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Integration by Residue Calculus

  1. Dec 8, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm trying to evaluate the integral

    [itex]I(a)=\int\frac{cos(ax)}{x^{4}+1}[/itex]
    from 0 to ∞

    2. Relevant equations

    To do this I'm going to consider the complex integral:

    [itex]J=\oint\frac{e^{iaz}}{z^{4}+1}[/itex]

    Over a semi-circle of radius R in the upper half plane, then let R-->∞

    There are 2 curves, the straight curve along the real axis, and the semi-circular arc.

    3. The attempt at a solution
    Along the first curve (along the real axis) the line integral is just

    [itex]I_{1}=\int\frac{e^{iax}}{x^{4}+1}[/itex]

    from 0 to infinity

    Now the real part of this is 2*I(a)

    The second line integral goes to zero. So 2*I(a)=Re(J).

    So I need to evaluate J by residue calculus and take the real part to get I(a).

    So there are 2 simple poles inside of the semi-circle, namely
    [itex]z=e^{i\frac{\pi}{4}}[/itex] and [itex]z=e^{i\frac{3\pi}{4}}[/itex]

    So [itex]J=2i\pi(Res(f,e^{i\frac{\pi}{4}})+Res(f,e^{i\frac{3\pi}{4}}))[/itex]

    But I'm having trouble computing the Residues.

    I know that [itex]Res(f,e^{i\frac{\pi}{4}})=\frac{e^{iaz}}{4z^{3}} z=e^{i\frac{\pi}{4}}[/itex]

    But I'm struggling with the exponential to an exponential, I can't get a "clean" answer. I'm not sure if theres a better way I should go about this. Or if theres something I'm doing wrong.

    Please Help!!
     
  2. jcsd
  3. Dec 8, 2012 #2

    Mute

    User Avatar
    Homework Helper

    Are you sure about that result? I don't think there should be a z in the numerator. It looks like you calculated the residue using L'Hopital's rule, is that right? i.e.,

    $$\lim_{z\rightarrow z_1} (z-z_1)\frac{e^{i\alpha z}}{z^4+1} = \lim_{z\rightarrow z_1} \frac{(z-z_1)'}{(z^4+1)'}e^{i\alpha z}.$$
    When you apply the rule, you will get ##4z_1^3## in the denominator, but your numerator should just be a constant.

    You can always write the roots in cartesian form, ##z = a + ib##.
     
  4. Dec 8, 2012 #3
    Thanks for replying

    Thats what I got for my residue, Sorry that z is just saying plug in e^i...

    I try writing it in that form (a+bi), but its so messy!

    What I get is

    [itex]J=\frac{\pi\sqrt{2}}{2}e^{-a\frac{\sqrt{2}}{2}}(cos(a\frac{\sqrt{2}}{2})+sin(a\frac{\sqrt{2}}{2}))[/itex]

    Does anyone else think I'm crazy? I really feel like somethings wrong here!
     
  5. Dec 8, 2012 #4

    Dick

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    Science Advisor
    Homework Helper

    If that's your answer for the whole integral, I'll agree with it, at least for the case of a>=0. If you aren't give that, I'd write |a| instead of a. I'll agree it's not a "really clean" answer. But it's about as clean as you are going to get.
     
  6. Dec 8, 2012 #5
    Ya I only need to solve for a>0

    Thanks! It gives me some confidence!
     
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