8.（泊松上半平面积分公式）（注：这个上半平面不包含实轴）
若f(z)=u(x,y)+iv(x,y)f(z)=u(x,y)+iv(x,y)f(z)=u(x,y)+iv(x,y)是上半平面解析函数，当z→∞z\to\inftyz→∞时f(z)f(z)f(z)一致趋于零，则在上半平面有
u(x,y)=yπ∫−∞∞u(ξ,0)(ξ−x)2+y2dξ u(x,y)=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{u(\xi,0)}{(\xi-x)^2+y^2}d\xiu(x,y)=πy∫−∞∞(ξ−x)2+y2u(ξ,0)dξ
v(x,y)=yπ∫−∞∞v(ξ,0)(ξ−x)2+y2dξv(x,y)=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{v(\xi,0)}{(\xi-x)^2+y^2}d\xi v(x,y)=πy∫−∞∞(ξ−x)2+y2v(ξ,0)dξ
证： 如下图，在上半平面
f(z)=u(x,y)+iv(x,y)=12πi∮l1+l2f(ξ)ξ−zdz=12πi∮l2f(ξ)ξ−(x+iy)dξ+→R→∞时此项为零12πi∮l1f(Reiθ)Reiθ−ziReiθdθ=12πi∫−∞∞f(ξ,0)ξ−(x+iy)dξ...①\begin{aligned} f(z)=u(x,y)+iv(x,y)&=\frac{1}{2\pi i}\oint_{l_1+l_2}\frac{f(\xi)}{\xi-z}dz\\ &=\frac{1}{2\pi i}\oint_{l_2}\frac{f(\xi)}{\xi-(x+iy)}d\xi+\\ \xrightarrow{R\to\infty时此项为零}&\frac{1}{2\pi i}\oint_{l_1}\frac{f(Re^{i\theta})}{Re^{i\theta}-z}iRe^{i\theta}d\theta\\ &= \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{f(\xi,0)}{\xi - (x+iy)} d\xi \quad \dots \quad ① \end{aligned}f(z)=u(x,y)+iv(x,y)R→∞时此项为零 =2πi1∮l1+l2ξ−zf(ξ)dz=2πi1∮l2ξ−(x+iy)f(ξ)dξ+2πi1∮l1Reiθ−zf(Reiθ)iReiθdθ=2πi1∫−∞∞ξ−(x+iy)f(ξ,0)dξ...①
构造一个函数，自变量是ξ\xiξ。
g(ξ)=f(ξ,0)ξ−(x−iy),在上半平面 f(ξ) 解析。∵ξ≠z‾ g(\xi) = \frac{f(\xi,0)}{\xi - (x-iy)} \quad , \text{在上半平面 } f(\xi) \text{ 解析。} \because \xi \neq \overline{z} g(ξ)=ξ−(x−iy)f(ξ,0),在上半平面 f(ξ) 解析。∵ξ=z
∴12πi∫−∞∞f(ξ,0)ξ−(x−iy)dξ=0(柯西定理)...② \therefore\frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{f(\xi,0)}{\xi - (x-iy)} d\xi = 0 \quad (\text{柯西定理}) \quad \dots \quad ② ∴2πi1∫−∞∞ξ−(x−iy)f(ξ,0)dξ=0(柯西定理)...②
① - ②，得：
12πi∫−∞∞(1ξ−(x+iy)−1ξ−(x−iy))f(ξ,0)dξ \frac{1}{2\pi i} \int_{-\infty}^{\infty} \left( \frac{1}{\xi - (x+iy)} - \frac{1}{\xi - (x-iy)} \right) f(\xi,0) d\xi 2πi1∫−∞∞(ξ−(x+iy)1−ξ−(x−iy)1)f(ξ,0)dξ
=12πi∫−∞∞2iy(ξ−x)2+y2⋅(u+iv)dξ = \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{2iy}{(\xi - x)^2 + y^2} \cdot (u + iv) d\xi =2πi1∫−∞∞(ξ−x)2+y22iy⋅(u+iv)dξ
=yπi∫−∞∞u(ξ,0)+iv(ξ,0)(ξ−x)2+y2dξ = \frac{y}{\pi i} \int_{-\infty}^{\infty} \frac{u(\xi,0) + iv(\xi,0)}{(\xi - x)^2 + y^2} d\xi =πiy∫−∞∞(ξ−x)2+y2u(ξ,0)+iv(ξ,0)dξ
∴u(x,y)+iv(x,y)=yπ∫−∞∞u(ξ,0)+iv(ξ,0)(ξ−x)2+y2dξ \therefore u(x,y) + iv(x,y) = \frac{y}{\pi} \int_{-\infty}^{\infty} \frac{u(\xi,0) + iv(\xi,0)}{(\xi - x)^2 + y^2} d\xi ∴u(x,y)+iv(x,y)=πy∫−∞∞(ξ−x)2+y2u(ξ,0)+iv(ξ,0)dξ
对比两边，得证。