Both answers are actually the same. Can you show that using a double angle formula for cosh?
     
  6. Jan 26, 2010 #5
    I am not able to show that they are just the same. I did tried to compare my final answer to that of the reference by substituting a value for t (i.e t = 1),

    pi^2/ ( 1 + cosh( pi) ) =? pi / (4*cosh^2(pi*/2) )

    The difference of the two expression is 0.659 which suggest that they are not just equal.

    Actually this is part of the thesis that my friend in physics is working on and i am just trying to help. Thank you very much for the reply.
     
  7. Jan 27, 2010 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You have some typos in the expressions you are comparing. There should be a pi^2 in the numerator of the cosh^2 part and you missed a 1/2 in the cosh part. The identity you need to compare them is (cosh(x)+1)/2=cosh(x/2)^2 (analogous to a half angle formula in trig).
     
  8. Jan 27, 2010 #7
    You can avoid having to sum an infinite series by considering a different contour.
     
  9. Jan 27, 2010 #8
    Actually this integeral has been solved by researchers Khandekar and Wiegel. I've attached a page of their paper and kindly take a look at equation (22) and (23) in their work. I really have a hard time arriving at their answer. Thank you for the help. I just badly need it.
     
  10. Jan 27, 2010 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok, I think you are right for having a hard time arriving at their answer. As near as I can tell pi^2/(1+cosh(pi*t)) is correct, unless I've made another mistake. pi^2/(2*cosh(pi*t/2)^2) is also correct which is close to what I got but I wasn't really paying enough attention to factors of 2. The answer in the paper of pi/(4*cosh(pi*t/2)^2) is simply wrong. I thought the missing stuff was simply typos in your presentation. But it actually appears to be a typo in the paper. I'd also like to hear about Count Iblis' contour???
     
    Last edited: Jan 27, 2010
  11. Jan 27, 2010 #10
    Thank you very much!!! Could they be wrong with their answer? I have this doubt because a lot of their succeeding paper presentation use same value of their answer above. Attached is another paper that displays the same.


    Thank you for your time guys.

    Thanks for extending your help.
     

    Attached Files:

  12. Jan 28, 2010 #11
    You can also consider the integral of exp(i t x)/sinh(x) from minus to plus infinity where you take the principal part by omitting the interval from minus to plus epsilon and take the limit of epsilon to zero. The derivative w.r.t. t will then yield the desired integral. The contour can be taken to be a rectangle of width pi in the imaginary direction, which you deform so as to avoid the poles. Let's do this so that there no poles inside the contour. We avoid the origin using a semi-circle of radius epsilon in the upper half plane. We avoid the point z = i pi using a semi-circle of radius epsilon that lies below the line z = x + i pi.

    Let's denote the pricipal value of the integral of exp(i t x)/sinh(x) by I. Then the contour integral in the limit that the rectangle becomes infinitely long and epsilon goes to zero becomes:

    I - pi i + I exp(-pi t) + pi i exp(-pi t)

    This has to be zero, since there are no poles inside the contour. Therefore:

    I = pi i [1-exp(-pi t)]/[1+exp(-pi t)] = pi i tanh(pi t/2)


    The derivative w.r.t. t is i times the desired integral, so this is:

    1/2 pi^2/cosh^2(pi t/2)
     
  13. Jan 28, 2010 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Tough to say if the final answer is wrong. Were they sloppy when they wrote down the expression for P(A,N) preceding the integral? As for the integral if you ask Wolfram Alpha to integrate u/sinh(u) it will tell you it's pi^2/2. Not the pi/4 you get if you put zero into their expression.
     
  14. Jan 28, 2010 #13
    Thank you for the help Dick and Count Iblis. I did verify the result by substituting values for t and let Wolfram evaluate the integral and I have concluded that 1/2 pi^2/cosh^2(pi t/2) is indeed the correct answer.

    More power.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook