The integral of the Poisson kernel is a constant function in $\zeta$?

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For the ball of radius 1, $B$, in $\mathbb R$, the Poisson kernel takes the form

$$P(x,\zeta) = \frac{1-|x|^2}{\omega _{n-1}|x-\zeta|^n}$$

where $x\in B$, $\zeta\in \mathbb S$ (the surface of $B$), and $\omega _{n-1}$ is the surface area of the unit n−1-sphere.

I would like to know why, for $x=|x|\omega\in [0,1[\times \,\mathbb S$, $\,$ the function $$\zeta \mapsto \int_{\mathbb S} P(|x|w,\zeta) \, d\sigma(\omega)$$ is constant ?

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If $\xi$ is another point of $\mathbb{S}$, there exists a matrix $M\in SO(n)$ mapping $\zeta$ to $\xi$. Note that $$ P(x, \zeta) = P(Mx, M\zeta) $$ because $|Mx|=|x|$ and $|Mx-M\zeta| = |x-\zeta|$. So,

$$ \int_{\mathbb S} P(|x|\omega,\zeta) \, d\sigma(\omega) = \int_{\mathbb S} P(|x|M \omega , \xi) \, d\sigma(\omega) = \int_{\mathbb S} P(|x|\omega', \xi) \, d\sigma(\omega') $$ by the orthogonal change of variable $\omega'=M\omega$ (the Jacobian determinant of which is $1$.)