# Gaussian Integral

1. Nov 29, 2005

### bigplanet401

Evaluate:
$$\frac{1}{\sqrt{2\pi} \sigma} \int_{-\infty}^{\infty} \, dx \, exp\left[-\frac{(x - \mu)^2}{2\sigma^2}\right] \, ,$$
where $$\mu$$ and $$\sigma$$ are complex numbers.
I tried writing
\begin{align} \sigma &= s_1 + is_2 \,\\ \mu &= m_1 + i m_2 \, . \end{align}
The integral turned into
$$\int_{-\infty}^{\infty} \, dx \, e^{x(A + iB)} e^C \, ,$$
where A, B and C are constants. But then things got dark.

2. Nov 29, 2005

### saltydog

Convergence appears to me to be dependent upon the value of sigma as this will determine the sign of the exponential coefficient in the Euler expansion of the integrand. One would think that when it is less than zero, the integral converges but diverges when the exponent is greater than or equal zero. That is, when is:

$$Re[-\frac{1}{2\sigma^2}]< 0;\quad \sigma\in \mathbb{C}$$

and when is it greater than or equal to zero?

3. Nov 30, 2005

### saltydog

You guys mind if I run with this? It's very interesting and I need the practice too as I'm just learning Complex Analysis. Seems to me the integral can be characterized by the attached plot of the complex-sigma plane (mu has no effect on the value of the integral). That is, the integral has one value in red, another value in blue and something else in clear.

Bigplanet, would you kindly provide a proof or contradiction of such.

Perhaps someone in here more qualified than me could comment as well.

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4. Nov 30, 2005

### bigplanet401

Hi Salty,
I took another crack at the integral last night and got a closed-form result by using the prescription (1) and (2), and completing the square:
$$\text{(Integral)} = \frac{1}{\sqrt{2\pi} \sigma} \times \sqrt{2\pi}\frac{| \sigma |^2}{\sigma^*} = 1 .$$
(Normalization is preserved even though $$\mu$$ and $$\sigma$$ live in $$\mathbb{C}$$).

Last edited: Nov 30, 2005
5. Nov 30, 2005

### saltydog

Ok Bigplanet. It's not happening for me then. Thanks for replying.