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. 



Do these kernel functions satisfy the semigroup property?
Let $H\_a(t,\xi) = \hat G\_a(t,\xi)$ be the spatial Fourier transform of the kernel under study and let $h\_a(\xi) = H\_a(1,\xi)$. Then $H\_a(t,\xi) = H\_a(1,t^{1/a}\xi) = h\_a(t^{1/a}\xi)$, from the scaling property stated in your question. The postulated semigroup property now implies that for all $\xi$ and all $t, s > 0$
$$h\_a(t^{1/a}\xi)h\_a(s^{1/a}\xi) = h\_a((t+s)^{1/a}\xi)
$$
and therefore
$$
h\_a(t^{1/a}\xi) = e^{A(\xi)t}
$$
for some function $A(\xi)$ that may still depend on $a$. Since your kernel is assumed to be isotropic, $A$ can depend only on $|\xi|$. Another scaling argument then shows that $A(\xi) \propto |\xi|^a$ with a negative proportionality constant. Now if $a = 1$, then $A(\xi) \propto -|\xi|$, leading to the Poisson semigroup, which indeed has a kernel of this form. If $a = 2$, then you obtain the heat semigroup, with a very different kernel. 


Your question boils down to asking whether for $a \ne 1$, the inverse Fourier transform of $e^{-|\xi|^a}$ is of the form $(1+|x|^{1+a})^{-1}$, up to scaling. This is not true for $a = 2$ and it cannot be true for positive even values of $a$, since in these cases $h\_a(\xi)$ is analytic at $\xi = 0$ and hence its inverse Fourier transform must have moments of infinite order. I very much doubt that it is true for other $a \ne 1$. 


Consult the book by Zolotarev on stable distributions for more detail. As Dr. Kochubei already stated, the book and papers by Eidelman are also helpful. Finally, there is a website on fractional differentiation and applications at <http://www.fracalmo.org/>, maintained by F. Mainardi, that may be helpful.


