Fourier transformation of complex exponential proof

Question

How do I prove that the Fourier transformation of the complex exponential

$$\exp\left[ i \pi(( a^2 x^2) +(b^2 y^2))\right]$$ is $$\exp\left[i {\pi}\left( {fx^2 \over a^2} + {fy^2 \over b^2}\right)\right]$$

Answer 1

Careful analysis of the Cauchy principle value integral for the Fourier transform gives $$ \begin{align} &\mbox{c.p.v.}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i\pi a^{2}x^{2}}e^{-ixs}dx \\ & = \mbox{c.p.v.}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i\pi(ax-s/2\pi a)^{2}}dx\, e^{-is^{2}/4\pi a^{2}} \\ & = \left(\mbox{c.p.v.}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i\pi a^{2}x^{2}}dx\right)e^{-ia^{2}s^{2}/4\pi} = Ce^{-is^{2}/4\pi a^{2}}. \end{align} $$ The constant $C$ comes from a Fresnel integral , at least after a linear change of variable. $$ C=\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}cos(\pi a^{2} x^{2})dx. $$