I'm working with the isotropic $\alpha$-stable Lévy process in $\mathbb{R}^d$ $(\alpha \in (0,1) \text{ and } d \geq 2)$. I know that the distribution of the first hitting of this process into the half-space $\mathbb{H}_-:=\{x\in \mathbb{R}^d: x^{(1)}<0\}$ is given by the following expression
$$\mathbb{P}_x(X_{\tau_{\mathbb{H}_-}}\in {\rm d}y)=C(x^{(1)})^{\alpha/2}(-y^{(1)})^{-\alpha/2}|x-y|^{-d}{\rm d}y \quad x \notin \mathbb{H}_-, \ y\in \mathbb{H}_-,$$
where $x^{(1)}$ is the first coordinate of the vector $x\in \mathbb{R}^d$.
Via the Markov property I am able to show that this kernel has to satisfy the following equation (known as the Désiré-André eqn) $$|x-y|^{\alpha-d}=\int _{\mathbb{H}_-} \mathbb{P} _x(X_{\tau_{\mathbb{H}_-}}\in {\rm d}u)|u-y|^{\alpha-d} \quad x \notin \mathbb{H}_-, \ y\in \mathbb{H}_-$$
(The quantity on the left-hand side of the previous equation is the potential measure of the $\alpha$-stable process)
My question is: how to verify this is indeed the case?
The integral on my last equation, by separating my integration variables is clearly
\begin{align*} \int_{\mathbb{R}^{d-1}}\int_{(-\infty,0)}{\rm d}u^{(1)}{\rm d}u^{(2:d)}C(x^{(1)})^{\alpha/2}(-u^{(1)})^{-\alpha/2}|(x^{(1)}-u^{(1)})^2+|x^{(2:d)}-u^{(2:d)}|^2|^{-d/2}|(u^{(1)}-y^{(1)})^2+|u^{(2:d)}-y^{(2:d)}|^2|^{\frac{\alpha-d}{2}} \end{align*}
My first step was trying to solve the inner integral, which is roughly
$$\int_{(0,\infty)}u^{\alpha/2}|(u+x)^2+A|^{-d/2}|(u+y)^2+B|^{\frac{\alpha-d}{2}}{\rm d}u.$$
I haven't been able to reduce it to a simpler Beta/Hypergeometric function (perhaps is not one of these). Do you know any integration trick that could be useful?
Many thanks!