1. Limited time only! Sign up for a free 30min personal 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!

Integral using residue theorem

  1. Sep 23, 2011 #1
    The question asks to show using the residue theorem that

    [itex]\int[/itex]cos(x) / (x2 +1)2 dx = [itex]\pi[/itex] / e

    (the terminals of the integral are -∞ to ∞ but i didnt know the code to write that)

    I found the singularities at -i and +i

    so i think we change the function inside the integral to cos(z) / (z2 +1)2

    i expanded the cos(z) as cosh(1) - isinh(1)(z-i) -0.5cosh(1)(z-i)2 +...

    and i expanded (z2+1)2 as -(1/4)(z-i)2 - i/4(z-i) + 3/16 +...

    I did the same for the singularity at x=-i and when i added both the residues together i got

    (9/16e + e/16) (this is multiplied by 2[itex]\pi[/itex]i to find residues)

    this doesnt seem right? i dont know if what ive done is the right method. please help, ive spent soooo many hours on this one stupid question :(
     
  2. jcsd
  3. Sep 23, 2011 #2

    lanedance

    User Avatar
    Homework Helper

    why do you add singularities together? not too sure what you're attempting...

    i think the idea is you need to set up a closed contour in complex space and integrate around it. Then teh vlaue of teh integral will be related to singularities contained within the contour.

    a good one would be along the x axis for -R to R, and a semi-circle in the upper half of the complex plane to connect the contour. Then take the limit r->infinity. Hopefully the semi-cirlce contribution tends to zero, then you can relate the x axis component directly to the original integral
     
  4. Sep 23, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Wouldn't you have to do that first, in order to find the singularities?

    What path are you integrating over? Does it include both singularities? Obviously to do this real integral as a path integral, you need a closed path that includes the real axis. I don't see how you can do that and get both i and -i inside the path.
     
  5. Sep 23, 2011 #4
    oh yeah... that makes sense...

    your right we cant get i and -i in on a path integral across the real axis

    maybe i need to build a circle or something? i guess i have less of an idea now?

    thanks for your help?

    any idea where i should go from here?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integral using residue theorem
Loading...