Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cauchy theorem for integrals

  1. May 22, 2009 #1
    i want to perform the following integrals

    [tex] \int_{-\infty}^{\infty}dx \frac{f(x)}{x^{2}-a^{2}} [/tex]

    the problem is that the integral has poles at x=a and x=-a , could we apply

    i think this is the definition of Hadamard finite part integral, performing an integral with singularities by means of Cauchy's theorem.
  2. jcsd
  3. May 22, 2009 #2
    You can compute the Principal Part of the Integral. If you can apply the Residue therorem, then you find that the residues at x = a and x = -a count for half. This is known as Plemelj's formula.
  4. May 22, 2009 #3


    User Avatar
    Science Advisor

    Cauchy's theorem only applies to integrals around a closed path in the complex plane. That path also cannot include any singularities of the function. You could do this by integrating along the real axis, from -R to R, with small half circles, radius [itex]\epsilon[/itex] around -a and a in the upper half plane, then by a half circle from R to -R. Since the function is analytic inside that path, Cauchy's theorem gives 0 for the integral around the entire path. I think it would be easy to show that the limit for the upper half circle, as R goes to infinity, is 0. Thus the problem reduces to determining the integral around those small half circles around -a and a.
  5. May 22, 2009 #4
    yes of course but since there are real poles at x=a and x=-a then the integral would be infinite , this is a singular integral, from the beginnig and this can not be 'regularized' to give a finite real value isn't it ??
  6. May 22, 2009 #5
    I think it depends on what [tex]f(x)[/tex] is. For example, compute:

    [tex]P.V.\int_{-\infty}^{\infty} \frac{f(x)}{x^2-a^2}dx[/tex]


  7. May 22, 2009 #6
    nope due to the factor [tex] 2\pi i [/tex] the integral will be an imaginary number.
  8. May 22, 2009 #7

    Which is reasonable. As you move from one side of the 1/(x-a) and
    1/(x+a) singularities at x = a and -a to the other side, the argument of the integrals in the neighborhood of the singularities ( Log(x+a) and Log(x-a) ) pick up a factor exp(i pi).
  9. May 22, 2009 #8
    Hi. Here's what I calculated for two integrals:

    [tex]P.V. \int_{-\infty}^{\infty}\frac{x}{(x-1)(x+1)}dx=\pi i-\pi i=0[/tex]


    [tex]P.V. \int_{-\infty}^{\infty}\frac{1}{(x-i)(x-1)(x+1)}dx=-\pi i+\pi i(1/2)=-\pi i/2[/tex]
    Last edited: May 22, 2009
  10. May 24, 2009 #9


    User Avatar
    Homework Helper

    If f(x) is a real function then your integral will be real. Any [itex]2\pi i[/itex] terms that appear will necessarily be eliminated, either by some other factor of i appearing to cancel that i or the integral being zero.

    Since the poles of the integral are simple poles, you can apply the half residue theorem, and so

    [tex]P.V.\int_{-\infty}^{\infty}dx~\frac{f(x)}{(x-a)(x+a)} = \pi i\left[f(-a) + f(a)\right] [/tex]

    AS LONG AS f(z), where z = x + iy, is analytic within the upper half plane, AND AS LONG AS [itex]|f(Re^{i\theta})| < R[/itex] as [itex]R \rightarrow \infty[/itex], otherwise the contribution from the large arc will not go to zero. Note that the analyticity condition must be taken into account for the second integral squidsoft did, which has a pole at z = i, which is inside the contour. Accordingly, looking at squidsoft's results,

    [tex]P.V. \int_{-\infty}^{\infty}\frac{x}{(x-1)(x+1)}dx=\pi i-\pi i=0[/tex]

    is correct, but

    [tex]P.V. \int_{-\infty}^{\infty}\frac{1}{(x-i)(x-1)(x+1)}dx=-\pi i+\pi i(1/2)=-\pi i/2[/tex]

    is not quite what I seem to get. I find

    [tex]\int_\gamma dz~\frac{1}{(z-i)(z-1)(z+1)} = \frac{2\pi i}{((i)^2-1)} = P.V.\int_{-\infty}^{\infty}dx~\frac{1}{(x-i)(x-1)(x+1)} - \pi i \left[\frac{1}{2(1-i)} + \frac{1}{2(1+i)} \right][/tex]

    where the "half residue terms" on the RHS came from the small arcs, traversed clockwise, hence the minus sign. Solving,

    [tex]P.V.\int_{-\infty}^{\infty}dx~\frac{1}{(x-i)(x-1)(x+1)} = -\frac{3\pi i}{4}[/tex]

    The discrepancy appears to be accidentally dropping a factor of 1/2 on the half residue terms.
    Last edited: May 24, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook