a generalization of normal distribution to the complex case: complex integral over the real line How to prove $\int_{\mathbb{R}} e^{-\frac{(x+it)^2}{2}}dx=\sqrt{2\pi}$ for any $t\in \mathbb{R}$? I only obtained the case that $t=0$, $\int_{\mathbb{R}} e^{-\frac{x^2}{2}}dx=\sqrt{2\pi}$. 
Thanks. 
 A: $$
I = \int_{\mathbb{R}} \exp\left(-\frac{(x+it)^2}{2}\right)~dx=\exp\left(\frac{t^2}{2}\right)\int_{\mathbb{R}} \exp\left(-\frac{x^2+2itx}{2}\right)~dx
$$
so
$$
I = \exp\left(\frac{t^2}{2}\right)\int_{\mathbb{R}} (\cos(tx)-i\sin(tx))\exp\left(-\frac{x^2}{2}\right)~dx
$$
It is easy to show that $\int_{\mathbb{R}} \sin(tx)\exp(-\frac{x^2}{2})~dx=0$ since $x\mapsto\sin(tx)\exp(-\frac{x^2}{2})$ is an odd function.
Using the cosine's power series:
$$
I = \exp\left(\frac{t^2}{2}\right)\int_{\mathbb{R}} \sum_{n=0}^{\infty}(-1)^n \frac{(tx)^{2n}}{(2n)!} \exp\left(-\frac{x^2}{2}\right)~dx
$$
Swapping integral and summation order is justified by considering the absolute value or the dominated convergence theorem. Moreover, the common result:
$$
\int_{\mathbb{R}} \frac{(tx)^{2n}}{(2n)!} \exp\left(-\frac{x^2}{2}\right)~dx=t^{2n}\sqrt{2\pi}\frac{(2n)!}{n!2^n}
$$
is admitted ($n$ integrations by parts). Hence
$$
I = \exp\left(\frac{t^2}{2}\right)\sqrt{2\pi}\sum_{n=0}^{\infty}(-1)^n \frac{(t)^{2n}(2n)!}{(2n)!n!2^n}\\
I= \exp\left(\frac{t^2}{2}\right)\sqrt{2\pi}\sum_{n=0}^{\infty} \frac{\left(\frac{-t^{2}}{2}\right)^n}{n!}
$$
We recognize the power series of the exponential function, hence
$$I=\sqrt{2\pi}$$
