Integrating a Poisson kernel in $n$ dimensional unit sphere Let 
\begin{equation*}
P(x,y)=\frac{1}{\omega_n R} \frac{R^2-|x|^2}{|x-y|^n}
\end{equation*}
be a Poisson kernel where $x$, $y$ are in $R^n$, $|x|<R$, $|y|=R$, $\omega_n$ is area of n dimensional unit sphere. 
Then what is $$\int_{\partial B_1(0)}P(x,y)dA_y=?$$
 A: Hint: The function 
$$g(x) = \int_{\partial B_1(0)} P(x, y) dA_y = \int_{\partial B_1(0)}  1 \cdot P(x, y) dA_y$$
satisfies $\Delta g = 0$ in $B_1(0)$ and $g(y) = 1$ on $\partial B_1(0)$. What can $g$ be?
A: "P(x,y)= (omega[n]R ())^(-1)(R^(2)-|x|( )^(2))|x-y|( )^(^(-n) ) , 
  |x-y(|)^(2)=|x|^(2)-2|x|R costheta +R^(2), for x*y =|x|R costheta "

"dsigma[n]=sin^(n-2)theta dtheta dsigma[n-1] for functions depending only on 
theta, where dsigma[n] denotes Lebesgue measure on S( )^(n) , thus get 
integral of type ∫(a+b*costheta)^(-n)sin^(n-2)theta dtheta from 0 to 
Pi,, which can be evaluated by well known substitution tan(theta/(2))=u, 
get ∫(1+u^(2))^(2) u^(n-2)(A+u^(2))^(-n)du from 0 to 1 which can be 
evaluated by calculus of residues,  also note Claus Mueller, Spherical 
harmonics, Lecture Notes v.42 ("?) ""
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$P(x,y)=(\omega_n R)^{-1}(R^2-|x|^2)|x-y|^{-n}$,$x*y=|x|R\cos\theta$,$ |x-y|^2=|x|^2-2|x|R\cos\theta+R^2$, $d\sigma_n=\sin^{n-2}\theta d\theta d\sigma_{n-1}$ for integrands depending on $\theta$ only, where $\sigma_n$ denotes Lebesgue measure on $S^{n-1}$, thus get integral of type $\int^\pi_0 (a+b\cos\theta)^{-n}\sin^{n-2}\theta d\theta$, (wlog $R=1$, and x in same direction as North pole of coordinate system (turn coordinate system around) so that $\theta$ also is angle between $x$ and $y$) which can be evaluated by $\tan \theta/ 2=u$, get $\int_0^\infty(1+u^2)^2u^{n-2}(A+u^2)^{-n}du$ which can be evaluated by residue calculus, also see Claus M\''uller, Spherical Harmonics (search Internet).
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