Characteristic Function of Gaussian Distribution

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


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

Lemma 1
Let:


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

Then:


 * $\map \phi t = c \dfrac 1 {\sqrt {2 \pi \sigma^2} } \ds \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x$

Proof
The characteristic function is defined as

Begin by verifying that:


 * $i t x - \dfrac {\paren {x - \mu}^2} {2 \sigma^2} = -\dfrac {\paren {x - k}^2 + 2 \mu i t \sigma^2 - t^2 \sigma^4} {2 \sigma^2}$

We can then simplify the integral in $(1)$:

Lemma 2

 * $\ds \lim_{\alpha \mathop \to \infty} \int_{\frac{-\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} }^{\frac {\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} } e^{-z^2} \rd z = \sqrt {2 \pi \sigma^2}$

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

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 a similar argument as for between $c_2$ and $c_3$, the integral between $c_4$ and $c_1$ is also $0$.

Between $c_1$ and $c_2$
and therefore:

By Lemma 1:


 * $\map \phi t = c \dfrac 1 {\sqrt {2 \pi \sigma^2} } \ds \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x$

Let $z = \paren {\dfrac {x - k} {\sqrt 2 \sigma} }$.

Then: