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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?

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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

$$ \int_{-\infty}^{+\infty} e^{-2\pi^2(x + iu/\pi)^2}{\rm d}x = \frac{1}{\sqrt{2\pi}} $$

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  • $\begingroup$ So we will end up with $\frac{A}{2}e^{-\frac{u^2+v^2}{2}}$? $\endgroup$ – gbox Jan 24 '19 at 17:18
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    $\begingroup$ @gbox $(A/2\pi)e^{-(u^2 + v^2)/2}$ $\endgroup$ – caverac Jan 24 '19 at 17:20
  • $\begingroup$ yes sorry, so we still miss a sqrt in $2\pi$ $\endgroup$ – gbox Jan 24 '19 at 17:53
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    $\begingroup$ @gbox That could be the definition of Fourier transform you are applying. There's always a factor of $2\pi$ in there, authors use different conventions. You need to figure out which one is the one you're using math.stackexchange.com/questions/301860/… $\endgroup$ – caverac Jan 24 '19 at 18:12

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