← 学习库 复变函数与积分变换(第五版) 本册目录

部分习题答案

原书第 226 页

习题一

1.1 (1) $ -2 + 3\mathrm{i} $; (2) $ a^{3} - 3ab^{2} + \mathrm{i}(b^{3} - 3a^{2}b) $; (3) $ -\frac{3}{10} + \frac{\mathrm{i}}{10} $; (4) $ \frac{x^{2} + y^{2} - 1 + 2\mathrm{i}y}{(x + 1)^{2} + y^{2}} $.

1.3 $ z_{1} = -\frac{3}{5} - \frac{6}{5} \mathrm{i} $, $ z_{2} = -\frac{6}{5} - \frac{17}{5} \mathrm{i} $.

1.4 $ \overline{A}z + \overline{A}\overline{z} + B = 0 $,其中 $ A = a + ib $, $ B = 2C $(实数).

1.5 $ Az\bar{z}+\bar{B}z+B\bar{z}+C=0 $,其中 A=2a,C=2d 均为实数,B=b+ic.

1.6 (1) $ 2, \frac{\pi}{6} $; (2) $ \sqrt{2}, -\frac{3}{4}\pi $; (3) $ \sqrt{5}, -\arctan\frac{1}{2} $; (4) $ \sqrt{10}, \pi - \arctan 3 $.

1.7 (2) 此式表示:平行四边形对角线平方和等于各边平方和.

1.8 (1) $ \sqrt{13}\left[\cos\left(\pi-\arctan\frac{2}{3}\right)+\mathrm{i}\sin\left(\pi-\arctan\frac{2}{3}\right)\right] $;

(2) $ \cos\left(\frac{\pi}{2}-\alpha\right) + i \sin\left(\frac{\pi}{2}-\alpha\right) $; (3) $ \cos\left(-\frac{2}{3}\pi\right) + i \sin\left(-\frac{2}{3}\pi\right) $.

1.9 (1) 2; (2) i; (3) -1; (4) $ \sqrt[8]{8}\left(\cos\frac{3\pi+8k\pi}{16}+i\sin\frac{3\pi+8k\pi}{16}\right) $, k=0,1,2,3.

1.10 $ z_0 = \frac{1}{2} + \frac{\sqrt{3}}{2} \mathrm{i} $, $ z_1 = -1 $, $ z_2 = \frac{1}{2} - \frac{\sqrt{3}}{2} \mathrm{i} $.

1.11 (1)圆环、有界、多连通域;

(2) 以原点为中心,以 $ \frac{1}{3} $为半径的圆的外部、多连通域、无界;

(3)圆环的一部分、单连通域、有界;

(4)圆内一部分、有界、单连通域;

(5) $ x^{2}-y^{2}<1 $ 无界、单连通域;

(6)椭圆的内部及椭圆的边界、有界、闭区域;

(7)从原点出发的两条半射线所成的区域、无界、单连通域;

(8)分三种情况:01 为圆内有界单连通域.

1.12 (1) 圆;(2) b>2a>0,椭圆,2a>b>0,双曲线,2a=b,无意义;(3) a=b, y=0; a>b,抛物线;a<b,无意义;

(4) $ \left|a\right|^{2}=b $,点; $ \left|a\right|^{2}>b $,圆; $ \left|a\right|^{2}

1.13 (1) $ z = 1 + \mathrm{i} + (-2 - 5\mathrm{i})t $ (0 ≤ t ≤ 1);

(2) $ z = a \cos t + i b \sin t $ (0 ≤ t < 2π).

原书第 227 页

1.14 $ \frac{z^{2}}{4}+\frac{3\bar{z}^{2}}{4}+\frac{\mathrm{i}z}{2}+\frac{\mathrm{i}\bar{z}}{2} $

习题二

2.1 (I) $ \left(\frac{1}{z}\right)^{\prime} = -\frac{1}{z^{2}} $; (2) z=0 处 $ f^{\prime}(0) = 0, z \neq 0 $ 处 $ f^{\prime}(z) $ 不存在.

2.2 (1) z=0 可导, $ z \neq 0 $ 不可导, 复平面上处处不解析;

(2) x=y 上可导,其余点均不可导,复平面上处处不解析;

(3)复平面上处处解析;

(4)复平面上处处解析.

2.3 (1)除 $ z=\pm1 $ 外在复平面上处处解析, $ z=\pm1 $ 为奇点, $ f'(z)=-\frac{2z}{(z^{2}-1)^{2}} $;

(2) 除 $ z = -\frac{d}{c} $ ( $ c \neq 0 $) 外在复平面上处处解析, $ z = -\frac{d}{c} $ 为奇点, $ f'(z) = \frac{ad - bc}{(cz + d)^2} $.

2.9 (1) $ (1-\mathrm{i})z^{3}+\mathrm{c}\mathrm{i} $; (2) $ x^{2}-y^{2}-3y+c+\mathrm{i}(2xy+3x) $; (3) $ -\mathrm{i}(1-z)^{2} $; (4) $ ze^{z} $.

2.10 p = 1, $ e^{z} + c $; p = -1, $ -e^{-z} + c $.

2.13 (1) $ z = \ln 2 + i\left(\frac{\pi}{3} + 2k\pi\right) $, $ k = 0, \pm 1, \pm 2, \cdots $; (2) $ z = i $;

(3) $ z = 2k\pi + \mathrm{i} $ 或 $ z = (2k-1)\pi - \mathrm{i} $, k 为整数;

(4) $ z = k\pi - \frac{\pi}{4} $, $ k = 0, \pm 1, \pm 2, \cdots $.

2.14 (1) $ \frac{e^{-1}+e}{2}; $

(2) $ \ln 5 - i \arctan \frac{4}{3} + (2k+1)\pi i $, $ k=0, \pm1, \pm2, \cdots $;

(3) $ \sqrt{2}e^{\frac{\pi}{4}-2k\pi}\left[\cos\left(\ln\sqrt{2}-\frac{\pi}{4}\right)+\mathrm{i}\sin\left(\ln\sqrt{2}-\frac{\pi}{4}\right)\right] $, k 为整数;

(4) $ 27e^{2k\pi}(\cos \ln 3 - i \sin \ln 3) $.

2.19 否.

2.20(2)不成立;(1),(3),(4)均成立.

习题三

3.1 (1) $ -\frac{1}{3}+\frac{\mathrm{i}}{3} $; (2) $ -\frac{1}{2}+\frac{5}{6}\mathrm{i} $; (3) $ -\frac{1}{2}-\frac{\mathrm{i}}{6} $.

3.2 (1) $ 4\pi i $; (2) $ 8\pi i $.

3.4 (1) 0; (2) 0; (3) $ 2\pi i $.

3.50.

3.6 0.

3.70.

原书第 228 页

3.8 (1) $ 1 - \cos \pi i $; (2) $ ie^{1 + i} = ie(\cos 1 + i \sin 1) $; (3) $ 3e^i - 4 $.

3.9 函数 $ \frac{1}{z^{2}} $ 在全平面除去 z=0 的区域内为解析. 我们考虑一个单连通区域, 例如 D: $ \mathrm{Re} $ $ z > -\frac{1}{4} $, $ \left|z\right| > \frac{1}{2} $, 则 $ \frac{1}{z^{2}} $ 在 D 内解析, 于是取 $ \frac{1}{z^{2}} $ 的一个原函数 $ -\frac{1}{z} $, 则 $ \int_{C}\frac{\mathrm{d}z}{z^{2}} = -\frac{1}{z}\bigg|_{-3i}^{i} = \frac{4}{3}i $.

3.10 (1) $ 2\pi\mathrm{ie}^{2} $; (2) $ 4\pi\mathrm{i} $; (3) $ \pi\mathrm{e}^{\frac{\pi}{4}}=\pi\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\mathrm{i}\right) $; (4) r<1 时为 0; r>1 时 n=1 为 $ 2\pi\mathrm{i} $, n\neq1 为 0.

3.11 (1) $ \frac{\pi}{5}\mathrm{i} $; (2) $ \frac{4\pi}{5}\mathrm{i} $; (3)0; (4) $ \pi\mathrm{i} $.

3.13 (1) $ \frac{2\pi i}{99!} $; (2) 0; (3) 0.

习题四

4.1 (1) 无;(2) 0;(3) 无.

4.2 (1)发散;(2)绝对收敛;(3)发散。

4.4 (1) 1; (2) $ \frac{1}{e} $; (3) $ \infty $.

4.5 (1) $ 1-z^{3}+z^{6}-\cdots $, |z|<1;

(2) 当 a=b 时,级数为 $ \sum_{n=1}^{\infty}\frac{nz^{n-1}}{a^{n+1}} $, $ \left|z\right|<\left|a\right| $;

当 $ a \neq b $ 时,级数为 $ \frac{1}{b-a} \sum_{n=0}^{\infty} \left( \frac{1}{a^{n+1}} - \frac{1}{b^{n+1}} \right) z^{n} $, $ \left| z \right| < \min \left| a \right| $, $ \left| b \right| $。

(3) $ 1 - 2z^{2} + 3z^{4} - 4z^{6} + \cdots $, $ \mid z\mid < 1 $;

(4) $ 1 + \frac{z^2}{2!} + \frac{z^4}{4!} + \cdots $, $ \quad |z| < \infty $;

(5) $ -\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n}\frac{(2z)^{n}}{(2n)!},\quad|z|<\infty $;

(6) $ 1 - z - \frac{1}{2!} z^2 - \frac{1}{3!} z^3 - \cdots $, \quad | z | < 1.

4.7 (1) $ \sum_{n=0}^{\infty}(-1)^{n}(n+1)(z-1)^{n} $, $ \left|z-1\right|<1 $;

(2) $ \sum_{n=0}^{\infty} \sin\left(\frac{n\pi}{2}+1\right)\frac{(z-1)^n}{n!} $, $ \left|z-1\right|<\infty $;

(3) $ \sum_{n=0}^{\infty}\frac{3^{n}}{(1-3\mathrm{i})^{n+1}}[z-(1+\mathrm{i})]^{n},\quad\left|z-(1+\mathrm{i})\right|<\frac{\sqrt{10}}{3}; $

原书第 229 页

(4)

$$ 1+2\left(z-\frac{\pi}{4}\right)+2\left(z-\frac{\pi}{4}\right)^{2}+\frac{8}{3}\left(z-\frac{\pi}{4}\right)^{3}+\cdots,\mid z-\frac{\pi}{4}\mid<\frac{\pi}{4}. $$

4.8 (1) $ \frac{1}{z^2} - 2 \sum_{n=0}^{\infty} z^{n-2} $, $ 0 < |z| < 1 $, $ \frac{1}{z^2} + 2 \sum_{n=0}^{\infty} \frac{1}{z^{n+3}} $, $ 0 < |z| < \infty $;

(2) $ \sum_{n=-2}^{\infty}\frac{1}{(n+2)!}\cdot\frac{1}{z^{n}}, 0<|z|<\infty $;

(3) $ -\sum_{n=0}^{\infty}\frac{z^{n}}{2^{n+1}}+\sum_{n=0}^{\infty}(-1)^{n+1}\frac{1}{z^{2n+2}}, 1<|z|<2; $

(4) $ \sum_{n=0}^{\infty}\frac{1}{(2n)!(z-1)^{2n}} $, $ 0<|z-1|<\infty $.

4.9 (1) 在 $ 0 < |z - 2| < 1 $ 内, $ f(z) = -\sum_{n=0}^{\infty} (z - 2)^{n-1} $;

(2) 在 $ 1 < |z - 2| < +\infty $ 内, $ f(z) = \sum_{n=0}^{\infty} \frac{1}{(z - 2)^{n+2}} $;

(3) 在 $ 0 < |z - 3| < 1 $ 内, $ f(z) = \sum_{n=0}^{\infty} (-1)^{n}(z - 3)^{n-1} $;

(4) 在 $ 1 < |z - 3| < +\infty $ 内, $ f(z) = \sum_{n=0}^{\infty} (-1)^{n} \frac{1}{(z - 3)^{n+2}} $.

4.10 $ \sum_{n=0}^{\infty}(-1)^{n}(n+1)\frac{(z-\mathrm{i})^{n-2}}{(2\mathrm{i})^{n+2}},0<\mid z-\mathrm{i}\mid<2. $

习题五

5.1 (1)是;(2)不是;(3)是。

5.2 (1) $ z = \pm 3i $,一阶;(2) $ z = 0 $,二阶, $ z = k\pi $ (k 为整数, $ k \neq 0 $),一阶;(3) z = 0,四阶, $ z = \sqrt{2k\pi i} $,一阶.

5.3 (1) z=0 为简单极点, $ z=\pm2i $ 为二阶极点;(2) z=0 为二阶极点;

(3) $ z = k\pi - \frac{\pi}{4} $ ( $ k = 0, \pm 1, \cdots $) 各为简单极点;

(4) z=0 为三阶极点, $ z=2k\pi\mathrm{i} $ ( $ k=\pm1,\pm2,\cdots $) 各为简单极点; (5) z=0 为可去奇点;

(6) z=0 为可去奇点, $ z=2k\pi i $ ( $ k=\pm1,\pm2,\cdots $) 为简单极点.

5.6 (1) 是;(2) 不是.

5.7 (1) Res[f(z),0]=0;

$$ Res[f(z),2]=\frac{128}{25},Res[f(z),\pm i]=-\frac{56\pm33i}{100}; $$

(3) $ \mathrm{Res}[f(z),-1]=2\sin 2;\quad(4)\mathrm{Res}[f(z),0]=-\frac{1}{6}; $

(5) $ \mathrm{Res}[f(z),0]=0 $, $ \mathrm{Res}[f(z),k\pi]=\frac{(-1)^n}{k\pi}, k\neq0 $; (6) $ \frac{\sinh z}{\cosh z} $ 处处解析.

原书第 230 页

5.8 (1) $0$; (2) $\frac{1}{8}\pi i e$; (3) $4\pi e^{2}i$; (4) $-2\pi i$;

(5) 当 $ |a| < |b| $ 时,积分为零;当 $ |a| < 1 < |b| $ 时,积分为 $ (-1)^{n-1}\frac{2\pi(2n-2)!i}{[(n-1)!]^{2}(a-b)^{2n-1}} $;当 $ 1 < |a| < |b| $ 时,积分为零。

5.9 (1)本性奇点,0;(2)可去奇点,0;(3)简单极点,-1.

5.10 (1) $ -\frac{2}{3}\pi i $; (2) $ 2\pi i $.

5.12 (1) $ \frac{2\pi}{\sqrt{a^{2}-1}} $; (2) $ \frac{\pi}{2} $; (3) $ \frac{\pi}{2a} $; (4) $ \pi e^{-1}\cos2 $; (5) $ \frac{\sqrt{2}}{2}\pi $; (6) $ \pi e^{-a b} $.

5.14 在 $ \left|z\right|<1 $ 内有一个根. 在 $ 1<\left|z\right|<2 $ 内有三个根.

习题六

6.1 (1) 旋转角 $ \theta=0 $,伸缩率为 $ r=2 $;(2) $ \theta=\pi $, $ r=\frac{1}{2} $;(3) $ \theta=\frac{\pi}{4} $, $ r=2\sqrt{2} $;

(4) $ \theta = \pi - \arctan \frac{4}{3} $, r = 10.

6.2 (1) $ u^{2}+v^{2}=\frac{1}{4} $; (2) v=-u; (3) $ \left(u-\frac{1}{2}\right)^{2}+v^{2}=\frac{1}{4} $; (4) $ u=\frac{1}{2} $.

6.3 (1)下半个单位圆域;(2) $ \mathrm{Im} w > \mathrm{Re} w $;

$$ \operatorname{Im}w<0,\ \left|\ w-\frac{1}{2}(1-i)\right|\geq\frac{1}{\sqrt{2}};\left(4\right)\operatorname{Re}w>0,\ \left|\ w-\frac{1}{2}\right|>\frac{1}{2},\ \operatorname{Im}w>0. $$

6.4 $ |w| < 1 $ 且沿0到1的半径有割痕.

6.5 (1) $ w = \frac{z + 2 + i}{z + 2 - i} $; (2) $ w = \frac{iz + 2 + i}{z + 1} $; (3) $ w = \frac{1 - i}{2}(z + 1) $.

6.6 (1) $ w = -\left(\frac{z + \sqrt{3}}{z - \sqrt{3}}\right)^{3} $; (2) $ w = -\mathrm{i}\left(\frac{z + 1}{z - 1}\right)^{2} $.

6.7 (1) $ w = -\mathrm{i} \frac{z - \mathrm{i}}{\mathrm{i} + z} $; (2) $ w = \mathrm{i} \frac{z - \mathrm{i}}{\mathrm{i} + z} $; (3) $ w = \frac{3z + (\sqrt{5} - 2\mathrm{i})}{(\sqrt{5} - 2\mathrm{i})z + 3} $; (4) $ w = \frac{\mathrm{i} - z}{\mathrm{i} + z} $.

6.8 (1) $ w = \frac{2z - 1}{z - 2} $; (2) $ w = \frac{\mathrm{i}(2z - 1)}{2 - z} $.

6.9 (1) $ w = -\left(\frac{z + \sqrt{3} - \mathrm{i}}{z - \sqrt{3} - \mathrm{i}}\right)^3 $; (2) $ w = \left(\frac{z^2 + 4}{z^2 - 4}\right)^2 $; (3) $ w = \mathrm{e}^{\frac{\pi \mathrm{i}}{b - a}(z - a)} $;

$$ w=-\left(\frac{z^{2/3}+2^{2/3}}{z^{2/3}-2^{2/3}}\right)^{2};(5)w=\left(\frac{\sqrt{z}+1}{\sqrt{z}-1}\right)^{2}. $$

原书第 231 页

习题七

7.1 (1) $ v(z) = 2(\bar{z} - i) $,流线: $ x(y + 1) = C_{1} $,等势线: $ x^{2} - (y + 1)^{2} = C_{2} $;

(2) $ v(z) = -\frac{2\bar{z}}{(\bar{z}^{2}+1)^{2}} $,流线: $ \frac{xy}{(x^{2}-y^{2}+1)^{2}+4x^{2}y^{2}}=C_{1} $,等势线: $ \frac{x^{2}-y^{2}+1}{(x^{2}-y^{2}+1)^{2}+4x^{2}y^{2}}=C_{2} $;

(3) $ v = \overline{\omega}' = 1 - \frac{1}{z^2} $,流线: $ y - \frac{y}{x^2 + y^2} = C_1 $,等势线: $ \frac{x}{x^2 + y^2} + x = C_2 $;

(4) $ v(z)=\frac{1-i}{\bar{z}} $,流线: $ \rho=C_{1}e^{-\varphi} $,等势线: $ \rho=C_{2}e^{\varphi} $

7.2 (1)0,0; (2)0,0; (3)0,0.

$$ \omega=\frac{2}{\pi}\ln\frac{z^{2}-1}{z^{2}+1}. $$

7.4 $ E_{1} $: $ E_{2}=1:2 $.

7.5 $ \omega = v_{\infty} \sqrt{z^{2} + h^{2}} $, $ v(z) = \overline{f'(z)} = \frac{v_{\infty} \bar{z}}{\sqrt{z^{2} + h^{2}}} $.

习题八

8.3

$$ \omega_{0}=2,\ F(n\ \omega_{0})=\frac{-2}{(4n^{2}-1)\pi}--(n=0,\ \pm1,\ \pm2,\ \cdots),f(t)=\frac{-2}{\pi}\sum_{n=-\infty}^{\infty}\frac{1}{4n^{2}-1}\mathrm{e}^{\mathrm{j}n\omega_{0}t}. $$

8.4 (1) $ F(\omega) = -\frac{2\mathrm{j}}{\omega}[1 - \cos\omega] $; (2) $ F(\omega) = \frac{1}{1 - \mathrm{j}\omega} $; (3) $ F(\omega) = \frac{4}{\omega^3}(\sin\omega - \omega\cos\omega) $;

(4)

$$ \begin{array}{r l}{F(\omega)}&{=\frac{2}{4+\left(1+\mathrm{j}\omega\right)^{2}}.}\end{array} $$

8.5 (1) $ F(\omega) = \frac{2\sin\omega}{\omega} $; (2) $ F(\omega) = \frac{-2\mathrm{j}}{1 - \omega^{2}} \sin\omega\pi $.

8.6 (1)

$$ F(\omega)=\frac{2}{\mathrm{j}\omega};\quad(2)\quad F(\omega)=\frac{\pi}{2}\mathrm{j}\left[\delta(\omega+2)-\delta(\omega-2)\right]; $$

(3)

$$ F(\omega)=\frac{\pi}{4}\mathrm{j}\left[\delta(\omega-3)-3\delta(\omega-1)+3\delta(\omega+1)-\delta(\omega+3)\right]; $$

(4)

$$ F\left(\omega\right)=\frac{\pi}{2}\left[\left(\sqrt{3}+\mathrm{j}\right)\delta\left(\omega+5\right)+\left(\sqrt{3}-\mathrm{j}\right)\delta\left(\omega-5\right)\right]. $$

3.10 (1) $ \omega_{0} = \frac{\pi}{2} $, $ F(n\omega_{0}) = \begin{cases} \frac{4}{n^{2}\pi^{2}}, & n = \pm1, \pm3, \cdots, \\ 1, & n = 0, \\ 0, & n = \pm2, \pm4, \cdots \end{cases} $

$$ F\left(\omega\right)=2\pi\delta\left(\omega\right)+\sum_{n=-\infty\atop n\neq0}^{+\infty}\frac{\sin^{2}n\omega_{0}}{n^{2}\omega_{0}^{2}}\delta\left(\omega-n\omega_{0}\right); $$

原书第 232 页

$$ \omega_{0}=\frac{2\pi}{T},F(n\omega_{0})=\{\begin{aligned}&\frac{h\mathrm{j}}{2n\pi},&n=-1,-2,\cdots,\\&\frac{h}{2},&n=0,\\&\frac{h\mathrm{j}}{2n\pi},&n=1,2,\cdots,\end{aligned}. $$

$$ F\left(\omega\right)=\pi h\delta\left(\omega\right)+\sum_{n=-\infty\atop n\neq0}^{+\infty}\frac{h\mathrm{j}}{n}\delta\left(\omega-n\omega_{0}\right). $$

8.11 $ f(t) = \cos \omega_0 t $.

8.12 $ F(\omega) = \cos \omega a + \cos \frac{\omega a}{2} $

8.14 $ (1 - e^{-t}) u(t) $

8.16 (1) $ F(\omega) = \frac{\omega_0}{\omega_0^2 - \omega^2} + \frac{\pi}{2\mathrm{j}} \left[ \delta(\omega - \omega_0) - \delta(\omega + \omega_0) \right] $; (2) $ F(\omega) = \frac{-1}{(\omega - \omega_0)^2} + \pi\mathrm{j} \delta'(\omega - \omega_0) $.

习题九

9.1 (1) $ F(s) = \frac{1}{s} (3 - 4e^{-2s} + e^{-4s}) $; (2) $ F(s) = \frac{3}{s} (1 - e^{-\frac{1}{2}\pi s}) - \frac{1}{s^2 + 1} e^{-\frac{1}{2}\pi s} $;

(3)

$$ F(s)=\frac{1}{s-2}+5\;;\quad(4)\quad\frac{s^{2}}{s^{2}+1}. $$

9.2 (1)

$$ \frac{2}{4s^{2}+1};(2)\frac{1}{s+2};(3)\frac{2}{s^{3}};(4)\frac{1}{s^{2}};(5)\frac{1}{s^{2}+4};(6)\frac{s^{2}+2}{s(s^{2}+4)}. $$

9.3 (1) $ \frac{1}{s^{3}}(2s^{2}+3s+2) $; (2) $ \frac{1}{s}-\frac{1}{(s+1)^{2}} $; (3) $ \frac{s^{2}-4s+5}{(s-1)^{3}} $; (4) $ \frac{10-3s}{s^{2}+4} $; (5) $ \frac{s^{2}-a^{2}}{(s^{2}+a^{2})^{2}} $;

$$ \frac{s+4}{\left(s+4\right)^{2}+16}. $$

9.4 (1) $ F(s) = \frac{4(s+3)}{\left[(s+3)^2+4\right]^2} $; (2) $ F(s) = \frac{2(3s^2 + 12s + 13)}{s^2\left[(s+3)^2 + 4\right]^2} $.

9.5 (1) $ f(t) = -2\sinh t $; (2) $ f(t) = \frac{2\sinh t}{t} $; (3) $ f(t) = t\sinh t $.

(4) $ f(t) = \frac{1}{2} \int_{0}^{t} t \sinh t \, dt = \frac{t}{2} \cosh t - \frac{1}{2} \sinh t $.

9.6 (1) $ F(s)=\frac{\pi}{2}-\arctan\frac{s}{k};(2) $ $ F(s)=\frac{1}{s}\left(\frac{\pi}{2}-\arctan\frac{s+3}{2}\right) $.

9.7 (1) $ \ln 2 $; (2) $ \frac{1}{4} $.

9.8 (1) $ \frac{1}{a}\sin at $; (2) $ \frac{1}{a-b}(ae^{at}-be^{bt}) $; (3) $ \frac{c-a}{(b-a)^2}e^{-at}+\left[\frac{c-b}{a-b}t+\frac{a-c}{(a-b)^2}\right]e^{-bt} $;

原书第 233 页

(4)

$$ \frac{1}{3}(\cos t-\cos2t)\;;\;(5)\frac{1}{3}\sin t-\frac{1}{6}\sin2t;(6)\frac{1}{9}\left(\sin\frac{2}{3}t+\cos\frac{2}{3}t\right)\mathrm{e}^{-\frac{1}{3}t}; $$

(7) $ f(t)=\{\begin{array}{ll}t, & 0 \leq t < 2, \\ 2(t-1), & t > 2;\end{array}. $ (8) $ \frac{2}{t}(1-\cosh t) $.

9.9 $ \mathcal{L}\left[f(t)\right]=\frac{1}{\left(s^{2}+1\right)\left(1-\mathrm{e}^{-\pi s}\right)}. $

9.10 (1) $ t $; (2) $ \frac{m!\,n!\,t^{m+n+1}}{(m+n+1)!} $; (3) $ \frac{1}{2k}\sin kt - \frac{1}{2}t\cos kt (k \neq 0) $; (4) $ \sinh t - t $;

(5) $ \{\begin{aligned}&0,&t

9.12 (1) $ y = \frac{1}{2} t^{2} e^{t} $; (2) $ y = e^{t} + 1 $; (3) $ y = \sin t $;

(4) $ y = e^{-t} - e^{-2t} + u(t-1)\left[\frac{1}{2}e^{-2(t-1)} - e^{-(t-1)} + \frac{1}{2}\right] $;

(5) $ y = -1 + t + \frac{c}{2} t^{3} + \frac{1}{2} e^{-t} + \frac{1}{2} (\cos t - \sin t) $.

9.13 (1) $ \begin{cases} x(t) = -t + t\mathrm{e}^t, \\ y(t) = 1 + t\mathrm{e}^t - \mathrm{e}^t \end{cases} $ (2) $ \begin{cases} x(t) = u(t-1), \\ y(t) = 0. \end{cases} $

原书第 234 页
原书第 235 页

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