# Gaussian Integral Using Residues

## Main Question or Discussion Point

I am trying to to the Gaussian integral using contour integration.

What terrible mistake have I made.

$$I = \int_{-\infty}^\infty \mathrm{e}^{-x^2} \mathrm{d}x$$

I consider the following integral:

$$\int_C \mathrm{e}^{-z^2} \mathrm{d}z$$

where $C$ is the half-circle (of infinite radius) in the upper-half plane.
There are no singularities in the upper-half plane. So,

$$\int_C \mathrm{e}^{-z^2} \mathrm{d}z = 0$$

Now, I can break this integral up into two parts.

$$0 = \int_C \mathrm{e}^{-z^2} \mathrm{d}z = \int_{-\infty}^\infty \mathrm{e}^{-x^2} \mathrm{d}x + \int_{\theta=0}^{\theta=\pi} \mathrm{e}^{-z^2} \mathrm{d}z$$

Or....

$$\int_{-\infty}^\infty \mathrm{e}^{-x^2} \mathrm{d}x &= - \int_{\theta=0}^{\theta=\pi} \mathrm{e}^{-z^2} \mathrm{d}z = -\lim_{R\rightarrow \infty} \int_{0}^{\pi} \mathrm{e}^{-R^2 \exp(2\mathrm{i}\theta)}\cdot \mathrm{i} R \mathrm{e}^{\mathrm{i}\theta} \mathrm{d}\theta \stackrel{?}{=} \sqrt{\pi}$$

I know the answer should be $\sqrt{\pi}$...so the RHS must give this result! But it does not...at least, I have convinced myself that the RHS actually goes to zero. The exponential dominates the linear term, so in the limit as R goes to infinity, the RHS equals zero.

How can this be...it seems I have shown that the gaussian integral is zero.

I should cry.

lurflurf
Homework Helper
The terrible mistake have you made is the Gaussian diverges to infinity on the arc. You could show that all the integrals of the Gaussian from 0 to points r*exp(i theta) are equal when sin(theta)<=cos(theta). This won't answer any thing though as you would still need to calculate one such integral. I feel that this integral is not best done by contour integration (though it is possible and I have seen it done). There are other easier ways.

lurflurf said:
The terrible mistake have you made is the Gaussian diverges to infinity on the arc.
...which becomes painfully obvious when I include the effect of the oscillations. It is true that R->infinity makes exp(-R^2 exp(2*I*theta)) very large...but the oscillation causes it to go back and forth between +infinity and -infinity.

It seems that no matter which contour I pick (as far as infinite semi-circles go), I will have this issue. At iR and -iR, e^(-z^2) will have trouble. Thus, it seems that my contour CANNOT be anything that goes to infinity in the imaginary part.

I know that the Guassian integral can be done easily in polar coordinates...that doesn't concern me. I _want_ to do it when complex integration. So how do you do it?

lurflurf
Homework Helper