TITLE: Characteristic function of a standard normal random variable
QUESTION [29 upvotes]: The characteristic function of a random variable $X$ is given by
$$\Phi_X(\omega) = \mathbb{E}e^{j\omega X}=\int_{-\infty}^\infty e^{j\omega x}f_X(x) dx.$$
One can easily capture the similarity between this integral and the Fourier transform.
For a standard normal random variable, the characteristic function can be found as follows:
$$\Phi_X(\omega)=
\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}e^{j\omega x} dx = 
\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{(x^2-2j\omega x)}{2}\right)dx
$$
I know that the answer must be $\Phi_X(\omega) = \exp(-\omega^2/2)$, but can you explain how to evaluate the integral with a complex number in the exponent?

REPLY [3 votes]: There is an alternative approach which is based on the relation between moment-generating function and characteristic function. More accurately, assume that $X$ is a random variable such that there exists $\delta\in(0,\infty]$ for which
$$M_X(t)\equiv Ee^{tX}<\infty\ , \ \forall t\in(-\delta,\delta)\,.$$
Then, it is known (e.g., see my answer to How to prove: Moment Generating Function Uniqueness Theorem) that
$$\phi_X(z)\equiv Ee^{zX}\ \ , \ \ \forall z\in \Omega\equiv\left\{x+iy\ ;\ x\in(-\delta,\delta)\right\}$$
is an analytic continuation of $M_X(\cdot)$ to the domain $\Omega$. This implies that for every $t\in\mathbb{R}$
$$M_X(t)=\phi_X(t)$$ 
and
$$\psi_X(t)\equiv Ee^{itX}=\phi_X(it)\,.$$ 
Now, it is straightforward that for every $z\in\Omega$ it is possible to derive an expresssion for $\phi_X(iz)$ by computing a formula for $\phi_X(z)$ and then insert to this formula $iz$ instead of $z$. Thus, as a special case, in order to derive an expression for $\psi_X(t)$ at some point $t\in(-\delta,\delta)$, it is possible to compute $M_X(t)$ and then to insert $it$ instead of $t$ into the resulting expression.
In particular, if $X\sim N(0,1)$, then it is straightforward to show that for every $t\in\mathbb{R}$
$$M_X(t)=e^{\frac{t^2}{2}}<\infty$$ (that is $\delta=\infty$) and hence, replacing $t$ by $it$ implies that
$$\psi_X(t)= e^{-\frac{t^2}{2}}\ , \ \forall t\in\mathbb{R}\,.$$