Does the Gaussian have complex poles?

In summary: So even if there are poles in the upper half of the complex plane, they would not cause the integral to be zero in the limit.
  • #1
mjordan2nd
177
1
I know that

[tex]
\int_{- \infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}.
[/tex]

What if I tried evaluating this over the complex plane? If I created a contour that ran along the real axis, then circled around from positive to negative infinity this should not change things as along the curve at infinity my complex function should be 0. So using that contour, I should have

[tex]
\oint_C e^{-z^2} dz = 0,
[/tex]

unless there are some poles in the upper half of the complex plane. But if there aren't, wouldn't this imply [itex]\int_{- \infty}^{\infty} e^{-x^2}dx = 0[/itex], since the integral over the other part of my contour is 0? I'm forced to conclude I am missing something here, but can't quite put my finger on what that might be.
 
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  • #2
So, your closed contour consists of a piece along the real axis, say from -R to R, as R tends to infinity, and then a circular contour of radius R. This circular arc part of the integral is

$$\int_0^\pi d\theta Ri e^{i\theta} \exp\left(-R^2(\cos(2\theta)+i \sin(2\theta))\right).$$

This integral is not zero, and your contour calculation shows that it must be ##-\sqrt{\pi}## as ##R \rightarrow \infty##. The reason this not zero in the limit as ##R## goes to infinity is that ##\cos(2\theta)## is not always positive on ##[0,\pi)##, so the ##\exp(-R^2\cos(2\theta))## does not drive the integral to zero.

Note: it is actually pretty tough to show that ##\int_{-\infty}^\infty dx \exp(-x^2) = \sqrt{\pi}## via contour integral methods. The only example that I've seen which actually accomplishes it had a rather non-trivial integrand.
 
  • #3
Really I think that would be a sufficient proof that:

[tex]\lim_{R\to\infty} \int_0^{\pi} e^{-z^2}dz=-\sqrt{\pi},\quad z=Re^{it}[/tex]

However, it is based on the fact that the function [itex]e^{-z^2}[/itex] is entire and therefore has no poles in the finite complex plane.
 

1. What is the Gaussian distribution?

The Gaussian distribution, also known as the normal distribution, is a probability distribution that is commonly used in statistics. It is a bell-shaped curve that is symmetrical and centered around the mean value, with the majority of data points falling within one standard deviation of the mean.

2. What are complex poles?

In mathematics, a complex pole is a point on the complex plane where a function is undefined or has an infinite value. In the context of the Gaussian distribution, complex poles refer to the points on the complex plane where the probability density function approaches infinity.

3. Can the Gaussian distribution have complex poles?

Yes, the Gaussian distribution can have complex poles. This occurs when the variance of the distribution is equal to zero, resulting in a singularity at the mean value. In this case, the probability density function will have a pole at the mean value on the complex plane.

4. What is the significance of complex poles in the Gaussian distribution?

Complex poles in the Gaussian distribution can affect the shape of the probability density function, causing it to deviate from the traditional bell-shaped curve. This can impact the accuracy of statistical analyses and predictions based on the Gaussian distribution.

5. How can complex poles in the Gaussian distribution be handled?

In most cases, complex poles in the Gaussian distribution can be avoided by ensuring that the variance is not equal to zero. However, if complex poles do occur, they can be addressed through techniques such as regularization, which adds a small constant to the variance to prevent it from being zero.

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