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

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.

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    $\begingroup$ Try computing the Fourier transform of the Gaussian $\exp (-\sigma |x|^2)$ where $\sigma$ is complex but with positive real part. Take the (formal) limit $\Re \sigma \to 0$. $\endgroup$ Apr 22, 2011 at 19:02
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    $\begingroup$ (BTW, this is also the rigorous answer if you consider tempered distributions. Take the action on both sides on Schwarz class functions.) $\endgroup$ Apr 22, 2011 at 20:19

1 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 $\mathcal{S}'(\mathbb{R})$.

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    $\begingroup$ On a rough glance, it looks fine. Just note that there's also the "physicist's method" of computing the Fourier transform of the complex Gaussian, via completing the square in the exponential. This way you reduce it to purely algebraic computation (modulo the Gaussian integral), without the ODE step. $\endgroup$ Apr 23, 2011 at 0:29
  • $\begingroup$ @Willie: Yes, I think I understand what you are talking about. For example I see it in Reed & Simon's Methods of modern mathematical physics, vol. I, §IX.1. $\endgroup$ Apr 23, 2011 at 11:11

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