Can anyone help with this integral?

Question

I need to evaluate this integral:

IntegralFromMinusInfinityToInfinity( e^(ax^2) )dx

Where a is an IMAGINARY constant.

I initially thought that the Gaussian integral, which gives the result
sqrt(pi / -a) would work, but it is now my understanding that it does not hold for complex or imaginary a.

Can anyone help?
If you could provide a proof, that would be great, but really I only need the answer for a physics calculation that i'm doing.

The definite integral is not impossible to find - it requires advanced techniques of integration, however.
I'm just wondering how having an imaginary constant affects the evaluation of the integral.

Answer 1

If you break it into real and imaginary parts, you get integrals of
cos(x^2) and sin(x^2) from -infty to infty. These integals are finite. The anti-derivatives are called Fresnel integrals and show up in studies of diffraction. The infinite integrals are sqrt(pi/2) each.

Answer 2

The reason that using the gaussian integral for the function doesn't work when a is imaginary is that the necessary substitution changes the limits of integration from real infinity to complex infinity, and the proof of the value of the gaussian integral doesn't work if the integral is free to assume negative values (as it will when being evaluated from complex infinity to complex infinity). I've been trying to figure out a method of transforming it into something that only involves real numbers, but thus far have had no success. I'll edit this answer if I find something.

Answer 3

As far as I've been instructed, the indefinite integral of e^(ax^2) is impossible to find, at least with elementary integration techniques. I've always encountered approximations of definite integrals.

Answer 4

Yeah. This one's kinda nasty because it walks you right into the imaginary error function integral defined as
erfi (z) = -i*erf(iz) (where erf is the error function integral that you're probably already familiar with)

The good news is that erfi is odd (just like erf is). The bad news is that it's entire (which means it's unbounded).

Chances are you can probably find something about it on the wolfram site ( http://mathworld.wolfram.com/ ) which is a good one to know about anyway ☺


Doug