=== Lemma 1 === Let: :$k = \mu + i t \sigma^2$ :$c = e^{\mu i t - \frac 1 2 t^2 \sigma^2}$ Then: :$\map \phi t = c \dfrac 1 {\sqrt {2 \pi \sigma^2} } \displaystyle \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x$ === Proof === The characteristic function is defined as {{begin-eqn}} {{eqn | l = \map \phi t       | r = \expect {e^{i t X} }       | c = where $\expect {\, \cdot \,}$ denotes expectation }} {{eqn | r = \int_{x \mathop \in \R} e^{i t x} \frac 1 {\sqrt {2 \pi \sigma^2} } e^{-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x }} {{eqn | r = \frac 1 {\sqrt {2 \pi \sigma^2} } \int_{x \mathop \in \R} e^{i t x} e^{-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x }} {{eqn | r = \frac 1 {\sqrt {2 \pi \sigma^2} } \int_{x \mathop \in \R} e^{i t x - \frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x       | n = 1 }} {{end-eqn}} 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 $(1)$: {{begin-eqn}} {{eqn | l = \int_{x \mathop \in \R} e^{i t x - \frac {\paren {x - \mu}^2} {2\sigma^2} } \rd x       | r = \int_{x \mathop \in \R} e^{-\frac {\paren {x - \mu}^2 + 2 \mu i t \sigma^2 - t^2 \sigma^4} {2 \sigma^2} } \rd x }} {{eqn | r = e^{\frac {2 \mu i t \sigma^2 - t^2 \sigma^4} {2 \sigma^2} } \int_{x \mathop \in \R} e^{-\frac {\paren {x - k}^2} {2 \sigma^2} } \rd x }} {{eqn | r = e^{\mu i t - \frac 1 2 t^2 \sigma^2} \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x }} {{eqn | r = c \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x }} {{end-eqn}} {{qed|lemma}} === Lemma 2 === :$\displaystyle \lim_{\alpha \mathop \to \infty} \int_{\frac{-\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} }^{\frac {\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} } e^{-z^2} \rd z = \sqrt {2 \pi \sigma^2}$ === Proof === Let $\Gamma \subset \C$ be the rectangular contour with corners: {{begin-eqn}} {{eqn | l = c_1       | r = \frac {-\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} }} {{eqn | l = c_2       | r = \frac {\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} }} {{eqn | l = c_3       | r = \frac {\alpha - \mu} {\sqrt 2 \sigma} }} {{eqn | l = c_4       | r = \frac {-\alpha - \mu} {\sqrt 2 \sigma} }} {{end-eqn}} Since $e^{-z^2}$ is holomorphic everywhere in the region bounded by $\Gamma$, the Cauchy Integral Theorem states that: {{begin-eqn}} {{eqn | l = \oint_{z \mathop \in \Gamma} e^{-z^2} \rd z       | r = \int_{c_1}^{c_2} e^{-z^2} \rd z + \int_{c_2}^{c_3} e^{-z^2} \rd z + \int_{c_3}^{c_4} e^{-z^2} \rd z + \int_{c_4}^{c_1} e^{-z^2} \rd z }} {{eqn | r = 0 }} {{end-eqn}} We now evaluate each linear contour integral in the limit as $\alpha$ goes to infinity: ==== Between $c_3$ and $c_4$ ==== {{begin-eqn}} {{eqn | l = \lim_{\alpha \mathop \to \infty} \int_{c_3}^{c_4} e^{-z^2} \rd z       | r = \lim_{\alpha \mathop \to \infty} \int_{\frac {\alpha - \mu} {\sqrt 2 \sigma} }^{\frac {-\alpha - \mu} {\sqrt 2 \sigma} } e^{-z^2} \rd z }} {{eqn | r = -\lim_{\alpha \mathop \to \infty} \int_{\frac {-\alpha - \mu} {\sqrt 2 \sigma} }^{\frac {\alpha - \mu} {\sqrt 2 \sigma} } e^{ -z^2 } \rd z }} {{eqn | r = -\sqrt {2 \pi \sigma^2}       | c = Integral of Gaussian Distribution }} {{end-eqn}} ==== Between $c_2$ and $c_3$ ==== {{begin-eqn}} {{eqn | l = \norm {\lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} e^{-z^2} \rd z}       | r = \norm {\lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} e^{-z^2} \rd z} }} {{eqn | o = \le       | r = \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \norm {e^{-z^2} } \rd z }} {{eqn | o = \le       | r = \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \max_{z \mathop \in \closedint {c_2} {c_3} } \norm {e^{-z^2} } \rd z }} {{eqn | o = \le       | r = \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \norm {e^{-\paren {\frac {\alpha - \mu} {\sqrt 2 \sigma} }^2} } \rd z }} {{eqn | o = \le       | r = \lim_{\alpha \rightarrow \infty} \norm {e^{-\paren {\frac {\alpha - \mu} {\sqrt 2 \sigma} }^2} } \int_{c_2}^{c_3} \rd z }} {{eqn | r = 0 }} {{end-eqn}} ==== 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 $c_2$ ==== {{begin-eqn}} {{eqn | l = \oint_{z \mathop \in \Gamma} e^{-z^2} \rd z       | r = \int_{c_1}^{c_2} e^{-z^2} \rd z + \int_{c_2}^{c_3} e^{-z^2} \rd z + \int_{c_3}^{c_4} e^{-z^2} \rd z + \int_{c_4}^{c_1} e^{-z^2} \rd z }} {{eqn | r = \int_{c_1}^{c_2} e^{-z^2} \rd z + 0 - \sqrt {2 \pi \sigma^2} + 0 }} {{eqn | r = 0 }} {{end-eqn}} and therefore: {{begin-eqn}} {{eqn | l = \int_{c_1}^{c_2} e^{-z^2} \rd z       | r = \sqrt {2 \pi \sigma^2} }} {{end-eqn}} {{qed|lemma}} By Lemma 1: :$\map \phi t = c \dfrac 1 {\sqrt {2 \pi \sigma^2} } \displaystyle \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x$ Let $z = \paren {\dfrac {x - k} {\sqrt 2 \sigma} }$. Then: {{begin-eqn}} {{eqn | l = \map \phi t       | r = c \frac 1 {\sqrt {2 \pi \sigma^2} } \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 }} {{eqn | r = c \frac 1 {\sqrt {2 \pi \sigma^2} } \sqrt {2 \pi \sigma^2}       | c = by Lemma 2 }} {{eqn | r = c }} {{eqn | r = e^{\mu i t - \frac 1 2 t^2 \sigma^2} }} {{end-eqn}} {{qed}} sakbvf605dw0zq05fjgh4yip9or3b0x