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Integral with Cauchy Prinicpal value

  1. Jul 23, 2012 #1
    Hi everyone,

    I would like to calculate the following integral:
    [itex]P\int_0^{\pi}\frac{1}{cos(x)-a}dx[/itex], with [itex]
    |a|\leq 1[/itex].
    The 'P' in front stands for the so-called Cauchy Principle value.
    Whenever a is not in the specified domain, the integrand does not have a pole and one can do the integration immediately (I have Maple computed it for me).
    However, when a is in the above domain, the denominator can become zero and one has to integrate through the pole, hence the P.
    But, I don't know how to do this in practice.
    They say just cut out a piece of size delta before and after the pole and just do the integration. Than take the proper limits.
    But, can the same rules be applied as wouldnt a have been in this domain. Is it allowed to use the same techniques as without any poles in the integrand?
    Otherwise, are there any other techniques, tricks ... ?

    Many thanks in advance.
     
  2. jcsd
  3. Jul 23, 2012 #2
    For starters,

    [tex]\int_0^{\pi} \frac{d\theta}{\cos(\theta)-a}=1/2 \int_{-\pi}^{\pi}\frac{d\theta}{\cos(\theta)-a}[/tex]

    and for [itex]|a|\leq 1[/itex] that always has a pole when a is real. Now what happens if we make the normal substitution with these types of integrals: [itex]z=e^{i\theta}[/itex]? Do that and you'll get a closed-contour integral for [itex]|z|=1[/itex] with the pole now always on the contour when a is real. But we can indent around the pole rather than going through it. In that case, the integral over this contour is the part not including the indentation, that's the principal value and the contribution from the indentation depending on which direction you go around it. You can calculate the integral over the entire contour (principal value+indentation) by using the Residue Theorem and you can calculate the part over the indentation by letting the radius of that indentation go to zero and using a theorem which describes the value of an integral over such a limited indentation when it's a simple pole. Get them two, bingo-bango, the principal value is left.
     
  4. Jul 24, 2012 #3
    When you do the typical substitution one gets
    [tex]\frac{2}{i}\int_{C_1}\frac{dz}{z^2-2za+1}[/tex]
    which results in 2 complex poles (a is real)
    [tex]z_\pm=a\pm i\sqrt{1-a^2}[/tex]
    which are indeed on the contour.
    One could indeed avoid the poles by for example going around them in a small circle but with the poles in the contour.
    The residue theorem than says that the contour integral is [itex]2 \pi i[/itex] times the sum of the residues. In this case adding the residus gives zero.
    So, what I have is
    [tex]\frac{2}{i}\int_{C_1}\frac{dz}{z^2-2za+1}=0=P\frac{2}{i}\int_{0}^{2\pi}\frac{dz}{z^2-2za+1}+\frac{2}{i}\int_{small circles}\frac{dz}{z^2-2za+1}[/tex]
    For the small circles (I have now 2 of them because I have 2 complex poles on the contour), I would parametrize them as follows (for the pole [itex]z_+[/itex])
    [tex]z=z_++Re^{i\phi}, dz=iRe^{i\phi}d\phi[/tex]
    and than try to do the integration and take the limit [itex]R\to 0[/itex]
    Is this the right track or not at all ... ? (I did not convince myself to this point really).

    Anyway, as a remark, the result should be real. Is this always for a principal value integral or is it just in this case because we are integrating a real function over a real domain?
     
  5. Jul 24, 2012 #4
    When it's a simple pole and you let the radius of the indentation, [itex]\rho[/itex], go to zero, you can use the theorem which states:

    [tex]\lim_{\rho\to 0} \mathop\int\limits_{\text{indentation}} fdz=\gamma i r[/tex]

    where [itex]\gamma[/itex] is the radian measure of the indentation and r is the residue of the pole. In your case, gamma is pi right. Also, principal value integrals can be real or complex. It's just real in this case because the integral is real.
     
  6. Jul 24, 2012 #5
    As mentioned before: I have now 2 complex poles.
    When I take pi x i x Residu in both and add them I got zero, meaning the Principle value would be zero ...
    That is just twice the same calculation, for the integration along the whole contour and for the poles on the contour. This seems strange to me ...

    I did know that poles lying on the contour count for one half of a pole inside the contour.

    Also, parametrizing such little circles seems fine to me as long as the pole is indeed on the real axis.
    but now, my 2 poles!, are complex ...?
     
  7. Jul 24, 2012 #6
    What's wrong with zero? What happens if you code the original principal-valued integral into Mathematica? For example, using a=9/10:

    Code (Text):

    In[17]:=
    NIntegrate[1/(Cos[z] - 9/10), {z, 0, ArcCos[9/10],
       Pi}, Method -> "PrincipalValue"]

    Out[17]=
    -1.1746159600534156*^-13
     
     
  8. Jul 24, 2012 #7
    There is apparently nothing wrong with zero,
    but to me it is more logical to use small circles on a straight line (e.g. for integrating functions over the whole real axis) than to parametrize them themselve on a circle, which is our contour in this case.

    Many thanks!
     
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