Prove: $$h(x,y)=A\cdot e^{-2\pi^2(x^2+y^2)}$$ is $$H(u,v)=\frac{A}{\sqrt{2}}\cdot e^{-\frac{u^2+v^2}{2}}$$ After fourier transform
$$F(u,v)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)e^{-i2\pi (ux+vy)}dxdy$$
$$H(u,v)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}A\cdot e^{-2\pi^2(x^2+y^2)}e^{-i2\pi (ux+vy)}dxdy$$
$$H(u,v)=A\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-2\pi^2(x^2+y^2)}e^{-i2\pi (ux+vy)}dxdy$$
$$H(u,v)=A\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-2\pi^2x^2-2\pi^2y^2-i2\pi ux -i2\pi vy}dxdy$$
How should I group $-2\pi^2x^2-2\pi^2y^2-i2\pi ux -i2\pi vy$ in order to integrate?
Complete squares
\begin{align} -2\pi^2 x^2 - 2\pi^2y^2 - i2\pi u x - i 2\pi v y &= -2\pi^2\left(x + \frac{i u}{\pi}\right)^2 - \frac{u^2}{2} -2\pi^2\left(y + \frac{i u}{\pi}\right)^2 - \frac{v^2}{2} \end{align}
So the integral becomes
$$ H(u, v) = A e^{-u^2/2 - v^2/2}\left(\int_{-\infty}^{+\infty} e^{-2\pi^2(x + iu/\pi)^2}{\rm d}x\right)\left(\int_{-\infty}^{+\infty} e^{-2\pi^2(y + iv/\pi)^2}{\rm d}y\right) $$
The integrals are fairly easy to calculate
$$ \int_{-\infty}^{+\infty} e^{-2\pi^2(x + iu/\pi)^2}{\rm d}x = \frac{1}{\sqrt{2\pi}} $$