Fourier Transform Identity I'm trying to verify the following:
$$ \int_{\mathbb{R}} e^{-z\xi^2} \hat{f}(\xi) \; d\xi = \sqrt{\frac{\pi}{z}} \int_{\mathbb{R}} e^{-\pi^2x^2/z} f(x) dx, $$
for $z = \alpha i$ purely imaginary and $f$ in the Schwartz class. 
For $z$ in the right half-plane $\{\Re( z) > 0\}$, the identity is clear using $\int g \hat{f} = \int \hat{g} f$ and the fact that the Fourier transform of $e^{-zx^2}$ is $\sqrt{\frac{\pi}{z}} e^{-\pi^2\xi^2/z}$. 
I was told that this can be extended to the imaginary axis by analytic continuation, but I can't see how to justify that. Any help would be appreciated. 
 A: Since the integrals may not exist for $\operatorname{Re} z < 0$, analytic continuation is not what one can do here. However, one can continuously extend both sides to $\operatorname{Re} z = 0$. The first integrand
$$e^{-z\xi^2}\hat{f}(\xi)$$
is uniformly dominated on $\operatorname{Re} z \geqslant 0$ by $\lvert \hat{f}(\xi)\rvert$, since
$$\left\lvert e^{-z\xi^2}\right\rvert = e^{-(\operatorname{Re} z)\xi^2} \leqslant 1,$$
and since $$\operatorname{Re} \frac{1}{z} = \frac{\operatorname{Re} z}{\lvert z\rvert^2},$$ the integrand of the second integrand is uniformly dominated by $\lvert f(x)\rvert$. Both, $f$ and $\hat{f}$, are integrable, and the integrands depend continuously on $z$, so
$$A(z) = \int_{\mathbb{R}} e^{-z\xi^2}\hat{f}(\xi)\,d\xi$$
and
$$B(z) = \int_{\mathbb{R}} e^{-\pi^2x^2/z}f(x)\,dx$$
are continuous functions on $H = \{ z : \operatorname{Re} z \geqslant 0\}$ resp. $H\setminus \{0\}$ by the dominated convergence theorem. Since also $\sqrt{\frac{\pi}{z}}$ is continuous on $H\setminus 0$, the equality
$$A(z) = \sqrt{\frac{\pi}{z}}\cdot B(z)$$
extends from $\operatorname{Re} z > 0$ to $H\setminus \{0\}$ by continuity.
The right hand side is a priori not defined for $z = 0$, but can be defined as the continuous extension of the right hand side to $z = 0$.
