What is the Gaussian Integral for negative $n$ exponents?

Question

If I have a Gaussian integral of the form $$\int_{0}^{\infty}x^{-2}e^{-ax^2}dx, a>0$$ do I use the expression $$\int_{0}^{\infty} x^{n}e^{-ax^2} dx=\frac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\frac{\pi}{a}},~~n=2k, k\in\mathbb Z, a>0$$ or is there another expression to be used for negative exponents? I am asking because this is the formula that I know, but when I evaluate the integral using Wolfram Alpha, it tells me that the integral does not converge, whereas, with the formula above, I get some finite value.

Thanks.

$x=0$ is a problem because it's a non-removable discontinuity — user170231, May 29 '20 at 18:57
Your expression is valid only for $n>0$. Regardless, $$\int\limits_0^\infty \frac{e^{- a x^2}}{x^2}\ dx$$ does not converge. — David G. Stork, May 29 '20 at 19:00

score 0 · Accepted Answer · answered May 30 '20 at 06:48

0

$$I(\epsilon)=\int_\epsilon^\infty\frac{e^{-a x^2}}{x^2}\,dx=\frac{e^{-a \epsilon ^2}}{\epsilon }-\sqrt{\pi a} \, \text{erfc}\left(\epsilon\sqrt{a} \right)$$ I suppose that you see what happens to the first term when $\epsilon \to 0$.

answered May 30 '20 at 06:48

Claude Leibovici

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What is the Gaussian Integral for negative $n$ exponents?

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