Fourier transform of Schrödinger kernel: how to compute it?

Question

Let

$$K_t(x)=\frac{1}{(4 \pi i t)^{\frac{n}{2}}}e^{i \frac{\lvert x \rvert^2}{4t}}\quad x \in \mathbb{R}^n,\ t \in \mathbb{R},\ t\ne 0.$$

Clearly this is not a $L^1$ or $L^2$ function with respect to the spatial variable $x$. Nevertheless, the paper I'm reading says:

"A direct computation shows that $\hat{K}_t(\xi)=e^{-it \lvert \xi \rvert^2}.$"

This mysterious direct computation leaves me puzzled. The Fourier transform must be taken in the distributional sense, this is clear to me; I just can't figure out how to carry the actual computation.

I'm not necessarily after a fully-rigorous answer. Even a morally correct (so as to quote Willie Wong) but not rigorous answer is fine.

Answer 1

@Willie: Here's my try at doing what you suggested. Thank you!

It is not a major loss to suppose $n=1$. Let $\sigma$ be a complex number with strictly positive real part and $g(x)=e^{-\sigma x^2}$. Our task is to compute

$$\hat{g}(\xi)=\int_{-\infty}^\infty g(x)e^{-i x \xi}\, dx.$$

To do so we take a derivative w.r.t. $\xi$:

$$\frac{d \hat{g}}{d \xi}(\xi)=\int_{-\infty}^{\infty} -i x e^{-\sigma x^2}e^{-i x \xi}\, dx = [\text{integrating by parts}] -\frac{\xi}{2 \sigma} \int_{-\infty}^{\infty}e^{-(i \xi x + \sigma x^2)}\, dx;$$

(the boundary terms of integration by parts vanish because of $\Re e(\sigma) > 0$) so $\hat{g}$ is subject to the differential condition $\frac{d \hat{g}}{d \xi}(\xi)=-\frac{\xi}{2 \sigma}\hat{g}(\xi)$. We compute

$$\hat{g}(0)=\int_{-\infty}^\infty e^{-\sigma x^2}\,dx=\sqrt{\frac{\pi}{\sigma}},$$

and infer that $\hat{g}(\xi)=\sqrt{\frac{\pi}{\sigma}}e^{-\frac{\xi^2}{4 \sigma}}$.

This last formula holds true if $\Re e(\sigma)=0$. In fact, if the sequence $\sigma_p \to \sigma$ and $\Re e(\sigma_p) >0$ for all $p\in \mathbb{N}$, then $e^{-i \sigma_p x} \to e^{-i \sigma x}$ and $\sqrt{\frac{\pi}{\sigma_p}}e^{-\frac{\xi^2}{4 \sigma_p}}\to \sqrt{\frac{\pi}{\sigma}}e^{-\frac{\xi^2}{4 \sigma}}$ in $\mathcal{S}'(\mathbb{R})$. The claim now follows from the continuity of the Fourier transform in $\mathcal{S}'(\mathbb{R})$.