Closed form of a Fourier transform

Question

I apologize for not being able to motivate the question below; it would go into technicalities.

Let $n=d+1\ge2$ be the space-time dimension, and $$H(y,t):=\left(\frac{t^2}{(t^2+|y|^2)^{1+d/2}}\right)^{\frac1d}.$$

Is there a closed form for the Fourier transform $\hat H(\xi,\tau)$ ?

Mind that $H$ is homogeneous of degree $-1$. The singularity at the origin being integrable, $H$ might be viewed, at worst, as a tempered distribution. Thus $\hat H$ is, at least, a tempered distribution, homogeneous of degree $1-n=-d$. Of course, it is isotropic in $\xi$.

Actually, I should be very happy if $H$ was known to be the solution of $$P(D_{y,t})H=\delta_{y=0,t=0},$$ where $P$ is a pseudo-differential operator, of course homogeneous of degree $d$. One has then $$P(\xi,\tau)=\frac1{\hat H(\xi,\tau)}\,.$$

Answer 1

For $d<4$ the Fourier transform with respect to the radial coordinate $y$ has a closed form expression in terms of a Bessel function:

$$\hat{G}(\xi,t)=(2\pi)^{d/2}\xi^{1-d/2}\int_0^\infty H(y,t)J_{d/2-1}(\xi y)y^{d/2}\,dy$$ $$\qquad=\frac{1}{\Gamma \left(\frac{1}{2}+\frac{1}{d}\right)}2^{\frac{d^2+d-2}{2 d}} \pi ^{d/2} (t/\xi)^{\frac{d-1}{2}} (t \xi)^{1/d} K_{-\frac{d}{2}+\frac{1}{2}+\frac{1}{d}}(t \xi).$$

This can then be Fourier transformed with respect to $t$, in terms of a hypergeometric function,

$$\hat{H}(\xi,\tau)=2\int_0^\infty \cos(t\tau)\hat{G}(\xi,t)\,dt=$$ $$\qquad = 2^{d} \pi ^{d/2} \xi^{-d} \Gamma \left(\frac{d}{2}\right) \, _2F_1\left(\frac{1}{2}+\frac{1}{d},\frac{d}{2};\frac{1}{2};-\frac{\tau^2}{\xi^2}\right).$$