How should I approach this integral? I am trying to work out the following integral:
$$
\int_{-\infty}^{\infty}
\left\vert\frac{1}{1 + \alpha\,\iota}\,
\exp\left(-\,\frac{\pi x^{2}}
{\,\sqrt{\,{1 + \alpha\,\iota}\,}\,}\right)
\,\right\vert\mathrm{d}x
$$
Edit: $\alpha$ is just some constant and $\iota$ is supposed to be the imaginary unit. I was thinking that I can take the absolute of $\frac{1}{1+\alpha\iota}$ and throw that outside the integral. Then for ${\,\sqrt{\,{1 + \alpha\,\iota}\,}\,}$ I could find the square root, find the reciprocal and exploit $\left|e^{c\iota}=1\right|$ for any constant c and the fact that what remains looks like the pdf of a standard normal? Does that sound reasonable?
Thanks.
 A: As a hint:
$$\left|\frac{1}{1+ a I}\right| = \frac{1}{\sqrt{1+a^2}}$$
and
$$\left|\exp\left(-\pi \frac{x^2}{\sqrt{1+ a I}}\right)\right|  = \exp\left(-\pi x^2 Re\left(\frac{1}{\sqrt{1 + a I}}\right)\right),$$
where
$$Re\left(\frac{1}{\sqrt{1 + a I}}\right) = \frac{\cos(\arctan(a)/2)}{(1 + a^2)^{1/4}}.$$
Now the integral can be solved as a usual squared exponential integral.
A: Let's take the square root which preserves the quadrant; i.e.
$$
\sqrt{1+\alpha i}=\sqrt{\frac{1+\sqrt{1+\alpha^2}}2}+i\operatorname{sgn}(\alpha)\sqrt{\frac{-1+\sqrt{1+\alpha^2}}2}
$$
Then, define $\beta$ as the real part of the reciprocal:
$$
\frac1{\sqrt{1+\alpha i}}=\underbrace{\sqrt{\tfrac{1+\sqrt{1+\alpha^2}}{2\left(1+\alpha^2\right)}}}_\beta-i\operatorname{sgn}(\alpha)\sqrt{\tfrac{-1+\sqrt{1+\alpha^2}}{2\left(1+\alpha^2\right)}}
$$
Furthermore, note that $\left|e^{x+iy}\right|=e^x$
$$
\begin{align}
\int_{-\infty}^\infty\left|\frac1{1+\alpha i}\exp\left(-\frac{\pi x^2}{\sqrt{1+\alpha i}}\right)\right|\,\mathrm{d}x
&=\frac1{\sqrt{1+\alpha^2}}\int_{-\infty}^\infty\exp\left(-\pi\beta x^2\right)\,\mathrm{d}x\\
&=\frac1{\sqrt{\beta\left(1+\alpha^2\right)}}\\
&=\sqrt[\large 4]{\frac2{\left(1+\alpha^2\right)\left(1+\sqrt{1+\alpha^2}\right)}}
\end{align}
$$
