Characteristic Function of Gaussian Distribution/Lemma 2

Lemma for Characteristic Function of Gaussian Distribution

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

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: