## Tempered distributions

Hi

Problem: Show that if $b\in \mathbb{R}$ and $b \neq 0 \; , e^{-a|\mathbf{x}|^{2}}$ converges temperately to
$e^{-ib|\mathbf{x}|^{2}}$ as $a$ approaches $ib$ from the right half-plane. Deduce that

$
\mathcal{F}\left[e^{-ib|\mathbf{x}|^{2}}\right]=e^{-in\pi(\mbox{sgn } b)/4} \left(\frac{\pi}{|b|}\right)^{\frac{n}{2}}e^{i|\bo ldsymbol{\xi}|^{2}/4b}
$

I have tried to use the fact that if Re $a >0$ then
$
\mathcal{F}\left[e^{-a|\mathbf{x}|^{2}}\right] = \left(\frac{\pi}{a}\right)^{\frac{n}{2}}e^{-\frac{|\boldsymbol{\xi}|^{2}}{4a}} \; \mbox{ for Re } a>0
$
is valid.

But I simply can´t plug in it in into the formula. Somehow a term involving sgn(b) should appear.
I guess this happens when we approach the imaginary axle.

Thanks