Cauchy trick for divergent integrals.

1. Oct 10, 2009

zetafunction

is this trick valid at least in the 'regularization' sense ?? for example

$$\int_{-\infty}^{\infty} \frac{dx}{x^{2}-a^{2}}$$

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

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

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

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. Oct 10, 2009

g_edgar

My opinion: what you wrote is worthless as mathematics. Divergent integrals (as in these two examples) should not be said to "have a value".

3. Oct 12, 2009

trambolin

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.

4. Oct 12, 2009

Landau

5. Oct 12, 2009

g_edgar

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.

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

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

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