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Trig contour integral

  1. Jun 15, 2011 #1
    1. The problem statement, all variables and given/known data

    I have to prove the following: [itex]\int_0^{2\pi} \frac{\mathrm{d}\theta}{(a + cos(\theta))^2} = \frac{2pia}{(a^{2}-1)^{3/2}}[/itex] for a > 1.

    2. Relevant equations

    I have an example at hand for [itex]\int_0^{2\pi} \frac{\mathrm{d}\theta}{a + cos(\theta)}[/itex] from which I know I have to substitute [itex]K = e^{i\theta} \rightarrow \mathrm{d}\theta = \frac{\mathrm{d}K}{iK}[/itex] and [itex]cos(\theta) = \frac{K + K^{-1}}{2}[/itex], use a keyhole contour based on the unit circle and the location of the singularity, use the residue theorem and use the fact that the cosine of a complex angle can apparently be bigger than 1 (otherwise there would be no singularities and the integral would be zero).

    3. The attempt at a solution

    Unlike in the example the denominator doesn't yield a nice quadratic function, instead it yields a cubic function and K^-1 hasn't been eliminated.

    How do I solve this?
     
  2. jcsd
  3. Jun 15, 2011 #2

    micromass

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    Hi Gulli! :smile:

    Just exchange

    [tex]\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}[/tex]

    and theta runs between 0 and 2pi. Thus your integral equals

    [tex]\int_\gamma{\frac{dz}{(a+\frac{1}{2}\left(z+\frac{1}{z}\right))^2}}[/tex]

    where gamma is the unit circle. This can be calculated through the residue theorem.
    I'll leave the details to you.
     
  4. Jun 15, 2011 #3
    Uhm... no, I'm pretty sure it becomes (I use K instead of z to avoid confusion):

    [itex]\int_{|K|=1} \frac{\mathrm{d}K}{iK(a + \frac{1}{2}(K + \frac{1}{K}))^2}[/itex], maybe I can find the singularities like this: [itex]a = - \frac{1}{2}(K + \frac{1}{K}) \rightarrow Ka + \frac{1}{2}K^2 - \frac{1}{2} = 0[/itex]...
     
    Last edited: Jun 15, 2011
  5. Jun 15, 2011 #4

    micromass

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    Yes, yes, of course, that's what I meant :blushing:

    Try to simplify that fraction first, before you search for the singularities.
     
  6. Jun 15, 2011 #5
    Do I have to simplify the fraction? The quadratic equation gives me two singularities, one inside the unit circle, on outside, just as you would expect for a solvable keyhole contour problem. Then again I did implicitly assume K is never zero (which is true for real thetas, but is it true in this context?)
     
  7. Jun 15, 2011 #6

    micromass

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    Yes it's best that you simplify the fraction. Especially if you want to calculate the residue later on. Start by multiplying numerator and denominator by K2, that gives you

    [tex]\frac{K^2}{iK(\frac{1}{2}K^2+aK+\frac{1}{2})^2}[/tex]

    You see easily that 0 will be a singularity. To find the other two, you must factorize the quadratic in the denominator. This will depend on a. (or did you already do this?)
     
  8. Jun 15, 2011 #7

    micromass

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    You're integrating on a circle with |K|=1. So all K's have norm 1. So they are never 0!
     
  9. Jun 15, 2011 #8
    Ah yes, of course, but that's on the unit circle, what about the keyhole?

    I found singularities at [itex]-a - (a^2 +1)^{1/2}[/itex] and [itex]-a + (a^2 +1)^{1/2}[/itex], should I include K = 0 (if this even exists) as well to get a keyhole contour with the keyhole itself at the origin and the positive singularity in the "tube" of the keyhole while the negative singularity falls outside the unit circle and can thus be ignored, so that the integral equals the sum of the residues of the two singularities?
     
    Last edited: Jun 15, 2011
  10. Jun 15, 2011 #9

    micromass

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    I don't see why you need to do something with a keyhole contour now. Can't you just apply the residue theorem to evaluate the integral? Just use the circle |K|=1 as your contour.
     
  11. Jun 15, 2011 #10
    Ok, I see what you mean (res = (1/(i2pi))*integral of f over unit circle), but do I have two singularities (K = 0 and [itex]-a + (a^2 +1)^{1/2}[/itex] or just one?
     
  12. Jun 15, 2011 #11

    micromass

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    Well, you'll need to figure out whether

    [itex]-a+(a^2+1)^{1/2}<1[/itex]

    or not. Remember that a>1.

    We do expect to have only 0 as singularity, because the solution of the integral should look much more complicated otherwise...
     
  13. Jun 15, 2011 #12
    It's definitely between 0 and 1, so it counts as a singularity. It's the K = 0 part I'm not so sure about. K = e^itheta and that's cos(theta) + isin(theta) and that's never zero for any real theta (don't know about complex thetas though, obviously the normal rules don't apply anymore because otherwise a + (cos(theta))^2 could never be zero and the integral would just be zero). I'm just really confused and my professor didn't count K = 0 as a singularity in the example, even though the denominator was of the form K times (stuff).
     
    Last edited: Jun 15, 2011
  14. Jun 15, 2011 #13

    micromass

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    Yes, it is indeed a singularity. And you're indeed correct about worrying about the K=0. It is indeed not a singularity as we can cancel it with the numerator:

    [tex]\frac{K}{i(\frac{1}{2}K^2+aK+\frac{1}{2})^2}[/tex]

    So it appears we only have two singularities!
     
  15. Jun 15, 2011 #14
    You mean the two singularities that follow from the quadratic equation? One of those lies outside the unit circle, so I only have to find the residue coming from the positive root -a+sqrt(a^2 + 1. I'm glad K = 0 cancels out, but that's pure luck and I wonder whether K = 0 actually exists and am still a bit confused as to why there would even be singularities since (a + cos(theta)^2) should never be zero for any real theta. I don't need to know these things to solve this particular problem but they still bug me.
     
    Last edited: Jun 15, 2011
  16. Jun 15, 2011 #15

    micromass

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    Yes, you're correct that (a+cos(theta)^2) is not zero, but that's not the integral wer're dealing with here. We have transformed this real integral into a complex integral. But while doing so, we've had to introduce some singularities.

    OK, now you only need to calculate the residue of the function at -a+sqrt(a^2+1)...
     
  17. Jun 16, 2011 #16
    Yes, now I "only" have to find the limit for [itex]K \rightarrow -a + (a^2 +1)^{1/2}[/itex] of [itex]\frac{K^2 + Ka - K(a^2 +1)^{1/2}}{i(K^{2}/2 + aK +1/2)2}[/itex]. Well, actually I think the denominator becomes i (which is nice because it cancels out the i from i*2*pi, just as it's supposed to), but I get zero for the numerator...
     
    Last edited: Jun 16, 2011
  18. Jun 16, 2011 #17

    micromass

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    You're supposed to get zero for the denominator too, because -a+(a^2+1)^{1/2} was chosen to be a root of the denominator...
     
  19. Jun 16, 2011 #18

    Gib Z

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    We can avoid a lot of mess by dealing with the specifics only at the very end.

    Right now, the problem is to find (with some constant factors out the front) [tex] \Res_{z=a} \frac{z}{(z-a)^2(z-b)^2} = \lim_{z\to a} \frac{d}{dz}\left( \frac{z}{(z-b)^2} \right) = \lim_{z\to a} \frac{ (z-b)^2 - 2z(z-b) }{(z-b)^4} = \frac{ (a-b)^2 - 2a(a-b) }{(a-b)^4}[/tex]

    Now put in the specific values of the roots of the denominator as your a and b. (a-b) should take a relatively simple form because they are complex conjugates.
     
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