Define the kernel functions for $a\ge 1$,

$$ G_a(t,x) := \frac{C_a t}{t^{1+1/a}+|x|^{1+a}}, \qquad \forall t>0,\: x\in \mathbb{R}\;, $$

where the constant $C_a$ is some normalization constant such that $\int_R G_a(t,x) d x =1$. Clearly, when $a=1$, this kernel function $G_1(t,x)$ is nothing but the Poisson kernel function (see e.g., Yoshida's Functional Analysis, p. 268) with $C_1=1/\pi$, which satisfies the semi-group property:

$$ \int_R G_1(t-s,x-y) G_1(s,y) d y = G_1(t,x) \;. $$

The problem is that whether this is true for general $a>1$? If it is not true in general, is there any possibility to handle the integral like the above convolution? More precisely, are there some ways to calculate or bound from below the following integral:

$$ \int_R G_a(t-s,x-y) G_a(s,y) d y \ge \;? \qquad \forall t>0, x\in R, $$

with $a>1$.

By the way, this kernel function has a nice scaling property:

$$ G_a(t,x) = \frac{1}{t^{1/a}} G_a\left(1,\frac{x}{t^{1/a}}\right)\;. $$

**Edit:** According to Prof. Hans Engler's comments, here is another related question: what is the inverse Fourier transform of the function

$$ \exp(-|\xi|^a),\qquad a>1. $$

I checked the book "Tables of Integral Transform V.I" (Erdelyi). There are only cases that $a=1$ and $a=2$. This is related to the symmetric stable law.