How should I calculate $\displaystyle\int_{-\infty}^\infty\exp\left\{-\frac{1}{2}(x-it)^2\right\}dx$? I've read that the residue theorem would help to calculate $$I:=\displaystyle\int_{-\infty}^\infty\underbrace{\exp\left\{-\frac{1}{2}(x-it)^2\right\}}_{=:f(x)}dx$$ Since $f$ is an entire function $\oint_\gamma f=0$ for every closed path $\gamma$ (Cauchy's integral theorem would also yield this information).
So, of course, we can consider the composition $\gamma$ of two paths $\gamma_1(s):=s$, $s\in[-R,R]$, and $\gamma_2(s):=Re^{is}$, $s\in[0,2\pi]$, for some $R>0$. It holds $$0=\oint_\gamma f(z)dz=\int_{-R}^Rf(s)\;ds+iR\int_0^{2\pi}\exp\left\{-\frac{1}{2}(Re^{is}-it)^2+is\right\}ds$$
If we could calculate the second integral, we could calculate $I$. However, it doesn't seem to be helpful to consider that integral instead of the original one.
So, do I miss anything? What's the easiest way to obtain $I$?

if $f(x,t) = \exp \left\{-\frac{1}{2}(x-it)^2\right\} $ then 
$$
\frac{\partial f}{\partial t} = -i \frac{\partial f}{\partial x}
$$
so
$$
\frac{\partial }{\partial t} \int_{-\infty}^{\infty} f dx =  \int_{-\infty}^{\infty} \frac{\partial f }{\partial t} dx = -i \int_{-\infty}^{\infty} \frac{\partial f }{\partial x} dx = \left[-if \right]_{-\infty}^{\infty} =0
$$
so we may set $t=0$ giving a well-known form
