Solving Complex Gaussian Integral Problems

[Pacopag]

Homework Statement

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

Does this hold even if a is complex?

Homework Equations

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.

 

[benorin]

It holds for Re(a)>0.

 

[tiny-tim]

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.

 

[Pacopag]

Great. Thank you for your replies.

 

[Pacopag]

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?