# Gaussian integral w/ imaginary coeff. in the exponential

So I've seen this type of integral solved. Specifically, if we have

∫e-i(Ax2 + Bx)dx then apparently you can perform this integral in the same way you would a gaussian integral, completing the square etc. I noticed on wikipedia it says doing this is valid when "A" has a positive imaginary part. I can see why that might be important, we would then have an overall decaying magnitude of the integrand, as opposed to purely oscillating.

What if A has no imaginary part, how would one go about doing such an integral?

## Answers and Replies

Ben Niehoff
Science Advisor
Gold Member
If A is real, then the integral does not strictly converge. However, sometimes you can get a finite answer by replacing ##A \to A + i \varepsilon## and then take ##\varepsilon \to 0## at the end of the calculation.

What you're doing here is interchanging the order of limits, which is not really allowed. The answer you get, although finite, might not be unique (i.e., there are many sequences of converging integrals that approach your diverging one, and we've only looked at one such sequence).

Svein
Science Advisor
You have not said anything about the curve you are integrating along or whether x is real or complex. For example: If x is complex and you integrate over a circle anywhere in the complex plane, the answer is 0 (since the integrand is analytic).