As part of an exercise in Fourier transforms, I need to evaluate the following integral: $$ \int_{-\infty}^{\infty}e^{-\frac{u^{2}-i4\sigma^{2}\pi ux}{2\sigma^{2}}}du $$ The exercise is to prove that for the the following Fourier transform: $$ \hat{f}(u,v)=\exp\left(-\frac{u^{2}+v^{2}}{2\sigma^{2}}\right) $$
we get: $$ f(x,y)=2\pi\sigma^{2}\exp\left(-2\pi^{2}\sigma^{2}\left(x^{2}+y^{2}\right)\right) $$
In my textbook they did: $$ \begin{align*} \int_{-\infty}^{\infty}e^{-\frac{u^{2}-i4\sigma^{2}\pi ux}{2\sigma^{2}}}du&=\int_{-\infty}^{\infty}e^{-\frac{u^{2}-i4\sigma^{2}\pi ux+\left(4\pi^{2}x^{2}\sigma^{4}-4\pi^{2}x^{2}\sigma^{4}\right)}{2\sigma^{2}}}du=\int_{-\infty}^{\infty}e^{-\frac{u^{2}-i4\sigma^{2}\pi ux-4\pi^{2}x^{2}\sigma^{4}}{2\sigma^{2}}}e^{\frac{4\pi^{2}x^{2}\sigma^{4}}{2\sigma^{2}}}du\\&=e^{\frac{4\pi^{2}x^{2}\sigma^{4}}{2\sigma^{2}}}\int_{-\infty}^{\infty}e^{-\frac{u^{2}-i4\sigma^{2}\pi ux-4\pi^{2}x^{2}\sigma^{4}}{2\sigma^{2}}}du=e^{\frac{4\pi^{2}x^{2}\sigma^{4}}{2\sigma^{2}}}\int_{-\infty}^{\infty}e^{-\frac{u^{2}-2u\left(i2\sigma^{2}\pi x\right)+\left(i2\sigma^{2}\pi x\right)^{2}}{2\sigma^{2}}}du\\&=e^{\frac{4\pi^{2}x^{2}\sigma^{4}}{2\sigma^{2}}}\int_{-\infty}^{\infty}e^{-\frac{\left(u-i2\sigma^{2}\pi x\right)^{2}}{2\sigma^{2}}}du \end{align*} $$ Then they stated that because of the normal distribution, you get: $$ \int_{-\infty}^{\infty}e^{-\frac{\left(u-i2\sigma^{2}\pi x\right)^{2}}{2\sigma^{2}}}du=\sigma \sqrt{2\pi} $$ I don't quite understand why. First of all, as I understand that, they stated that $U\sim N\left(i2\sigma^{2}\pi x,\sigma\right)$, but can $\mu$ (the first argument) be a non-real value? Secondly, how did they got to $\sigma\sqrt{2\pi}$?
If someone is interested in the original exercise, then they did: $$ \begin{align*} f(x,y)&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\hat{f}(u,v)e^{i2\pi(ux+vy)}dudv=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{u^{2}}{2\sigma^{2}}}e^{-\frac{v^{2}}{2\sigma^{2}}}e^{i2\pi ux}e^{i2\pi vy}dudv\\&=\int_{-\infty}^{\infty}e^{-\frac{u^{2}}{2\sigma^{2}}}e^{i2\pi ux}\int_{-\infty}^{\infty}e^{-\frac{v^{2}}{2\sigma^{2}}}e^{i2\pi vy}dudv=\int_{-\infty}^{\infty}e^{-\frac{u^{2}}{2\sigma^{2}}+i2\pi ux}du\int_{-\infty}^{\infty}e^{-\frac{v^{2}}{2\sigma^{2}}+i2\pi vy}dv\\&=\int_{-\infty}^{\infty}e^{-\frac{u^{2}-i4\sigma^{2}\pi ux}{2\sigma^{2}}}du\int_{-\infty}^{\infty}e^{-\frac{v^{2}-i4\sigma^{2}\pi vy}{2\sigma^{2}}}dv \end{align*} $$ Then they started talking about $\int_{-\infty}^{\infty}e^{-\frac{u^{2}-i4\sigma^{2}\pi ux}{2\sigma^{2}}}du$ (where I got stuck).