Characteristic function of a standard normal random variable 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?
 A: I will give two answers:
Do it without complex numbers, notice that
$$ \begin{eqnarray}
  \mathcal{F}(\omega) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{-\infty}^0 \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x \\
   &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{0}^{\infty} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{-j \omega x} \mathrm{d} x \\
   &=&  2 \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \cos(\omega x) \mathrm{d} x
  \end{eqnarray}
$$
Now, compute $\mathcal{F}^\prime(\omega)$, and integrate by parts:
$$\begin{eqnarray}
   \mathcal{F}^\prime(\omega) &=& -\frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} x \sin(\omega x) \mathrm{d} x = \frac{2}{\sqrt{2\pi}} \int_0^\infty \sin(\omega x) \mathrm{d} \left( \mathrm{e}^{-\frac{x^2}{2}} \right) \\
   &=& \frac{2}{\sqrt{2\pi}} \left. \mathrm{e}^{-\frac{x^2}{2}} \sin(\omega x) \right|_0^\infty - \frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} \omega \cos(\omega x) \mathrm{d} x \\ 
    &=& - \omega \mathcal{F}(\omega) 
  \end{eqnarray}
$$
The solution to so obtained ODE, $\mathcal{F}^\prime(\omega) = - \omega \mathcal{F}(\omega)$ is $\mathcal{F}(\omega) = c \exp\left( - \frac{\omega^2}{2} \right)$, and the integration constant is seen to be one from normalization requirement $\mathcal{F}(0)=1$ of the Gaussian probability density.
Complex integration: As you have started, complete the square: 
$$ \left( -\frac{x^2}{2} + j \omega x \right) = 
\left( -\frac{x^2}{2} + j \omega x + \frac{\omega^2}{2} \right) - \frac{\omega^2}{2} = 
  -\frac{1}{2} \left( x - j \omega \right)^2 - \frac{\omega^2}{2}
$$
We then have:
$$
   \mathcal{F}(\omega) = \mathrm{e}^{-\frac{\omega^2}{2}} \cdot \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x
$$
The integral $I = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x$ is indeed $1$. To see this, consider
$$
\begin{eqnarray}
  I_L &=& \int_{-L}^L \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x = \int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z \\
   &=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z\right) + \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z \\
   &=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z\right) + \mathcal{I}_L
 \end{eqnarray}
$$
Here we denoted $\mathcal{I}_L = \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z$. Notice that $\lim\limits_{L \to \infty} \mathcal{I}_L = 1$.
Consider a complex contour $\mathcal{C}$, $ -L \to L \to L - j \omega \to -L - j \omega \to -L$:
$$
\begin{eqnarray}
  I_L - \mathcal{I}_L &=&\left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z\right) \\ &=& -\int_\mathcal{C} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z
\end{eqnarray}
$$
The integral over $\mathcal{C}$ is zero, since the integrand is holomorphic. Therefore:
$$
   I-1 = \lim_{L \to \infty} (I_L-\mathcal{I}_L) = \lim_{L \to \infty} \left( - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2}  } \mathrm{d} z \right)
$$
And the limit above is easily seen to vanish. Indeed:
$$
 \lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(-L - j \omega t)^2}{2}} \right| = 
 \lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(L^2 + \omega^2 t^2)}{2}} \right| =0.
$$
A: 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
$$\psi_X(t)= e^{-\frac{t^2}{2}}\ , \ \forall t\in\mathbb{R}\,.$$
