I am attempting to compute the Quaternion Fourier transform (QFT) of the Poisson kernel in upper-half space. Specifically, I would like to compute
$$ \mathcal{QFT_r}\{p\}(u,v; s) = \int_{\mathbb{R}^2} \frac{s}{2\pi(x^2 + y^2 + s^2)^{\frac{3}{2}}} e^{-2\pi \mathbf{i} x u} e^{-2\pi \mathbf{j} y v} \;dx\;dy, $$ for $s > 0$, where $\mathbf{i}$ and $\mathbf{j}$ are the standard quaternions such that $\mathbf{i} \mathbf{j} = \mathbf{k}$.
Wolfram alpha has gotten me so far as $$ \int_{\mathbb{R}^2} \frac{s}{2\pi(x^2 + y^2 + s^2)^{\frac{3}{2}}} e^{-2\pi \mathbf{i} x u} e^{-2\pi \mathbf{j} y v} \;dx\;dy = \int_{\mathbb{R}} \frac{2s|u|K_1\left(2\pi |u|\sqrt{y^2 + s^2}\right)}{\sqrt{y^2 + s^2}} \cos(2\pi y v) \;dy, $$
where $K_1$ is the $1^{st}$ order modified Bessel function of the second kind, and may be written as the integral equation $$ K_1(z) = z \int_{0}^{+\infty} \frac{\cos(t)}{(t^2 + z^2)^{3/2}} \; dt. $$ It may also be worth noting the identity $$ K_0(2\pi s|v|) = \int_{\mathbb{R}} \frac{s}{2\sqrt{y^2 + s^2}} \cos(2\pi y v) \;dy, $$ where $K_0$ is the $0^{th}$order modified Bessel function of the second kind.
Unfortunately I do not have significant reason to believe that there will be a nice closed form of this integral, but I am hoping some further insight could be provided.
DISCLAIMER: This is nothing other than the Fourier transform of the standard Poisson kernel on $\mathbb{R}^3_{+}$. A standard identity of the modified Bessel function of the second kind, $$ K_{\frac{1}{2}}(\xi) = \left(\frac{2\xi}{\pi}\right)^{-\frac12}\exp(-\xi), $$
makes that clear at the end of the following computation. So the final answer is simply $\exp(-2\pi s \sqrt{u^2 + v^2})$. If you instead consider the un-factored Quaternion Fourier transform of the Poisson kernel, that is, $$ \int_{\mathbb{R}^2}p(\mathbf{x};s) \exp(-2\pi(\mathbf{i}ux_1 + \mathbf{j}vx_2))\;d\mathbf{x}, $$
then this should no longer be the standard FT of $p$.
All that being said, this is a computation of the Fourier transform of the Poisson kernel I haven't seen before, so perhaps it is worth leaving up here.
This integral is possible after some wrestling with integral tables and stumbling across this related post which makes heavy use of the integral representation:
$$ K_{\nu}(z) = \frac{1}{2}\left(\frac{z}{2}\right)^{\nu} \int_{t = 0}^{+\infty} \exp\left(-t - \frac{z^2}{4t}\right) \frac{d t}{t^{\nu+1}} \text{ when $|ph(z)| < \pi/4$} \label{eq:Knu1}, $$
specifically for $\nu = 1$ and $\nu = 1/2$, and $z = \sqrt{y^2 + 1}$.
Starting from $$ \int_{\mathbb{R}} \frac{2s|u|K_1\left(2\pi |u|\sqrt{y^2 + s^2}\right)}{\sqrt{y^2 + s^2}} \exp(\mathbf{j} 2\pi y v) \;dy, $$
we first let $a = 2\pi |u|$, $k = 2\pi v$, and set $s = 1$ for simplicity, to give $$ \frac{a}{\pi} \int_{\mathbb{R}} \frac{K_1\left(a\sqrt{y^2 + 1}\right)}{\sqrt{y^2 + 1}} \exp(\mathbf{j} ky) \;dy. $$
Now substitute the integral representation for $K_1$ given above and using the same technique as in the other post referenced we have:
\begin{align*} \frac{a}{\pi} \int_{y=-\infty}^{+\infty} \frac{K_1(a \sqrt{y^2 + 1})}{\sqrt{y^2 + 1}} e^{\mathbf{j} k y }\;dy &= \frac{a}{4\pi} \int_{y=-\infty}^{+\infty} \frac{\sqrt{y^2 + 1}}{\sqrt{y^2 + 1}} \left(\int_{t=0}^{+\infty} \exp\left(-t - \frac{a^2(y^2 + 1)}{4t}\right) \frac{dt}{t^2}\right) e^{\mathbf{j} k y }\;dy\\ &= \frac{a}{4\pi} \int_{t=0}^{+\infty} \exp\left(-t - \frac{a^2}{4t}\right) \frac{dt}{t^2} \int_{y = -\infty}^{+\infty} \exp\left(\mathbf{j} k y\right)\exp\left(-\frac{a^2y^2 }{4t}\right)\;dy \\ &= \frac{1}{\sqrt{\pi}} \int_{t=0}^{+\infty} \exp\left(-t\left(1 + \frac{k^2}{a^2}\right) - \frac{a^2}{4t}\right) \frac{dt}{t^{\frac{3}{2}}}, \end{align*}
where the last equality comes from computing the 1D Fourier transform of the Gaussian. Making the substitution $\omega = t\left(1 + \frac{k^2}{a^2}\right)$ and using the identity
\begin{align*} K_{\frac{1}{2}}\left(\sqrt{a^2 + k^2}\right) &= \frac{1}{\sqrt{2}}\left(a^2 + k^2\right)^{\frac{1}{4}} \int_{\omega = 0}^{+\infty} \exp\left(-\omega - \frac{a^2 + k^2}{4\omega}\right) \frac{d\omega}{\omega^{\frac{3}{2}}} \end{align*}
finally yields
\begin{align*} \frac{a}{\pi}\int_{\mathbb{R}} \frac{K_1(a \sqrt{y^2 + 1})}{\sqrt{y^2 + 1}} e^{\mathbf{j} k y }\;dy &= \frac{\sqrt{2}(a^2 + k^2)^{\frac{1}{4}}}{\sqrt{\pi}} K_{\frac{1}{2}}\left(\sqrt{a^2 + k^2}\right). \end{align*}
With some re-substitution and an easy u-sub to write this for any $s > 0$, the final result I computed is
\begin{align*} \mathcal{QFT}_r\{p\}(u,v;s) &= \int_{\mathbb{R}^2} \frac{s}{2\pi\left(x^2 + y^2 + s^2\right)^{\frac{3}{2}}} e^{-2\pi \mathbf{i} x u} e^{-2\pi \mathbf{j} y v}\;dx dy\\ &= 2s^{\frac{1}{2}} (u^2 + v^2)^{\frac{1}{4}} K_{\frac{1}{2}}\left(2\pi s \sqrt{u^2 + v^2}\right) \end{align*}