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Interesting contour integral

  1. Oct 12, 2011 #1
    For my research I am trying to solve a particular integral. I know that I have to transform it properly to use the residue theorem, but I am having difficulty attaining this form. Here is the integral:

    [itex]\int^{\infty}_{-\infty}\frac{e^{ix}dx}{\sqrt{a - x^2 + ib}}[/itex]

    where a and b are real constants. If you have any suggestions I would be more than grateful.

    Thanks in advance.
     
  2. jcsd
  3. Oct 14, 2011 #2
    Let a+ib =c^2, so that the integrand has poles at +-c.Integrate along a semicircle which contains a part of the real axis and loops around +-c.
     
  4. Oct 15, 2011 #3
    Then [itex]\pm c[/itex] would be branch points, not poles. In this case, the integral could be evaluated using Cauchy's Theorem but not the Residue Theorem. How about doing this first: Just let a+ib=1+i. Then in order to capture the entire real axis, let the branch cuts extend outward towards infinity from each branch-point. So the one in the upper half-plane would be [itex]2^{1/4}e^{\pi i/8}[/itex]. Now, let the branch-cut from that one extend along the line [itex]re^{\pi i/8}[/itex]. Now, close the contour in a semi-circular arc which traverses over that branch-cut and just compute the two terms along the branch-cut. Does it agree with direct numerical integration of the integral?
     
  5. Oct 17, 2011 #4
    I understand what you mean by which lines the cuts are on, but I am having trouble explicitly evaluating the contributions from either side of the cut in the upper half plane.

    For example, along the cut on the right side, x would have the value re[itex]^{i*\pi/8}[/itex]. Therefore x[itex]^{2}[/itex] is r[itex]^{2}[/itex]e[itex]^{i*\pi/4}[/itex]?

    Along the cut on the left side, x would have the value re[itex]^{-i*15\pi/8}[/itex]. Therefore x[itex]^{2}[/itex] is r[itex]^{2}[/itex]e[itex]^{-i*15\pi/4}[/itex]?

    Is this correct?
     
  6. Oct 17, 2011 #5
    The value on one side of a square-root branch-cut is just the negative of the other side. Remember, the multifunctioned square root is [itex]\pm [/itex]. So on the lower side, the integral would be:

    [tex]\int_{\infty}^{2^{1/4}} \frac{e^{iz}}{\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}[/tex]

    and on the other side, it would be:

    [tex]\int_{2^{1/4}}^{\infty} \frac{e^{iz}}{-\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}[/tex]

    and after analyzing all the other parts of the contour and simplifying, we can write:

    [tex]\int_{-\infty}^{\infty} \frac{e^{iz}}{\sqrt{z_0-z^2}}dz=2\int_{\infty}^{2^{1/4}} \frac{e^{iz}}{-\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}[/tex]

    at least for this problem. May have other issues if z0 is on the real axis or may have to modify the analysis somewhat.

    Not sure if that's any easier to evaluate analytically than the original problem. May be or probably is an easier way to evaluate it but it's good practice to do it any way possible to help understand and perhaps lead to a better way of doing it.
     
    Last edited: Oct 17, 2011
  7. Oct 17, 2011 #6
    Thank you very much!
     
  8. Oct 18, 2011 #7
    That's not correct. Remember it's the sum over all the contours and by Cauchy's Theorem, that's zero so I should have written:

    [tex]\int_{-\infty}^{\infty} \frac{e^{iz}}{\sqrt{z_0-z^2}}dz+\left(2\int_{\infty}^{2^{1/4}} \frac{e^{iz}}{-\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}\right)=0[/tex]

    so that:

    [tex]\int_{-\infty}^{\infty} \frac{e^{iz}}{\sqrt{z_0-z^2}}dz=\left(-2\int_{\infty}^{2^{1/4}} \frac{e^{iz}}{-\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}\right)[/tex]

    However when I compute them numerically in Mathematica, there is a sign difference. I suspect that's because Mathematica is integrating over two different branches but I'm not sure.
     
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