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!

Definite integrals: solving with residue theory and contour integration

  1. Oct 26, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to solve this integral for a>0:

    [itex]\int _0^{\infty }\frac{\text{Sin}[x]}{x}\frac{1}{x^2+a^2}dx[/itex]


    3. The attempt at a solution

    Using wolfram mathematica, I get that this integral is:

    [itex]\frac{\pi -e^{-x} \pi }{2x^2}=\frac{\pi (1-\text{Cosh}[a]+\text{Sinh}[a])}{2 a^2}[/itex]

    I cannot get to this result. I have tried to split the sine into exponentials, and use the upper/lower infinite semi-circle contours, which results in

    [itex]I =\frac{\pi }{2} \text{Res}\left[\frac{ e^{-i x}}{x}\frac{1}{x^2+a^2},\left\{x,a e^{i \frac{3\pi }{2}}\right\}\right]-\frac{\pi }{2} \text{Res}\left[\frac{ e^{i x}}{x}\frac{1}{x^2+a^2},\left\{x,a e^{i \frac{\pi }{2}}\right\}\right]=\frac{\pi \text{Sinh}[a]}{2 a^2}[/itex]

    I also tried to add a branch point and solve with that method but then I get the same answer...

    Does the "removable singularity" at z=0 have something to do with this? Please steer me in the right direction!
     
  2. jcsd
  3. Oct 26, 2012 #2
    If you split sine into exponentials and then split the integrals, then each integral has a pole at x=0 as well.
     
  4. Oct 26, 2012 #3
    But then, isn't my contour running over the pole in each integral?
     
  5. Oct 26, 2012 #4
    Tell you what, how about you forget that problem for now and just try and solve:

    [tex]\oint \frac{e^{iz}}{z}\frac{1}{z^2+a^2} dz[/tex]

    where the contour is the upper half-disc with an indentation around the pole at the origin. Take the limit as the radius of the indentation goes to zero and let the radius of the large arc go to infinity. Then you have:

    [tex]\int_{-\infty}^{\infty} f(x)dx+\lim_{\rho\to 0}\int_{\pi}^0 f(z)dz+\lim_{R\to\infty}\mathop\int_{\text{half-arc}} f(z)dz=2\pi i r[/tex]

    Won't that work? See if it does. I don't know for sure.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Definite integrals: solving with residue theory and contour integration
Loading...