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Cauchy trick for divergent integrals.

  1. Oct 10, 2009 #1
    is this trick valid at least in the 'regularization' sense ?? for example

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

    then we replace thi integral above by [tex] \int_{-\infty}^{\infty} \frac{dx}{x^{2}+ie-a^{2}} [/tex] for 'e' tending to 0

    using Cauchy residue theorem i get [tex] \int_{-\infty}^{\infty} \frac{dx}{x^{2}+ie-a^{2}}=(i\pi )/a [/tex]

    the same trick applied to [tex] \int_{-\infty}^{\infty} \frac{dx}{x+a}= 2\pi i [/tex]

    however we have obtained a complex regularization to a Real valued integral, i have seen this trick whenever dealing integrals in Quantum Physics but i want to hear the opinion of a Mathematician.
     
  2. jcsd
  3. Oct 10, 2009 #2
    My opinion: what you wrote is worthless as mathematics. Divergent integrals (as in these two examples) should not be said to "have a value".
     
  4. Oct 12, 2009 #3
    I am no mathematician, but when you "complex regularize" (whatever that means) I think you forgot to change your domain of integration. Probably that is why this result is finite.
     
  5. Oct 12, 2009 #4

    Landau

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  6. Oct 12, 2009 #5
    The Cauchy principal value for a the integral of a real function is (if it exists) real. But of course, it doesn't obey the usual rules for change of variables in an integral.

    [tex]
    {\rm P.V.}\int_{-\infty}^{+\infty} \frac{dx}{x^2-a^2} =
    \lim _{u\rightarrow 0^{+}} \left( \int _{a+u}^{\infty }\! \left( {x}^{
    2}-{a}^{2} \right) ^{-1}{dx}+\int _{-a+u}^{a-u}\! \left( {x}^{2}-{a}^{
    2} \right) ^{-1}{dx}+\int _{-\infty }^{-a-u}\! \left( {x}^{2}-{a}^{2}
    \right) ^{-1}{dx} \right) =0
    [/tex]

    The second example is improper (and divergent) both at [tex]a[/tex] and at [tex]\pm\infty[/tex]. Principal value computation yields zero also in that case.
     
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