Characteristic Function of Gaussian Distribution

Theorem
The characteristic function of the normal distribution with mean $\mu$ and variance $\sigma^2$ is


 * $\phi(t) = e^{ i t \mu - \frac{1}{2} t^2 \sigma^2 }$

Statement
Let


 * $k = \mu + it\sigma^2$
 * $c = e^{ \mu it - \frac{1}{2}t^2\sigma^2 }$

Then

Proof
The characteristic function is defined as

Begin by verifying that

We can then simplify the integral in (1):

Lemma 1

Proof
Let $\Gamma \subset \C$ be the rectangular contour with corners


 * $c_1 = \frac{-\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} }$
 * $c_2 = \frac{\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} }$
 * $c_3 = \frac{\alpha-\mu}{\sqrt{2}\sigma}$
 * $c_4 = \frac{-\alpha-\mu}{\sqrt{2}\sigma}$

Since $e^{-z^2}$ is holomorphic everywhere in the region bounded by $\Gamma$, the Cauchy Integral Theorem states that

We now evaluate each linear contour integral in the limit as $\alpha$ goes to infinity:

Between $c_4$ and $c_1$
By similar argument, the integral between $c_4$ and $c_1$ is also 0.

Between $c_1$ and $c_2$
and therefore

Lemma 2

Proof
By Lemma 1


 * $\phi(t) = c \frac{1}{ \sqrt{2\pi \sigma^2} } \displaystyle \int_{x \in \R} e^{ - \left( \frac{ x - k } {\sqrt{2}\sigma } \right)^2} \d x$

Let $z = \left( \frac{ x - k } {\sqrt{2}\sigma } \right)$, then