数学物理方法 02 解析函数积分

$\color{blue}{第二章 解析函数积分}$

$\color{blue}{\S 2.1 复变函数的积分}$

$\color{blue}{2.1.1 复积分的定义}$

$记\int _ l f(z) dz = \lim \limits _ {\left. \begin{array}{c}n \to \infty \ \max|\Delta z_k| \to 0 \end{array} \right.} \sum \limits _ {k = 0} ^n f(\zeta_k) \Delta z_k – 称f(z)沿l从A到B的积分。\\ \Delta z_k = z_k - z_{k-1} \\ f(z) – 被积分函数\\ l – 积分路径$

$\color{blue}{2.1.2 复积分存在条件}$

$令 \zeta _k = \xi_k + i \eta_k, \Delta z_k = \Delta x_k + i \Delta y_k \\ 则 f(\zeta_k) = u(\xi_k, \eta_k) + i v(\xi_k, \eta_k) = u_k + i v_k \\ f(\zeta_k) \Delta z_k = (u_k + i v_k)(\Delta x_k + i \Delta y_k) \\ = (u_k \Delta x_k - v_k \Delta y_k) + i(v_k \Delta x_k + u_k \Delta y_k)$
$\sum \limits _ {k=1}^n f(\zeta_k) \Delta z_k = \sum \limits _ {k=1}^n(u_k \Delta x_k - v_k \Delta y_k) + i \sum \limits _ {k=1}^n(v_k \Delta x_k + u_k \Delta y_k) \\ 则有: \int _l f(z) dz = \int _l u dx - v dy + i \int _l v dx + u dy$
$存在条件:(1)l分段光滑;(2)f(z)在l上连续.$

$\color{blue}{2.1.3 复积分性质}$

$\left. \begin{array}{l}1.\int _ l \sum \limits _ {k=1}^n c_k f_k(z) dz = \sum \limits _ {k=1}^n c_k \int _ l f_k(z) dz \\ 2.\int _ l f(z) dz = \sum \limits _ {k=1}^n \int _ {l_k} f(z) dz, l = \sum \limits _ {k=1}^n l_k \\ 3.\int _ {l _ {AB}} f(z) dz = - \int _ {l _ {BA}} f(z) dz \end{array} \right \} \longleftarrow 积分定义$
$4.|\int _l f(z) dz | \leq \left \{ \begin{array}{l} \int _l |f(z)| \cdot |dz|, |dz| = ds \\ Ms, M \geq |f(z)|, s = l的长度。 \end{array} \right. \leftarrow 积分定义+|z_1 + z_2| \leq |z_1| + |z_2|$

$\left | \sum \limits _ {k=0} ^n f(\zeta_k) \Delta z_k \right | \leq \sum _ {k=0}^n |f(\zeta _k) | \cdot | \Delta z_k| \leq M \sum \limits _ {k=0}^n | \Delta z_k |$

$例1.\lim \limits _ {r \to 0} \int _ {|z| = r} \dfrac{z^3}{1 + z^2} dz = ?$
$\left | \int _{|z| = r} \dfrac{z^3}{1 + z^2} dz \right | \leq \int _ {|z| = r} \left | \dfrac{z^3}{1 + z^2} \right | |dz| \leq \dfrac{r^3}{1 - r^2} \cdot 2 \pi r \stackrel{r \to 0}{\longrightarrow} 0$

$\color{blue}{2.1.4 计算方法}$

$1.用定义计算$
$例2.计算:I = \int _l z dz, l: z_0 \to z_n$
$1) 选\zeta_k = z_ {k-1}, f(z) = z, \Delta z_k = z_k - z_ {k-1} \\ I = \int _ l z dz = \lim \limits _ {n \to \infty} \sum \limits _ {k=1}^n [z_ {k-1}(z_k - z_{k-1})] \\ 2)选\zeta_k = z_k \\ I = \int _ l z dz = \lim \limits _ {n \to \infty} \sum \limits _ {k=1}^n[z_k(z_k - z_{k-1})] \\ I = \int _ l z dz = \dfrac{1}{2} \lim \limits _ {n \to \infty} \left \lbrace \sum \limits _ {k=1}^n [ z_{k-1}(z_k - z_{k-1})] + \sum \limits _ {k=1}^n [z_k(z_k - z_{k-1})] \right \rbrace \\ = \dfrac{1}{2} \lim \limits _ {n \to \infty} \sum \limits _ {k=1}^n [z_k^2 - z_{k-1}^2] \\ = \dfrac{1}{2} [ z_n^2 - z_0^2]$

$2.通过计算实线积分来计算$
$例3.计算: I = \int _ l Re \; z \; dz \\ l:1) 0 \to 2 + i; 2) 0 \to 2 \to 2 + i$
$I = \int _ l Re z dz = \int _ l x d(x + iy) = \int _ l x dx + i \int _ l x dy \\ l:1) 0 \to 2 + i: x = 2y, y:0 \to 1 \\ I = \int _ l Re z dz = \int _ 0 ^ 1 2y d(2y) + i \int _0 ^ 1 2y dy = 2 + i \\ l:2) 0 \to 2 \to 2 + i: l_1:y = 0, x:0 \to 2; l_2: x=2, y:0 \to 1 \\ I = \int _ l Re z dz = \int _ {l_1 + l_2} x dx + i \int _ {l_1 + l_2} x dy = \int_0^2 x dx + i \int _0^1 2 dy = 2 + 2i$

$例4.计算：I = \int _l z dz, l:1) 0 \to 2 + i; 2) 0 \to 2 \to 2+i$
$1),2) I = \dfrac{3}{2} + 2i$

$3.用极坐标计算$
$例5.证明: \oint _ l \dfrac{dz}{(z - a)^n} = \left \{ \begin{array}{l} 2 \pi i, n = 1, \\ 0, n \neq 1, \end{array} \right. l:|z - a| = r$

$\color{blue}{\S 2.2 柯西定理}$

$\color{blue}{2.2.1 单连通区域的Cauchy定理}$

$设f(z)在单连通区域\sigma内解析，l为\sigma内的任意一条分段光滑的曲线，则\\ \oint _ l f(z) dz = 0\\ 注:Cauchy定理被人们称之为解析函数的基本定理。$
$例1.\oint _ l \dfrac{dz}{(z - 3)} = ? l:1)|z - 3| = 2; 2)|z| = 2$
$答:1) 2\pi i; 2) 0$
$例2.\oint _ {|z-3| = 2} \dfrac{dz}{(z-3)^2} = ?$
$答: 0$
$注：Cauchy定理逆定理不成立。$

$\color{blue}{2.2.2 推论}$

$解析函数的积分之值与路径无关。$
$例3.计算\int _l \sin z dz, l:|z-1| = 1的上半圆从0 \to 2$
$答:1 - \cos 2$

$\color{blue}{2.2.3 不定积分和原函数}$

$1.定理：若f(z)在\sigma内解析，则在\sigma内 F(z) = \int _{z_0}^z f(\zeta) d\zeta,单值、解析且F^{\prime}(z) = f(z)$
$证明：(1)\int_{l_{AB}} f(\zeta) d \zeta \\ = \int _ A ^ B f(\zeta) d \zeta \\ = \int_{z_0}^z f(\zeta) d \zeta (A = z_0, B = z) \\ = F(z) (单值)$
$(2) \dfrac{F(z + \Delta z) - F(z)}{\Delta z} = \dfrac{1}{\Delta z} [\int _ {z_0}^{z+ \Delta z} f(\zeta) d \zeta - \int _ {z_0}^z f(\zeta) d \zeta ] \\ = \dfrac {1}{\Delta z} [\int_ {z_0}^z f(\zeta) d \zeta + \int _ z ^ {z + \Delta z} f(\zeta) d \zeta - \int _ {z_ 0} ^z f(\zeta) d \zeta] \\ = \dfrac{1}{\Delta z} \int _ {z} ^ {z + \Delta z} f(\zeta) d \zeta (沿z \to z + \Delta z 直线)$
$f(z) = \dfrac{1}{\Delta z} \int _ {z}^{z + \Delta z} f(z) d \zeta (沿 z \to z + \Delta z直线)$

$证明：\left| \dfrac{F(z + \Delta z) - F(z)}{\Delta z} - f(z) \right| = \left| \dfrac{1}{\Delta z} \right| \left| \int_{z}^{z + \Delta z} [f(\zeta) - f(z)] d \zeta \right| \\ \leq \dfrac{1}{\Delta z}\int _z^{z + \Delta z} |f(\zeta) - f(z)||d \zeta|$
$\because f(z) \in H(\sigma) \therefore f(z)连续\\ \therefore \forall \varepsilon > 0, \exists \delta > 0, \epsilon|\zeta - z| < \delta \to |f(\zeta) - f(z) | < \varepsilon \\ \therefore \left| \dfrac{F(z + \Delta z) - F(z)}{\Delta z} - f(z) \right| < \left| \dfrac{1}{\Delta z} \right| \varepsilon | \Delta z| = \varepsilon \\ \therefore \lim \limits _ {\Delta z \to 0} \dfrac{F(z + \Delta z) - F(z)}{\Delta z} = f(z) \to F^{\prime}(z) = f(z)$

$注：上述定理条件可减弱为(1) \oint_l f(z) dz = 0; (2) f(z)在\sigma内连续。$

$2.原函数的定义：\\ 若\varphi^{\prime}(z) = f(z), 则称\varphi(z)为f(z)的原函数。\\ 显然，F(z) = \int_{z_0}^z f(\zeta) d \zeta 是f(z)的一个原函数。$
$3.\varphi(z) - F(z) = c$
$4.Newton-Leibniz公式:\\ 若f(z) \in H(\sigma),则有\int_{z_0}^z f(\zeta) d \zeta = \varphi(z) - \varphi(z_0)$
$例4.I = \int _0^{2 + i} z dz = \dfrac{z^2}{2} | _0^{2+i} = \dfrac{3}{2} + 2i$
$5.对于解析函数，也可采用分部积分法$
$如，\int _1^{1 + \frac{\pi}{2} i} z e^z dz$
$答：-\dfrac{\pi}{2} e$

$例5.已知u = x^2 - y^2 - x 为f(z) \in H(\sigma)的实部,求f(z).$
$\because f(z) \in H(\sigma) \to \left \{ \begin{array}{l} \dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}, \dfrac{\partial u}{\partial x} = 2x -1 \\ \dfrac{\partial v}{\partial x} = - \dfrac{\partial u}{\partial y}, \dfrac{\partial u}{\partial y} = -2y \end{array} \right.$
$f^{\prime}(z) = \dfrac{\partial u}{\partial x} - i \dfrac{\partial u}{\partial y} = 2x - 1 + i 2y = 2z - 1$
$\to f(z) = \int_{z_0}^z (2 \zeta - 1) d \zeta = z^2 - z + c$

$柯西定理：设f(z)在单连通区域\sigma内解析，l为\sigma内的任意一条分段光滑的曲线,$
$\qquad \qquad \Updownarrow$
$设f(z)在单连通区域\sigma内及\bar \sigma = \sigma + l上均解析,$
$则 \oint _l f(z) dz = 0$

$\color{blue}{2.2.4 推广的柯西定理}$

$若f(z) \in H(\sigma),在\bar \sigma = \sigma + l上连续，则\oint _l f(z) dz = 0$

$2\color{blue}{.2.5 复连通区域的柯西定理}$

$设L = l + \sum \limits _ {k=1}^n l_k 为\sigma的复围线,f(z) \in H(\sigma)在 \bar \sigma = \sigma + L上连续, 则\\ \oint _ l f(z) dz = \sum \limits _ {k=1}^n \oint _ {l_k} f(z) dz$
$例6. \oint _ {|z| = a} \dfrac{1}{z^2 - 1} dz = ?, a > 2$
$\dfrac{1}{z^2-1} = \dfrac{1}{2}[\dfrac{1}{z-1} - \dfrac{1}{z+1}], 答:0$

$例7.计算I = \oint_ l \dfrac{1}{(z-a)^n} dz; l: 包围a的围道$
$答:\oint _ l \dfrac{dz}{(z - a)^n} = \left \{ \begin{array}{l} 2 \pi i, n = 1, \\ 0, n \neq 1 \end{array} \right.$

$\color{blue}{\S 2.3 柯西公式}$

$\color{blue}{2.3.1 Cauchy 公式}$

$1.单连通区域的Cauchy公式:$
$设f(z) \in H(\sigma),在\bar \sigma = \sigma + l 上连续, a \in \sigma,则 f(a) = \dfrac{1}{2\pi i} \oint _ l \dfrac{f(z)}{z - a} dz$

$证明：\oint _ l \dfrac{f(z) - f(a)}{z - a} dz = \oint _ {l_{\rho}} \dfrac{f(z) - f(a)}{z - a} dz$
$\to \left| \oint _{l_{\rho}} \dfrac{f(z) - f(a)}{z - a} dz \right| \leq \oint _ {l_{\rho}} \left| \dfrac{f(z) - f(a)}{z - a} \right| | dz | \leq \dfrac{\max{|f(z) - f(a)|}}{\rho} \cdot 2 \pi \rho$
$\because f(z)在a点连续, \therefore \forall \varepsilon > 0, \exists \delta > 0, \epsilon|z - a| < \delta \to |f(z) - f(a)| < \varepsilon$
$\oint _ {l_{\rho}} \left| \dfrac{f(z) - f(a)}{z - a} \right| | dz| < 2 \pi \varepsilon = \varepsilon^{\prime}$
$\therefore \oint _ {l} \dfrac{f(z) - f(a)}{z - a} dz = 0$
$注：1)更一般：f(z) = \dfrac{1}{2 \pi i} \oint _ l \dfrac{f(\zeta)}{\zeta - z} d \zeta$
$2)意义：解析函数在区域内的值由边界上的积分值确定。$
$3)可用来计算围道积分$
$\oint _ l \dfrac{f(\zeta)}{\zeta - z} d \zeta = 2 \pi i f(z)$
$\oint _ {|z| = 1} \dfrac{e^z}{2} dz = 2 \pi i$

$例1.计算\oint _{|z| = 2} \dfrac{3z - 1}{z(z-1)} dz$
$答：6\pi i$
$例2.计算\oint _ {|z-i| = \frac{1}{2}} \dfrac{e^z}{z(z^2+1)} dz$
$答：\pi(\sin 1 - i \cos 1)$

$2.复通区域的柯西公式$
$设L = l + \sum \limits _ {k=1}^n l_k为\sigma的边界复围线，f(z) \in H(\sigma)在\bar \sigma = \sigma + L上连续,则$
$f(z) = \dfrac{1}{2 \pi i} \left[ \oint _ l \dfrac{f(z)}{\zeta - z} d \zeta - \sum \limits _ {k=1}^n \oint _ {l_k} \dfrac{f(z)}{\zeta - z} d \zeta \right]$

$3.无界区域的柯西公式$
$设f(z)在围道l外单值解析，在l外至l上连续，且当z \to \infty时, f(z) 一致区域零, 则\\ f(z) = \dfrac{1}{2 \pi i} \oint _ l \dfrac{f(\zeta)}{\zeta - z} d \zeta$

$\color{blue}{2.3.2 柯西公式的推论}$

$1.解析函数的任意阶导数$
$设l, \sigma, f(z)满足柯西定理存在的条件,则在\sigma内有:$
$f^{(n)}(z) = \dfrac{n!}{2 \pi i} \oint _ l \dfrac{f(\zeta)}{(\zeta - z)^{n+1}} d \zeta$
$当n = 1时有：f^{\prime}(z) = \dfrac{1}{2 \pi i} \oint _ l \dfrac{f(\zeta)}{(\zeta - z)^2} d \zeta$
$证明:\dfrac{\Delta f}{\Delta z} = \dfrac{f(z + \Delta z) - f(z)}{\Delta z} \\ = \dfrac{1}{2 \pi i} \dfrac{1}{\Delta z} \oint _ l [\dfrac{f(\zeta)}{\zeta - z - \Delta z} - \dfrac{f(\zeta)}{\zeta - z}] d \zeta \\ = \dfrac{1}{2 \pi i} \oint _ l \dfrac{f(\zeta)}{(\zeta - z - \Delta z)(\zeta - z)} d \zeta$
$\left| \dfrac{\Delta f}{\Delta z} - \dfrac{1}{2 \pi i} \oint _ l \dfrac{f(\zeta)}{(\zeta - z)^2} d \zeta \right| = \left| \dfrac{1}{2 \pi i} \oint _ l \dfrac{f(\zeta) \Delta z}{(\zeta - z - \Delta z)(\zeta - z)^2} d \zeta \right| \leq \dfrac{1}{2 \pi i} \oint _ l \dfrac{|f(\zeta)| \cdot |\Delta z|}{|\zeta - z - \Delta z||\zeta - z|^2} |d \zeta|$
$\because f(z)在\bar \sigma上连续, \therefore 有上界, \\ 设\max|f(\zeta)| = M, d = \min|\zeta - z|, s - l 长, |\Delta z| < \dfrac{d}{2} \\ 则|\zeta - z| \geq d, |\zeta - z - \Delta z| \geq |\zeta - z| - |\Delta z| > \dfrac{d}{2}$
$\therefore \left| \dfrac{\Delta f}{\Delta z} - \dfrac{1}{2 \pi i} \oint _ l \dfrac{f(\zeta)}{(\zeta - z)^2} d \zeta \right| \leq \dfrac{1}{2 \pi i} \oint _ l \dfrac{|f(\zeta)| \cdot | \Delta z|}{|\zeta - z - \Delta z||\zeta - z|^2} |d \zeta| < \dfrac{1}{2 \pi} \dfrac{M|\Delta z|}{\frac{d^3}{2}} s = \dfrac{|\Delta z| Ms}{ \pi d^3}$
$取\delta = \min[\dfrac{d}{2}, \dfrac{\varepsilon \pi d^3}{Ms}], 则当|\Delta z| < \delta时有:$
$\left| \dfrac{\Delta f}{\Delta z} - \dfrac{1}{2 \pi i} \oint _ l \dfrac{f(\zeta)}{(\zeta - z)^2} d \zeta \right| < \varepsilon$
$即f^{\prime} = \dfrac{1}{2 \pi i} \oint _ l \dfrac{f(\zeta)}{(\zeta - z)^2} d \zeta$

$注意:(1)上述公式成立,实际上只用到条件:\\ 1) f(z) = \dfrac{1}{2 \pi i} \oint _ l \dfrac{f(\zeta)}{(\zeta - z)} d \zeta, \\ 2) f(z)连续$
$(2) 对复变函数,若一阶导数存在，则任意阶导数存在;对实变函数则不然。$
$(3)柯西导数公式可用来计算积分:\\ \oint _ l \dfrac{f(z)}{(z - a)^{n+1}} dz = \dfrac{2 \pi i}{n!} f^{(n)}(a)$
$例3.\oint _ {|z|=1} \dfrac{e^z}{z^n} dz = ?$
$答:\left \{ \begin{array}{l} \dfrac{2 \pi i}{(n-1)!}, n \geq 1 \\ 0, n < 1 \end{array} \right.$
$(4)推论：若\varphi(z)在l上连续,f(z) = \dfrac{1}{2 \pi i} \int _l \dfrac{\varphi(\zeta)}{(\zeta - z)} d \zeta -- 柯西型积分$
$则f^{(n)}(z) = \dfrac{n!}{2 \pi i} \int _ l \dfrac{\varphi(\zeta)}{(\zeta - z)^{n+1}} d \zeta \quad (z \notin l)$
$(5) 复通区域的柯西导数公式仍然成立。$

$2.柯西不等式：$
$设f(z) \in H(\sigma),在\bar \sigma = \sigma + l 上连续,则有|f^{(n)}(z)| \leq \dfrac{n! Ms}{2 \pi d^{n+1}}$
$其中, M = \sup|f(z)|, s - l 的长,d = \min|\zeta - z|$
$特别是，当l:|\zeta - z| = R时，有:|f^{(n)}(z)| \leq \dfrac{n!M}{R^n}$

$3.刘维尔定理：$
$设f(z)在复平面解析，且当z \to \infty时 |f(z)| \leq M, 则f(z)必为常数.$
$注：(1)在复变函数中,函数可导有界必为常数;在实变函数中则不然.$
$(2)P_n(z), e^z, \sin z等不为常数，所以均无界。$

$4.模数原理:$
$若f(z) \in H(\sigma),在 \bar \sigma = \sigma + l 上连续, 则f(z)只能在边界上取最大值。$

$5.平均值定理:$
$若f(z)在|z - a| < R内解析,在|z - a| \leq R上连续, 则$
$f(a) = \dfrac{1}{2 \pi} \int _0^{2 \pi} f(a + R e^{i \varphi}) d \varphi$

$6.摩勒纳定理:$
$设f(z)在区域\sigma内连续,且对\sigma内任意围线l都有\oint _ l f(z) dz = 0,则f(z)在\sigma 内解析。$