数学

Let R+n={(x1,⋯ ,xn−1,xn)∣xn>0}\mathbb{R}^n_+=\left\{(x_1,\cdots,x_{n-1},x_n)\mid x_n > 0\right\}.The Poisson\mathrm{Poisson}'s kernel is

K(x,y)=2xnnα(n)1∣x−y∣nK(x,y)=\frac{2x_n}{n\alpha(n)}\frac{1}{|x-y|^n}

with x∈R+nx\in \mathbb{R}^n_+ and y∈∂R+ny\in \partial \mathbb{R}^n_+,then

∫∂R+nK(x,y)dy=1\int_{\partial \mathbb{R}^n_+}K(x,y)\mathrm{d}y=1

Proof\textbf{Proof}.Some notation:Given a vector xx I will write x=(x′,xn)x=(x',x_n) where x′=(x1,⋯ ,xn−1)x'=(x_1,\cdots,x_{n-1}).Since ∂R+n=Rn−1×{0}\partial \mathbb{R}^n_+=\mathbb{R}^{n-1}\times \left\{0\right\},

∫∂R+nK(x,y)dy=2xnnα(n)∫Rn−11(∣x′−y′∣2+xn2)n2dy′\int_{\partial \mathbb{R}_+^n}K(x,y)\mathrm{d}y=\frac{2x_n}{n\alpha(n)}\int_{\mathbb{R}^{n-1}}\frac{1}{(|x'-y'|^2+x_n^2)^{\frac{n}{2}}}\mathrm{d}y'

Making the change of coordinates z′=x′−y′xn\displaystyle z'=\frac{x'-y'}{x_n} gives

2xnnα(n)∫Rn−11(∣x′−y′∣2+xn2)n2dy′=2xnnαn∫Rn−1xnn−1xnn(∣z′∣2+1)n2dz′=2nα(n)∫Rn−11(∣z′∣2+1)n2dz′\begin{aligned} \frac{2x_n}{n\alpha(n)}\int_{\mathbb{R}^{n-1}}\frac{1}{(|x'-y'|^2+x_n^2)^{\frac{n}{2}}}\mathrm{d}y'&=\frac{2x_n}{n\alpha_n}\int_{\mathbb{R}^{n-1}}\frac{x_n^{n-1}}{x_n^n(|z'|^2+1)^{\frac{n}{2}}}\mathrm{d}z'\\ &=\frac{2}{n\alpha(n)}\int_{\mathbb{R}^{n-1}}\frac{1}{(|z'|^2+1)^{\frac{n}{2}}}\mathrm{d}z' \end{aligned}

To calculate this final integral you will need to use polar coordinates in Rn−1\mathbb{R}^{n-1}.After changing to polar coordinates,let τ=ρ2\tau=\rho^2 to get

∫Rn−11(∣z′∣2+1)n2dz′=(n−1)α(n−1)∫0∞ρn−2(ρ2+1)n2dρ=12(n−1)α(n−1)∫0∞τn−32(τ+1)n2dτ=12(n−1)α(n−1)B(n−12,12)\begin{aligned} \int_{\mathbb{R}^{n-1}}\frac{1}{(|z'|^2+1)^{\frac{n}{2}}}\mathrm{d}z'&=(n-1)\alpha(n-1)\int_0^{\infty}\frac{\rho^{n-2}}{(\rho^2+1)^{\frac{n}{2}}}\mathrm{d}\rho\\ &=\frac{1}{2}(n-1)\alpha(n-1)\int_0^{\infty}\frac{\tau^{\frac{n-3}{2}}}{(\tau+1)^{\frac{n}{2}}}\mathrm{d}\tau\\ &=\frac{1}{2}(n-1)\alpha(n-1)\mathrm{B}\left(\frac{n-1}{2},\frac{1}{2}\right) \end{aligned}

where B\mathrm{B} is the Beta function.Thus,

∫∂R+nK(x,y)dy=(n−1)α(n−1)nα(n)B(n−12,12)\int_{\partial \mathbb{R}^n_+}K(x,y)\mathrm{d}y=\frac{(n-1)\alpha(n-1)}{n\alpha(n)}\mathrm{B}\left(\frac{n-1}{2},\frac{1}{2}\right)

Using that α(n)=πn2Γ(n2+1),B(x+y)=Γ(x)Γ(y)Γ(x+y)\displaystyle\alpha(n)=\frac{\pi^{\frac{n}{2}}}{\displaystyle \Gamma \left(\frac{n}{2}+1\right)},\mathrm{B}(x+y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} and Γ(x+1)=xΓ(x)\Gamma(x+1)=x\Gamma(x) we can verify that

(n−1)α(n−1)nα(n)B(n−12,12)=1\frac{(n-1)\alpha(n-1)}{n\alpha(n)}\mathrm{B}\left(\frac{n-1}{2},\frac{1}{2}\right)=1

BIBLIOGRAPGY

1.L.C.Evans，Partial Differential Equations(Second Edition),Graduate Studies in Math.19,AMS,2010

2.Integral of the Poisson’s kernel: ∫∂RnK(x,y) dy=1\int_{\partial\mathbb{R}^n} K(x,y)\,dy = 1