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Complex analysis definite integral involving cosine

  1. Apr 14, 2009 #1
    1. The problem statement, all variables and given/known data
    integral 1/(a+cos(t))^2 from 0 to pi.


    2. Relevant equations

    cos(t)=1/2(e^it+e^-it)
    z=e^it
    dz/(ie^it)=dt


    3. The attempt at a solution

    int dt/(a+cos(t))^2 = int dz/iz(a2+az+az-1+z2/4 +1/2 +z-2/4)

    so with these types of problems I normally can factor this guy some how and get a nice looking quadratic to find the roots and calculate the residue and I'm done. I don't know what to do with this thing in the denominator how can I find the poles of this guy?
     
  2. jcsd
  3. Apr 14, 2009 #2

    Dick

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    That doesn't smell like a contour integral. I don't see a closed contour in there. It looks like a job for a u=tan(t/2) substitution. If the integrand were 1/(a+cos(t)^2), I might think differently.
     
    Last edited: Apr 14, 2009
  4. Apr 16, 2009 #3
    I think this is ok guys:

    [tex]\int_0^{\pi} \frac{1}{(a+\cos(t))^2}=-\frac{i}{2}\mathop\oint\limits_{|z|=1}\frac{4z}{(z^2+2za+1)^2}dz;\quad |a|>1[/tex]

    ok, you can factor the denominator and get:

    [tex]\frac{4z}{(z-z_0)^2(z-z_1)^2}[/tex]

    That immediately tells you poles of second order. Solve for the poles, only one is inside the contour, need to calculate the residue for that pole and you should get after some messy algebra:

    [tex]\int_0^{\pi} \frac{1}{(a+\cos(t))^2}=\frac{\pi a}{(a^2-1)^{3/2}}[/tex]
     
  5. Apr 16, 2009 #4
    if |a|>1 how can you have a pole?

    [itex]\frac{1}{a+cos(t)}[/itex] will be holomorphic on all of [itex]\mathbb{C}[/itex] and Cauchy's theorem will give it integrating to 0 will it not?
     
  6. Apr 16, 2009 #5
    squidsoft, thanks, that factorization is what I was really looking for, it's lovely.

    latentcorpse, that may be true, but when you make the change of variable then integrate over the unit circle one pole will be in the unit circle. I think the pole was -a+(a^2-1)^1/2 or something like that.

    Dick, I'm not about to break out the old single variable calculus book, but does your substitution work? I sincerely doubt it after scribbling out what d/du looks like.
     
  7. Apr 16, 2009 #6

    Dick

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