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Math Help - Residue theorem troubles

  1. #1
    Member Mentia's Avatar
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    Residue theorem troubles

    Hi. I am trying to evaluate the following integral:
    I = \int_{0}^{\infty } \frac{x^{-p}}{1-2x\cos{\theta}+x^{2} }

    where -1 < p < 1 and 0 < \theta < 2*Pi. One can easily show the following:

    1-2x\cos{\theta}+x^{2}=(x-e^{i \theta } )(x - e^{-i \theta })

    When we transfer this over to the complex plane this means we have a branch point at x = 0 and a couple of simple poles at x = e^{i \theta } and x = e^{-i \theta }

    I am attempting to apply the pacman or keyhole contour here where the branch cut is made along the positive real axis.

    I end up with the following:

    A = \oint_{pacman} \frac{z^{-p}}{1-2z\cos{\theta}+z^{2} }dz = \int_{0}^{\infty } \frac{r^{-p}}{1-2r\cos{\theta}+r^{2} }dr - \int_{0}^{\infty } \frac{e^{-i2\pi p}r^{-p}}{1-2r\cos{\theta}+r^{2} }dr = (1 - e^{-i2 \pi p}) I = e^{-i \pi p}2i \sin (\pi p) I

    But by the residue theorem we have:

    A = 2 \pi i \sum Residues = 2 \pi i \frac{- \sin (\theta p)}{\sin (\theta )}

    which implies:

    I = \int_{0}^{\infty } \frac{x^{-p}}{1-2x\cos{\theta}+x^{2} } = \frac {-e^{i \pi p} \pi \sin (\theta p)}{ \sin (\theta ) \sin (\pi p) }

    but this cannot be right since we are integrating a real valued function over the real line... it shouldnt have any imaginary components. What have I done wrong?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    I didn't check all the details, but you probably know that e^{i\pi}=-1?
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  3. #3
    Member Mentia's Avatar
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    yes, but p is a fraction or irrational or zero so making that substitution doesnt help really. Youll have some fractional or irrational root of -1.
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  4. #4
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    I believe you're using the wrong residue for the pole at e^{-i\theta} based on the branch you're using: You define:

    z^{-p}=e^{-p\log(z)}=e^{-p(\ln(r)+i\arg(z))},\quad 0\leq \arg(z)\leq 2\pi

    Then for z=e^{-i\theta}

    \arg(z)=2\pi-\theta

    Also, try it first for p=1/4 and \theta=\pi/4 or something easy like this. Do it numerically first in Mathematica, then do it numerically via the Residue Theorem:

    Code:
    In[56]:=
    p = 1/4; 
    t = Pi/4; 
    NIntegrate[1/(x^p*(1 - 2*x*Cos[t] + 
         x^2)), {x, 0, Infinity}]
    r1 = Exp[-7*Pi*(I/16)]/
        (Exp[(-Pi)*(I/4)] - Exp[Pi*(I/4)]); 
    r2 = Exp[(-Pi)*(I/16)]/(Exp[Pi*(I/4)] - 
         Exp[(-Pi)*(I/4)]); 
    N[((2*Pi*I)/(1 - Exp[-2*Pi*I*p]))*
       (r1 + r2)]
    
    Out[58]=
    3.4907507252185646
    
    Out[61]=
    3.4907507252151033 + 
      3.487868498008632*^-16*I
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  5. #5
    Member Mentia's Avatar
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    Correct correct! My understanding of branch cuts was flawed. The final answer comes out to be:

    \frac {\pi \sin (p (\pi - \theta ))}{\sin (\pi p) \sin (\theta )}
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