Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Improper integral using the Residue Theorem

  1. Dec 13, 2005 #1


    User Avatar
    Homework Helper

    I'm trying to find

    [tex]\int\limits_0^{ + \infty } {\frac{{\sin x}}
    {{x^3 + x}}dx} [/tex]

    Since the function is even, I can compute it as

    {2}\int\limits_{ - \infty }^{ + \infty } {\frac{{\sin x}}
    {{x^3 + x}}dx} [/tex]

    To use the residue theorem, I construct a large semi-circle C with center O and radius R. To avoid the real singularity at x = 0, we construct another (small) semi-circle c with radius r. The closed path is then given by C + [-R,-r] + c + [r,R]. We then let R go to infinity and r to 0. So we have:

    [tex]\oint {\frac{{\sin x}}
    {{x^3 + x}}dx} = \int\limits_C {\frac{{\sin x}}
    {{x^3 + x}}dx} + \int\limits_c {\frac{{\sin x}}
    {{x^3 + x}}dx} + \int\limits_{ - R}^{ - r} {\frac{{\sin x}}
    {{x^3 + x}}} dx + \int\limits_r^R {\frac{{\sin x}}
    {{x^3 + x}}} dx[/tex]

    From the residue theorem, I found the LHS to be [itex]\pi \sinh 1[/itex], by applying the theorem to the only enclosed pole i.
    Then, taking the limits or R and r to inf and 0, I found the integral over C to be 0 and over c to be [itex]-\pi[/itex] which results in:

    [tex]\int\limits_{ - \infty }^{ + \infty } {\frac{{\sin x}}
    {{x^3 + x}}} dx = \pi \sinh 1 + \pi [/tex]

    Now that appears to be too large, does anyone know what mistake(s) I made or what I should do? Thanks!
  2. jcsd
  3. Dec 13, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    A couple mistakes, the integral over C does not go to zero and there's no pole at x=0, so the integral over c will go to zero.

    It's probably easiest to replace sin(x) with [tex]e^{ix}[/tex], which is bounded in the upper half plane, making the integral over C much easier to deal with, though you'll now have a pole at z=0.
  4. Jun 7, 2010 #3
    There is indeed a pole at z=0, by Jordan's... but you only need to worry about when doing exactly that kind of problem. If I'm not mistaken you have to set the LHS to Im(2*pi*i*Resi (eiz/(z3+z))
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook