# Gaussian integral

## Homework Statement

We know that
$$\int_{-\infty}^\infty e^{-ax^2}dx = \sqrt{\pi \over a}$$.

Does this hold even if $$a$$ is complex?

## The Attempt at a Solution

In the derivation of the above equation, I don't see any reason why we must assume that $$a$$ be real. So I think it does hold for complex $$a$$.

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benorin
Homework Helper
It holds for Re(a)>0.

tiny-tim
Homework Helper
It holds for Re(a)>0.
Yes … it relies on $$e^{-ax^2}\arrowvert_{\infty}\,=\,0$$ .

If a = b + ic, then $$e^{-ax^2}=\,e^{-bx^2}e^{-icx^2}$$ , which is 0 if b > 0, and really wobbly if b ≤ 0.

Great. Thank you for your replies.

Hi again. I'd just like to make a remark that is bothering me. I created this thread because I was trying to find a Green function and arrived at
$$G(x,x',t) = {1\over{2\pi}}e^{im(x-x')\over{2t}}\int_{-\infty}^\infty e^{-{it\over{2m}}\left(p-{m(x-x')\over t}\right)^2}$$.
Please ignore everything except the integral.
When I naiively use the gaussian integral formula in my original post, I get the correct answer. But according to what you said, I should not be able to do this since (i.e. Re(a)<0) in this case. Can you see any reason for this?