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Complex Analysis Integration with Sin and Cos

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Compute the integral from 0 to 2∏ of:

    sin(i*ln(2e^(iθ)))*ie^(iθ)/(8e^(3iθ)-1) dθ

    (Sorry for the mess, I don't know how to use latex)



    2. Relevant equations

    dθ=dz/iz

    sinθ = (z - z^(-1))/2i



    3. The attempt at a solution

    So I tried to change it into a contour integral of a function of z with the substitutions dθ=dz/iz and sinθ = (z - z^(-1))/2i. The dθ part is nice because we have ie^(iθ) in the equation so it cancels out. However, the sin function is not a function of theta but of a complex variable. I tried to reduce the inside of the sin as follows:

    sin(i*ln(2e^iθ)) = sin(i*(ln2 + iθ)) = sin(i*ln2 - θ), but this doesn't really get me anywhere as the sin function is still not a function of theta.

    Am I on the right track? And can anyone point me in the right direction of how to continue from here? Any help is much appreciated.
     
    Last edited: Dec 7, 2011
  2. jcsd
  3. Dec 8, 2011 #2
    Here's the guide:

    https://www.physicsforums.com/showthread.php?t=546968


    Let's first write it as:

    [tex]\int_0^{2\pi}\frac{\sin[i\log(2e^{it})] ie^{it}}{8e^{3it}-1}dt[/tex]

    when you make the change of variables with z=e^it, it becomes (formally):

    [tex]\oint \frac{\sin[i\log(2z)]}{8z^3-1}dz[/tex]

    Now the question becomes is that analytic in the unit disc except for a finite number of (isolated ) singularities? If it is, then you need only use the Residue Theorem to compute the integral. Is it? What happens to the integrand when I allow z to make one full loop around the origin? Does the integrand come back to the same point, or is it at another branch of the integrand?
     
    Last edited: Dec 8, 2011
  4. Dec 8, 2011 #3
    Ok, so after factoring, we can see that there's a simple pole at z = 1/2. Also, f is undefined at z = 0, but analytic at every point in a neighborhood of z = 0. We also have a branch along the positive x axis, with z behaving like 1 on top and e^(2∏i) when approaching on the other side. So then we can integrate along a path starting on the top of the branch at p (which is arbitrarily small, along the branch, counterclockwise around the unit circle until we hit the bottom of the branch, back along the bottom of the branch, and clockwise around Cp until we hit the original point. Sorry if that was explained poorly, but does it sound somewhat correct?
     
  5. Dec 8, 2011 #4
    But then the singularity z = 1/2 is on the branch cut. So we can just make the branch cut on the negative x axis, with z behaving like e^(i*pi) along the top and behaving like e^(3i*pi) on the bottom right?
     
  6. Dec 8, 2011 #5
    Ok, so I have this so far:

    \int_-p^-1 \! [itex]\frac{sin(i[ln(2r)+i∏]}{8r3-1}[/itex] \, \mathrm{d} r + \oint_C f(z) dz - \int_-p^-1 \! [itex]\frac{sin(i[ln(2r)+3i∏]}{8r3-1}[/itex] \, \mathrm{d} r + \oint_Cp f(z) dz = 2∏i*Res f(z) at z = 1/2

    Does this look right? One more thing, what do you do with the contour integral CR? We've only used this with improper integrals to show it converges to zero as R approaches \infinity, but here we're not evaluating an improper integral.
     
  7. Dec 8, 2011 #6
    Crap that didn't work at all. Anyway, it's supposed to be the integral from -p to -1 of sin(i[ln(2r) + i*pi]/(8r^3 - 1) dr + the contour integral of f(z) around the unit circle - the integral from -p to -1 of sin(i[ln(2r) + 3*i*pi]/(8r^3 - 1) dr + the contour integral of f(z) around Cp.
     
  8. Dec 8, 2011 #7
    Ok, this is what I think: even though there is a log term in there, the [itex] i\log(2z)[/itex] term inside of sin eliminates the non-analyticity along a branch-cut of what we'd expect from just [itex]\sin(\log(2z)) [/itex] so that the integrand reduces to a meromorphic function analytic throughout the unit disc except for a set of isolated poles. Now just suppose that's right and if it were a test question, I'd be worried after I turned in my paper, if that's correct, can we compute the residues at those poles?

    But not entirely sure about that analysis ok. Maybe someone else can comment on this.
     
    Last edited: Dec 8, 2011
  9. Dec 8, 2011 #8
    Oh, good point. I'm a little fuzzy on meromorphic functions, I'll re-read that part of the book and see if it provides any insight. Thanks for your help jackmell, your input has been really helpful.
     
  10. Dec 8, 2011 #9
    Then I think you should ask, "Has the apparent (notational) discontinuity been removed by virtue of the intrinsic algebra and geometry of the underlying functions?"

    I'm kinda' struggling with this though . . .
     
  11. Dec 8, 2011 #10
    Yeah, me too. I re-read that section in the book and it wasn't really helpful. Im pretty stumped, I think I'm just going to ask my professor how to solve it and hope there are no problems like this on the final.
     
  12. Dec 8, 2011 #11
    [tex]\frac{\sin(i\log(2z))}{8z^3-1}=\frac{i(4z^2-1)}{4z(8z^3-1)}[/tex]

    (use the exponential version of sin to get that)

    So it's analytic except for four poles and so the Residue Theorem can be applied. When we do that, we get zero.

    Tell you what, that's what I'm putting on the test but that's just me.
     
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