第五章 多元函数微分学

例 1 设 \({x}_{n},{y}_{n} \in {\mathbf{R}}^{m}\left( {n = 1,2,\cdots }\right) ,\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \mathbf{a},\mathop{\lim }\limits_{{n \rightarrow \infty }}{y}_{n} = \mathbf{b}\) ,证明: \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{x}_{n} \cdot {y}_{n}}\right) = a \cdot b. \]

例 2 设 \(A \subset {\mathbf{R}}^{m},B \subset {\mathbf{R}}^{m}\) ,证明: (1) \({\left( A \cap B\right) }^{ \circ } = {A}^{ \circ } \cap {B}^{ \circ }\) ； (2) \(\overline{A \cup B} = \overline{A} \cup \overline{B}\) .

例 3 设 \(E \subset {\mathbf{R}}^{m}\) 是闭集, \(x \in {\mathbf{R}}^{m}\) ,求证: (1) \(\exists y \in E\) ,使得 \(\rho \left( {x,E}\right) = \rho \left( {x,y}\right)\) ； (2)若 \(x\bar{ \in }E\) ,则 \(\rho \left( {x,E}\right) > 0\) .

例 4 设 \(\Omega\) 是有界闭区域, \(G\) 是闭区域,且 \(G \subset \Omega\) . 求证: 3 闭区域 \(V\) ,使得 \(G \subset V \subset \bar{V} \subset \Omega\) .

例 5 画出集合 \(\Omega = \{ \left( {x,y}\right) \mid 0 < x < y\) \(< 1\}\) 的图形. \begin{center} <img src=\"/static/img/math_analysis/036.jpg\" class=\"math-img\" style=\"max-width:100%;margin:10px 0;\"> \end{center} \hspace*{3em} 图 5.1

例 6 确定并画出下列函数的定义域, 指出等位面是什么样的曲面 (或曲线): (1) \(u = \ln \left( {y - {x}^{2}}\right)\) ; (2) \(u = \arccos \left( {{x}^{2} + {y}^{2} - {z}^{2}}\right)\) .

例 7 求下列极限: (1) \(\mathop{\lim }\limits_{\substack{{x \rightarrow + \infty } \\ {y \rightarrow + \infty } }}{\left( \frac{xy}{{x}^{2} + {y}^{2}}\right) }^{x}\) ; (2) \(\mathop{\lim }\limits_{\substack{{x \rightarrow \infty } \\ {y \rightarrow a} }}{\left( 1 + \frac{1}{x}\right) }^{\frac{{x}^{2}}{x + y}}\) .

例 8 对下列函数 \(f\left( {x,y}\right)\) ,求证: \(\mathop{\lim }\limits_{{\left( {x,y}\right) \rightarrow \left( {0,0}\right) }}f\left( {x,y}\right)\) 不存在. (1) \(f\left( {x,y}\right) = {x}^{y}\left( {x > 0,y > 0}\right)\) ; (2) \(f\left( {x,y}\right) = \frac{{x}^{3} + {y}^{3}}{{x}^{2} + y}\left( {{x}^{2} + y \neq 0}\right)\) .

例 9 设 \(f\left( {x,y}\right)\) 在开半平面 \(x > 0\) 上二元连续,固定 \(y\) ,极限 \(\mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}f\left( {x,y}\right) = \varphi \left( y\right)\) 存在,在 \(y\) 轴上函数补充定义 \(f\left( {0,y}\right) = \varphi \left( y\right)\) 后, 问函数 \(f\left( {x,y}\right)\) 是否在 \(x \geq 0\) 上二元连续. 考虑例子: \[ f\left( {x,y}\right) = \frac{{x}^{2}y}{{x}^{4} + {y}^{2}}\;\left( {x > 0}\right) . \]

例 10 设 \(f\left( {x,y}\right)\) 定义在开集 \(\Omega\) 内,若 \(f\left( {x,y}\right)\) 对 \(x\) 连续,对 \(y\) 满足李普希兹条件,即 \(\forall \left( {x,{y}^{\prime }}\right) ,\left( {x,{y}^{\prime \prime }}\right) \in \Omega\) ,有 \[ \left| {f\left( {x,{y}^{\prime }}\right) - f\left( {x,{y}^{\prime \prime }}\right) }\right| \leq L\left| {{y}^{\prime } - {y}^{\prime \prime }}\right| \;\left( {L\text{ 为常数 }}\right) . \] 求证: \(f\left( {x,y}\right)\) 在 \(\Omega\) 上连续.

例 11 设 \(f\left( x\right)\) 在 \({\mathbf{R}}^{m}\) 上连续,满足: (1) \(x \neq 0\) 时, \(f\left( x\right) > 0\) ； (2)对任意 \(x\) 和正常数 \(c,f\left( {cx}\right) = {cf}\left( x\right)\) . 求证: 存在 \(a > 0,b > 0\) ,使得 \(a\left| \mathbf{x}\right| \leq f\left( \mathbf{x}\right) \leq b\left| \mathbf{x}\right|\) . 思路 根据 \(f\) 的性质,要证的结论可改写成 \[ a \leq f\left( {x/\left| x\right| }\right) \leq b\;\left( {\forall x \in {\mathbf{R}}^{m}\smallsetminus \{ 0\} }\right) . \]

例 1 设 \(f\left( {x,y}\right) = {x}^{2}{\mathrm{e}}^{y} + \left( {x - 1}\right) \arctan \frac{y}{x}\) ,求它在(1,0)点的偏导数.

例 2 设 \(\Omega \subset {\mathbf{R}}^{2}\) 为区域. 在 \(\Omega\) 内 \({f}_{x}^{\prime }\left( {x,y}\right) \equiv 0,{f}_{y}^{\prime }\left( {x,y}\right) \equiv 0\) . 求证: \(f\left( {x,y}\right)\) 在 \(\Omega\) 内为常数.

例 3 设 \[ f\left( {x,y}\right) = \left\{ \begin{array}{ll} {xy}\sin \frac{1}{\sqrt{{x}^{2} + {y}^{2}}}, & {x}^{2} + {y}^{2} \neq 0, \\ 0, & {x}^{2} + {y}^{2} = 0. \end{array}\right. \] 求证: (1) \({f}_{x}^{\prime }\left( {0,0}\right) ,{f}_{y}^{\prime }\left( {0,0}\right)\) 存在; (2) \({f}_{x}^{\prime }\left( {x,y}\right)\) 与 \({f}_{y}^{\prime }\left( {x,y}\right)\) 在(0,0)点不连续; (3) \(f\left( {x,y}\right)\) 在(0,0)点可微.

例 4 设 \(u = f\left( {x,\frac{x}{y}}\right)\) ,求 \(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}\) .

例 5 设 \(u = u\left( {x,y}\right)\) 在 \({x}^{2} + {y}^{2} > 0\) 上可微,令 \(x = r\cos \theta ,y =\) \(r\sin \theta\) . 在(x, y)点作单位向量 \({\mathbf{e}}_{r},{\mathbf{e}}_{\theta }\) . 向量 \({\mathbf{e}}_{r}\) 表示 \(\theta\) 固定沿 \(r\) 增加的方向, \({\mathbf{e}}_{\theta }\) 表示 \(r\) 固定沿 \(\theta\) 增加的方向. 证明: \[ \frac{\partial u}{\partial {\mathbf{e}}_{r}} = \frac{\partial u}{\partial r},\;\frac{\partial u}{\partial {\mathbf{e}}_{\theta }} = \frac{1}{r}\frac{\partial u}{\partial \theta }. \]

例 6 设 \(n\) 为整数,若 \(\forall t > 0,f\left( {{tx},{ty}}\right) = {t}^{n}f\left( {x,y}\right)\) ,则称 \(f\) 是 \(n\) 次齐次函数. 证明: \(f\left( {x,y}\right)\) 是零次齐次函数的充要条件是 \[ x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = 0. \]

例 5 知 \[ \frac{\partial f}{\partial r} = \frac{\partial f}{\partial x} \cdot \cos \theta + \frac{\partial f}{\partial y}\sin \theta = \frac{1}{r}\left( {x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}}\right) = 0. \] 上式说明 \(f\) 在极坐标系里只是 \(\theta = \arctan \left( {y/x}\right)\) 的函数,这等价于 \(f\) 只是 \(y/x\) 的函数,不妨记 \(f\left( {x,y}\right) = \varphi \left( {y/x}\right)\) . 显然 \(\varphi\) 是零次齐次函数.

例 7 求函数 \(u = f\left( {{xy},\frac{y}{x}}\right)\) 的二阶偏导数.

例 8 设 \(u = f\left( r\right) ,r = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}\) ,若 \(u\) 满足调和方程 \[ {\nabla }^{2}u = \frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} + \frac{{\partial }^{2}u}{\partial {z}^{2}} = 0, \] 试求出函数 \(u\) .

例 9 求出函数 \(f\left( {x,y}\right) = \frac{x}{y}\) 在(1,1)点邻域带皮亚诺余项的泰勒公式.

例 10 设 \(f,g : {\mathbf{R}}^{m} \rightarrow {\mathbf{R}}^{n}\) 是可微函数. 试用复合函数求导法则来证明向量内积的求导公式: \[ \mathrm{D}\left( {\mathbf{f}\left( \mathbf{x}\right) \cdot \mathbf{g}\left( \mathbf{x}\right) }\right) = {\mathbf{f}}^{\mathrm{T}}\left( \mathbf{x}\right) \mathrm{D}\mathbf{g}\left( \mathbf{x}\right) + {\mathbf{g}}^{\mathrm{T}}\left( \mathbf{x}\right) \mathrm{D}\mathbf{f}\left( \mathbf{x}\right) . \]

例 11 设 \(\Omega \subset {\mathbf{R}}^{m}\) 是凸域, \(f\left( \mathbf{x}\right) \in {C}^{2}\left( {\Omega ,\mathbf{R}}\right)\) ,且满足 \[ f\left( \mathbf{x}\right) \geq f\left( {\mathbf{x}}_{0}\right) + \mathrm{D}f\left( {\mathbf{x}}_{0}\right) \left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) \;\left( {\forall \mathbf{x},{\mathbf{x}}_{0} \in \Omega }\right) , \]

例 1 设 \(x = x\left( {y,z}\right) ,y = y\left( {z,x}\right) ,z = z\left( {x,y}\right)\) 为由方程 \(F(x,y\) , \(z) = 0\) 所确定的隐函数. 证明: \[ \frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z} \cdot \frac{\partial z}{\partial x} = - 1 \]

例 2 求由方程 \(f\left( {x - y,y - z,z - x}\right) = 0\) 所确定的函数 \(z =\) \(z\left( {x,y}\right)\) 的微分.

例 3 变换 \(x + y = u,y = {uv}\) 把区域 \(\{ \left( {u,v}\right) \mid u > 0,v > 0\}\) 变为区域 \(\{ \left( {x,y}\right) \mid x + y > 0,y > 0\}\) . 试求雅可比行列式 \[ \frac{\partial \left( {x,y}\right) }{\partial \left( {u,v}\right) },\;\frac{\partial \left( {u,v}\right) }{\partial \left( {x,y}\right) }. \]

例 5 取 \(y\) 为因变量,解方程 \[ {\left( \frac{\partial z}{\partial y}\right) }^{2}\frac{{\partial }^{2}z}{\partial {x}^{2}} - 2\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}\frac{{\partial }^{2}z}{\partial x\partial y} + {\left( \frac{\partial z}{\partial x}\right) }^{2}\frac{{\partial }^{2}z}{\partial {y}^{2}} = 0. \]

例 6 设函数 \(u = u\left( {x,y}\right)\) 由方程 \[ u = f\left( {x,y,z,t}\right) ,\;g\left( {y,z,t}\right) = 0,\;h\left( {z,t}\right) = 0 \] 定义,求 \(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}\) .

例 1 求曲线 \({x}^{2} + {y}^{2} + {z}^{2} = 6,x + y + z = 0\) 在(1, - 2,1)点的切线方程.

例 2 求椭球面 \(\frac{{x}^{2}}{{a}^{2}} + \frac{{y}^{2}}{{b}^{2}} + \frac{{z}^{2}}{{c}^{2}} = 1\) 在 \(M\left( {{x}_{0},{y}_{0},{z}_{0}}\right)\) 点的切平面方程.

例 3 给定曲面 \(F\left( {\frac{x - a}{z - c},\frac{y - b}{z - c}}\right) = 0\left( {a,b,c\text{ 为常数 }}\right)\) ,或由它确定的曲面 \(z = z\left( {x,y}\right)\) . 证明: (1)曲面的切平面通过一个定点; (2)函数 \(z = z\left( {x,y}\right)\) 满足方程 \(\frac{{\partial }^{2}z}{\partial {x}^{2}} \cdot \frac{{\partial }^{2}z}{\partial {y}^{2}} - {\left( \frac{{\partial }^{2}z}{\partial x\partial y}\right) }^{2} = 0\) .

例 4 给定 \(f\left( {x,y}\right) = 2{\left( y - {x}^{2}\right) }^{2} - \frac{1}{7}{x}^{7} - {y}^{2}\) . (1)求 \(f\left( {x,y}\right)\) 的极值； (2)证明:沿(0,0)点的每条直线,(0,0)点都是定义在该直线上的函数 \(f\left( {x,y}\right)\) 的极小值点.

例 5 求证: 锐角三角形内一点到三顶点连线成等角时, 该点到三顶点距离之和为最小.

例 6 求由方程 \(2{x}^{2} + {y}^{2} + {z}^{2} + {2xy} - {2x} - {2y} - {4z} + 4 = 0\) 所确定的函数 \(z = z\left( {x,y}\right)\) 的极值.

例 7 求椭圆 \(5{x}^{2} + {4xy} + 2{y}^{2} = 1\) 的长半轴 \(a\) 和短半轴 \(b\) .

例 8 给定 \(n\) 阶行列式 \(A = \left| {a}_{ij}\right|\) . (1)求证: 在行向量长有限条件下,即 \(\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}^{2} = {s}_{i}\left( {i = 1,\cdots ,n}\right)\) 条件下,使 \(A\) 达到最大值的列向量两两正交; (2) 求证: \[ {A}^{2} \leq \mathop{\prod }\limits_{{i = 1}}^{n}\left( {\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}^{2}}\right) . \]

例 1 设 \(f\left( x\right)\) 是周期为 \({2\pi }\) 的连续函数,令 \[ F\left( x\right) = \frac{1}{2h}{\int }_{x - h}^{x + h}f\left( t\right) \mathrm{d}t, \] 试求 \(F\left( x\right)\) 的傅里叶系数.

例 2 利用 \(\displaystyle{\frac{1}{2\pi }{\int }_{0}^{2\pi }\frac{1 - {r}^{2}}{1 - {2r}\cos x + {r}^{2}}\mathrm{\;d}x = 1}\) ,试计算积分 \[ I\left( r\right) = {\int }_{0}^{2\pi }\ln \left( {1 - {2r}\cos x + {r}^{2}}\right) \mathrm{d}x\;\left( {r > 0,r \neq 1}\right) . \]

解 \(\forall R > 0\) ,考虑矩形 \(\left| x\right| \leq R,\left| t\right| \leq R\) . 函数 \(\displaystyle{\int }_{t}^{x}{\mathrm{e}}^{-{s}^{2}}\mathrm{\;d}s}\) 及其对 \(x\) 的偏导数 \({\mathrm{e}}^{-{x}^{2}}\) 在矩形上连续, \(\varphi \left( x\right) = 0,\psi \left( x\right) = x\) 显然符合定理 1 中条件, 于是有 \[ {f}^{\prime }\left( x\right) = {\int }_{0}^{x}{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}t + {\int }_{x}^{x}{\mathrm{e}}^{-{s}^{2}}\mathrm{\;d}s = x{\mathrm{e}}^{-{x}^{2}}. \] 因 \(f\left( 0\right) = 0\) ,所以 \[ f\left( x\right) = {\int }_{0}^{x}t{\mathrm{e}}^{-{t}^{2}}\mathrm{\;d}t = \frac{1}{2} - \frac{1}{2}{\mathrm{e}}^{-{x}^{2}}. \]

解法 1 因 \[ \frac{{\mathrm{e}}^{-{ax}} - {\mathrm{e}}^{-{bx}}}{x} = {\int }_{a}^{b}{\mathrm{e}}^{-{tx}}\mathrm{\;d}t\;\left( {x \geq 0}\right) , \] 所以 \[ I = {\int }_{0}^{+\infty }\frac{{\mathrm{e}}^{-{ax}} - {\mathrm{e}}^{-{bx}}}{x}\mathrm{\;d}x = {\int }_{0}^{+\infty }{\int }_{a}^{b}{\mathrm{e}}^{-{tx}}\mathrm{\;d}t\mathrm{\;d}x. \] 由于函数 \({\mathrm{e}}^{-{tx}}\) 在 \(x \geq 0,a \leq t \leq b\) 上连续,又 \(\left| {\mathrm{e}}^{-{tx}}\right| \leq {\mathrm{e}}^{-{ax}}\) 及 \[ {\int }_{0}^{+\infty }{\mathrm{e}}^{-{ax}}\mathrm{\;d}x < + \infty , \] 由魏尔斯特拉斯判别法,积分 \(\displaystyle{\int }_{0}^{+\infty }{\mathrm{e}}^{-{tx}}\mathrm{\;d}x}\) 在 \(\left\lbrack {a,b}\right\rbrack\) 上一致收敛,故 \[ I = {\int }_{a}^{b}{\int }_{0}^{+\infty }{\mathrm{e}}^{-{tx}}\mathrm{\;d}x\mathrm{\;d}t = {\int }_{a}^{b}\frac{\mathrm{d}t}{t} = \ln \frac{b}{a}. \] 解法 2 引入参数 \(t\) , \[ I\left( t\right) = {\int }_{0}^{+\infty }\frac{{\mathrm{e}}^{-{tx}} - {\mathrm{e}}^{-{bx}}}{x}\mathrm{\;d}x, \] 函数 \(f\left( {t,x}\right) = \frac{{\mathrm{e}}^{-{tx}} - {\mathrm{e}}^{-{bx}}}{x}\) 及 \({f}_{i}^{\prime }\left( {t,x}\right) = - {\mathrm{e}}^{-{tx}}\) 在 \(x \geq 0,a \leq t \leq b\) 上连续,积分 \(\displaystyle{\int }_{0}^{+\infty }f\left( {t,x}\right) \mathrm{d}x\) 收敛,积分 \[ {\int }_{0}^{+\infty }{f}_{t}^{\prime }\left( {t,x}\right) \mathrm{d}x = - {\int }_{0}^{+\infty }{\mathrm{e}}^{-{tx}}\mathrm{\;d}x \] 在 \(\left\lbrack {a,b}\right\rbrack\) 上一致收敛,故 \[ {I}^{\prime }\left( t\right) = {\int }_{0}^{+\infty } - {\mathrm{e}}^{-{tx}}\mathrm{\;d}x = - \frac{1}{t}. \] 解出 \(I\left( t\right) = - \ln t + C\) . 令 \(t = b\) ,定出任意常数 \(C = \ln b\) ,故 \(I\left( t\right) = \ln \frac{b}{t}\) . 所求积分 \(I = I\left( a\right) = \ln \frac{b}{a}\) . 解法 3 令 \(x = - \frac{2}{a}\ln t\) ,把含参变量的广义积分变为含参变量的定积分: \[ I = {\int }_{0}^{+\infty }\frac{{\mathrm{e}}^{-{ax}} - {\mathrm{e}}^{-{bx}}}{x}\mathrm{\;d}x = {\int }_{0}^{1}\frac{{t}^{\frac{2b}{a} - 1} - t}{\ln t}\mathrm{\;d}t. \] 因 \[ \frac{{t}^{\frac{2b}{a} - 1} - 1}{\ln t} = {\int }_{1}^{\frac{2b}{a} - 1}{t}^{x}\mathrm{\;d}x, \] 所以 \[ I = {\int }_{0}^{1}{\int }_{1}^{\frac{2b}{a} - 1}{t}^{x}\mathrm{\;d}x\mathrm{\;d}t = {\int }_{1}^{\frac{2b}{a} - 1}{\int }_{0}^{1}{t}^{x}\mathrm{\;d}t\mathrm{\;d}x \] \[ = {\int }_{1}^{\frac{2b}{a} - 1}\frac{\mathrm{d}x}{x + 1} = \ln \frac{b}{a}. \] 解法 4 由广义积分定义与定积分第一中值定理得 \[ I = \mathop{\lim }\limits_{\substack{{N \rightarrow + \infty } \\ {\varepsilon \rightarrow {0}^{ + }} }}{\int }_{\varepsilon }^{N}\frac{{\mathrm{e}}^{-{ax}} - {\mathrm{e}}^{-{bx}}}{x}\mathrm{\;d}x \] \[ = \mathop{\lim }\limits_{\substack{{N \rightarrow + \infty } \\ {\varepsilon \rightarrow {0}^{ + }} }}\left\lbrack {{\int }_{a\varepsilon }^{aN}\frac{{\mathrm{e}}^{-t}}{t}\mathrm{\;d}t - {\int }_{b\varepsilon }^{bN}\frac{{\mathrm{e}}^{-t}}{t}\mathrm{\;d}t}\right\rbrack \] \[ = \mathop{\lim }\limits_{\substack{{N \rightarrow + \infty } \\ {\varepsilon \rightarrow {0}^{ + }} }}\left\lbrack {{\int }_{a\varepsilon }^{b\varepsilon }\frac{{\mathrm{e}}^{-t}}{t}\mathrm{\;d}t - {\int }_{aN}^{bN}\frac{{\mathrm{e}}^{-t}}{t}\mathrm{\;d}t}\right\rbrack \] \[ = \mathop{\lim }\limits_{\substack{{N \rightarrow + \infty } \\ {\varepsilon \rightarrow {0}^{ + }} }}\left\lbrack {{\mathrm{e}}^{-\xi }\ln \frac{b}{a} - {\mathrm{e}}^{-\eta }\ln \frac{b}{a}}\right\rbrack \] \[ \left( {{a\varepsilon } \leq \xi \leq {b\varepsilon },{aN} \leq \eta \leq {bN}}\right) \] \[ = \ln \frac{b}{a}. \]

解 (1) 令 \({x}^{1/4} = t\) ,有 \[ {\int }_{0}^{1}\frac{\mathrm{d}x}{\sqrt{1 - {x}^{1/4}}} = {\int }_{0}^{1}4{t}^{3}{\left( 1 - t\right) }^{-1/2}\mathrm{\;d}t = 4\mathrm{\;B}\left( {4,1/2}\right) \] \[ = 4\frac{\Gamma \left( 4\right) \Gamma \left( {1/2}\right) }{\Gamma \left( {4 + 1/2}\right) } \] \[ = 4\frac{3!\sqrt{\pi }}{\left( {3 + 1/2}\right) \left( {2 + 1/2}\right) \left( {1 + 1/2}\right) \sqrt{\pi }} \] \[ = \frac{64}{35}\text{ . } \] (2) \(\displaystyle{\int }_{0}^{\frac{\pi }{2}}\sqrt{\tan x}\mathrm{\;d}x = {\int }_{0}^{\frac{\pi }{2}}{\sin }^{\frac{1}{2}}x{\cos }^{-\frac{1}{2}}x\mathrm{\;d}x = \frac{1}{2}\mathrm{\;B}\left( {\frac{3}{4},\frac{1}{4}}\right)\) \[ = \frac{1}{2}\frac{\Gamma \left( {3/4}\right) \Gamma \left( {1/4}\right) }{\Gamma \left( 1\right) } = \frac{1}{2} \cdot \frac{\pi }{\sin \pi /4} = \frac{\pi }{\sqrt{2}}. \] (3)解法 1 令 \({x}^{2} = t\) ,我们有 \[ {\int }_{0}^{+\infty }{x}^{2n}{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x = {\int }_{0}^{+\infty }{t}^{n}{\mathrm{e}}^{-t}\frac{\mathrm{d}t}{2\sqrt{t}} = \frac{1}{2}{\int }_{0}^{+\infty }{t}^{n - \frac{1}{2}}{\mathrm{e}}^{-t}\mathrm{\;d}t \] \[ = \frac{1}{2}\Gamma \left( {n + \frac{1}{2}}\right) = \frac{\left( {{2n} - 1}\right) !!}{{2}^{n + 1}}\sqrt{\pi }. \] 解法 2 因 \[ {\int }_{0}^{+\infty }{\mathrm{e}}^{-a{x}^{2}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2}{a}^{-\frac{1}{2}}, \] 上式对 \(a\) 求导得 \[ {\int }_{0}^{+\infty }{x}^{2}{\mathrm{e}}^{-a{x}^{2}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2} \cdot \frac{1}{2}{a}^{-\frac{3}{2}}. \] 再对 \(a\) 微一次得 \[ {\int }_{0}^{+\infty }{x}^{4}{\mathrm{e}}^{-a{x}^{3}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2} \cdot \frac{3!!}{{2}^{2}}{a}^{-\frac{5}{2}}. \] 依次微下去,微到第 \(n\) 次得 \[ {\int }_{0}^{+\infty }{x}^{2n}{\mathrm{e}}^{-a{x}^{2}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2} \cdot \frac{\left( {{2n} - 1}\right) !!}{{2}^{n}}{a}^{-\frac{{2n} + 1}{2}}. \] 令 \(a = 1\) ,即得 \[ {\int }_{0}^{+\infty }{x}^{2n}{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x = \frac{\left( {{2n} - 1}\right) !!}{{2}^{n + 1}}\sqrt{\pi }. \]

证 (1) 因 \(\left| \frac{\cos {xt}}{1 + {x}^{2}}\right| \leq \frac{1}{1 + {x}^{2}}\) ,及 \(\displaystyle{\int }_{1}^{+\infty }\frac{\mathrm{d}x}{1 + {x}^{2}} < + \infty}\) ,所以积分在 \(\displaystyle{- \infty < t < + \infty}\) 上一致收敛. (2)因 \(\frac{\cos {xt}}{1 + {x}^{2}}\) 在 \(x \geq 1,\left| t\right| \leq R\) 上连续,由连续性定理知, \(f\left( t\right)\) 在 \(\left| t\right| \leq R\) 上连续,再由 \(R\) 的任意性,有 \(f\left( t\right) \in C\left( {-\infty , + \infty }\right)\) . (3)由积分一致收敛性, \(\forall \varepsilon > 0,\exists A,\forall t\) ,有 \[ \left| {{\int }_{A}^{+\infty }\frac{\cos {xt}}{1 + {x}^{2}}\mathrm{\;d}x}\right| < \varepsilon . \tag{6.1} \] 再由黎曼引理①知 \[ \mathop{\lim }\limits_{{t \rightarrow \infty }}{\int }_{1}^{A}\frac{1}{1 + {x}^{2}}\cos {xt}\mathrm{\;d}x = 0, \] 所以 \(\exists T > 0\) ,当 \(\left| t\right| > T\) 时,有 \[ \left| {{\int }_{1}^{A}\frac{\cos {xt}}{1 + {x}^{2}}\mathrm{\;d}x}\right| < \varepsilon \tag{6.2} \] 结合 (6.1) 与 (6.2) 式得到当 \(\left| t\right| > T\) 时,有 \[ \left| {{\int }_{1}^{+\infty }\frac{\cos {xt}}{1 + {x}^{2}}\mathrm{\;d}x}\right| < {2\varepsilon }, \] 即 \[ \mathop{\lim }\limits_{{t \rightarrow \infty }}f\left( t\right) = \mathop{\lim }\limits_{{t \rightarrow \infty }}{\int }_{1}^{+\infty }\frac{\cos {xt}}{1 + {x}^{2}}\mathrm{\;d}x = 0. \] (4)要得结论,只需证 \(f\left( \pi \right) \leq 0\) ,或证 \(\displaystyle{\int }_{0}^{\pi }f\left( t\right) \mathrm{d}t \leq 0\) ,或证 \[ {\int }_{0}^{\pi }\sin t \cdot f\left( t\right) \mathrm{d}t \leq 0. \] 我们来证最后一个不等式. 因函数 \(\frac{\sin \left( t\right) \cos \left( {xt}\right) }{1 + {x}^{2}}\) 在 \(x \geq 1,0 \leq t \leq \pi\) 上连续,积分 \[ {\int }_{1}^{\infty }\frac{\sin \left( t\right) \cos \left( {xt}\right) }{1 + {x}^{2}}\mathrm{\;d}x \] 对 \(t\) 一致收敛,所以可以交换求积次序,得 \[ {\int }_{0}^{\pi }\sin t \cdot f\left( t\right) \mathrm{d}t = {\int }_{1}^{\infty }{\int }_{0}^{\pi }\frac{\cos \left( {xt}\right) \sin \left( t\right) }{1 + {x}^{2}}\mathrm{\;d}t\mathrm{\;d}x \] \[ = \frac{1}{2}{\int }_{1}^{\infty }{\int }_{0}^{\pi }\frac{\sin \left( {x + 1}\right) t - \sin \left( {x - 1}\right) t}{1 + {x}^{2}}\mathrm{\;d}t\mathrm{\;d}x \] \[ = {\int }_{1}^{\infty }\frac{1 + \cos {\pi x}}{1 - {x}^{4}}\mathrm{\;d}x \leq 0. \] 注意负值函数 \(\frac{1 + \cos {\pi x}}{1 - {x}^{4}}\) 在 \(x = 1\) 点只要补充定义后即连续. 应用定积分第一中值定理, 得 \[ {\int }_{0}^{\pi }\sin f\left( t\right) \mathrm{d}t = f\left( \xi \right) {\int }_{0}^{\pi }\sin t\mathrm{\;d}t = {2f}\left( \xi \right) \leq 0.\;\left( {0 \leq \xi \leq \pi }\right) . \] 若 \(f\left( \xi \right) = 0\) ,命题得证. 若 \(f\left( \xi \right) < 0\) ,由 \(f\left( 0\right) = {\int }_{1}^{\infty }\frac{\mathrm{d}x}{1 + {x}^{2}} = \frac{\pi }{4} > 0\) 及连续函数的介值定理,知存在 \(\eta \left( {0 \leq \eta \leq \xi }\right)\) ,使 \(f\left( \eta \right) = 0\) . 命题得证.

5. 1.1 \(\forall x,y \in {\mathbf{R}}^{m}\) . 证明: \({\left| x + y\right| }^{2} + {\left| x - y\right| }^{2} = 2\left( {{\left| x\right| }^{2} + {\left| y\right| }^{2}}\right)\) ,并说明等式的几何意义.

5.1.2 证明下列三个命题等价: (1) \(x \cdot y = 0\) ; (2) \({\left| x - y\right| }^{2} = {\left| x\right| }^{2} + {\left| y\right| }^{2}\) ； (3) \(\left| {x - y}\right| = \left| {x + y}\right|\) .

5.1.3 设 \(\mathbf{x} \in {\mathbf{R}}^{m}\) 为常向量, \(c\) 为常数,证明: (1) \(H = \left\{ {\mathbf{x} \mid \mathbf{x} \in {\mathbf{R}}^{m},\mathbf{x} \cdot \mathbf{z} < c}\right\}\) 是开集； (2) \(\left\{ {x \mid x \in {\mathbb{R}}^{m},x \cdot z \geq c}\right\}\) 是闭集.

5.1.4 试画出下列集合 \(\mathbf{\Omega }\) 的图形: (1) \(\{ \left( {x,y}\right) \mid y > 0,x > y,x < 1\}\) ； (2) \(\{ \left( {x,y}\right) \mid 0 \leq y \leq 2,{2y} \leq x \leq {2y} + 2\}\) ； (3) \(\{ \left( {x,y}\right) \mid 1 \leq {xy} \leq 2,1/2 \leq y/x \leq 1\}\) ； (4) \(\{ \left( {x,y,z}\right) \mid 0 < x < y < z < 1\}\) .

5.1.5 证明: (1) \({\left( A \cup B\right) }^{ \circ } \supset {A}^{ \circ } \cup {B}^{ \circ }\) ； (2) \(\overline{A \cap B} \subset \overline{A} \cup \overline{B}\) .

5.1.6 设 \(A,B\) 为 \({\mathbf{R}}^{m}\) 中的有界集. 证明: (1) \(\partial \left( {A \cup B}\right) \subset \partial A \cup \partial B\) ； (2) \(\partial \left( {A \cap B}\right) \subset \partial A \cup \partial B\) .

5.1.7 设 \(A,B\) 是 \({\mathbf{R}}^{m}\) 中不相交的闭集,求证: 存在开集 \(W\) 和 \(V\) ,满足 \(A \subset\) \(W,B \subset V\) ,而 \(W \cap V = \varnothing\) .

5.1.8 设 \(E \subset {\mathbf{R}}^{m}\) ,证明: \(E = \{ x \mid \rho \left( {x,E}\right) = 0\}\) .

5. 1.9 对 \(\forall {x}_{n} \in {F}_{n}\) ,则 \(\left\{ {x}_{n}\right\}\) 为有界点列,证它是哥西序列.

5.1.10 设 \(E,F\) 为 \({\mathbf{R}}^{m}\) 中的闭集, \(E,F\) 中至少有一为有界集,求证: \(\exists x \in\) \(E,y \in F\) ,使得 \(\rho \left( {x,y}\right) = \rho \left( {E,F}\right)\) .

5.1.11 设 \(D\) 为 \({\mathbf{R}}^{m}\) 中的凸集,证明 \(\bar{D}\) 也是凸集.

5.1.12 证明: (1) \(\left| {x - 2\left( {x \cdot a}\right) \frac{a}{{\left| a\right| }^{2}}}\right| = \left| x\right| \left( {a \neq 0}\right)\) ； (2) \(\left| {\frac{x}{{\left| x\right| }^{2}} - \frac{y}{{\left| y\right| }^{2}}}\right| = \frac{\left| x - y\right| }{\left| x\right| \left| y\right| }\) ； (3) \(\left| \mathbf{x}\right| \left| {\mathbf{y} - \mathbf{x}/{\left| \mathbf{x}\right| }^{2}}\right| = \left| \mathbf{y}\right| \left| {\mathbf{x} - \mathbf{y}/{\left| \mathbf{y}\right| }^{2}}\right|\) .

5. 1.13 确定并画出下列函数的定义域, 指出后两题的等位面是什么曲面 (或曲线): (1) \(u = \sqrt{1 - {x}^{2}} + \sqrt{1 - {y}^{2}}\) ; (2) \(u = \sqrt{\frac{{2x} - {x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2} - x}}\) ; (3) \(u = \arcsin \frac{y}{x}\) ; (4) \(u = \ln \left( {-1 - {x}^{2} - {y}^{2} + {z}^{2}}\right)\) .

5.1.14 求下列函数的极限: (1) \(\mathop{\lim }\limits_{{\left( {x,y}\right) \rightarrow \left( {0,0}\right) }}\frac{{\mathrm{e}}^{x} + {\mathrm{e}}^{y}}{\cos x + \sin y}\) ; (2) \(\mathop{\lim }\limits_{{\left( {x,y}\right) \rightarrow \left( {0,0}\right) }}\frac{{x}^{2}{y}^{3/2}}{{x}^{4} + {y}^{2}}\) ; (3) \(\mathop{\lim }\limits_{\substack{{x \rightarrow + \infty } \\ {y \rightarrow + \infty } }}\left( {{x}^{2} + {y}^{2}}\right) {\mathrm{e}}^{-\left( {x + y}\right) }\) ; (4) \(\mathop{\lim }\limits_{{\left( {x,y}\right) \rightarrow \left( {0,0}\right) }}\frac{\sin \left( {{x}^{3} + {y}^{3}}\right) }{{x}^{2} + {y}^{2}}\) .

5. 1.15 对下列函数 \(f\left( {x,y}\right)\) ,证明 \(\mathop{\lim }\limits_{{\left( {x,y}\right) \rightarrow \left( {0,0}\right) }}f\left( {x,y}\right)\) 不存在: (1) \(f\left( {x,y}\right) = \frac{{x}^{2}}{{x}^{2} + {y}^{2}}\) ; (2) \(f\left( {x,y}\right) = \frac{{x}^{2}{y}^{2}}{{x}^{3} + {y}^{3}}\) .

5.1.16 问下列函数是否在全平面连续, 为什么? (1) \(f\left( {x,y}\right) = \left\{ \begin{matrix} \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}}, & {x}^{2} + {y}^{2} \neq 0, \\ 0, & {x}^{2} + {y}^{2} = 0; \end{matrix}\right.\) (2) \(f\left( {x,y}\right) = \left\{ \begin{array}{ll} \frac{\sin \left( {xy}\right) }{x}, & x \neq 0, \\ y, & x = 0; \end{array}\right.\) (3) \(f\left( {x,y}\right) = \left\{ \begin{array}{ll} \frac{{x}^{2}}{{y}^{2}}{\mathrm{e}}^{-\frac{{x}^{4}}{{y}^{2}}}, & y \neq 0, \\ 0, & y = 0; \end{array}\right.\) (4) \(f\left( {x,y}\right) = \left\{ \begin{array}{ll} {y}^{2}\ln \left( {{x}^{2} + {y}^{2}}\right) , & {x}^{2} + {y}^{2} \neq 0, \\ 0, & {x}^{2} + {y}^{2} =

0. \end{array}\right.\) 5.1.17 设函数 \(f\left( {x,y}\right)\) 在开半平面 \(x > 0\) 上连续,且对 \(\forall {y}_{0}\) ,极限 \[ \mathop{\lim }\limits_{\substack{{x \rightarrow {0}^{ + }} \\ {y \rightarrow {y}_{0}} }}f\left( {x,y}\right) = \varphi \left( {y}_{0}\right) \] 存在. 当函数 \(f\) 在 \(y\) 轴上补充定义 \(\varphi \left( y\right)\) 后,证明: 函数 \(f\left( {x,y}\right)\) 在闭半平面 \(x \geq 0\) 上连续.

5. 1.18 设函数 \(f\left( {x,y}\right)\) 在开半平面 \(x > 0\) 上一致连续. 证明: (1) \(\forall {y}_{0}\) ,极限 \(\mathop{\lim }\limits_{\substack{{x \rightarrow {0}^{ + }} \\ {y \rightarrow {y}_{0}} }}f\left( {x,y}\right) = \varphi \left( {y}_{0}\right)\) 存在; (2)函数在 \(y\) 轴上补充定义 \(\varphi \left( y\right)\) 后,所得函数 \(f\left( {x,y}\right)\) 在 \(x \geq 0\) 上一致连续.

5. 1.19 设 \(u = f\left( \mathbf{x}\right)\) 在 \({\mathbf{x}}_{0} \in {\mathbf{R}}^{m}\) 点连续,且 \(f\left( {\mathbf{x}}_{0}\right) > 0\) . 证明: 存在 \({\mathbf{x}}_{0}\) 的一个邻域 \(U\left( {{\mathbf{x}}_{0};\delta }\right)\) ,使得 \(f\left( \mathbf{x}\right)\) 在 \(U\left( {{\mathbf{x}}_{0};\delta }\right)\) 上取正值.

5.1.20 设 \(E\) 是 \({\mathbf{R}}^{m}\) 中任意点集,求证: \(\rho \left( {\mathbf{x},E}\right)\) 在 \({\mathbf{R}}^{m}\) 上一致连续.

5.1.21 设 \(f\left( \mathbf{x}\right) \in C\left( {{\mathbf{R}}^{m},\mathbf{R}}\right)\) ,对任意实数 \(\alpha\) ,作集合 \[ G = \{ \mathbf{x} \mid f\left( \mathbf{x}\right) > \alpha \} ,\;F = \{ \mathbf{x} \mid f\left( \mathbf{x}\right) \geq \alpha \} . \] 求证: \(G\) 是 \({\mathbf{R}}^{m}\) 中的开集, \(F\) 是 \({\mathbf{R}}^{m}\) 中的闭集.

5.1.22 设 \(x \in {R}^{m},x = \left( {{x}_{1},{x}_{2},\cdots ,{x}_{m}}\right)\) . 求证: (1) \(\exists a > 0,b > 0\) ,使得 \(\displaystyle{a\left| \mathbf{x}\right| \leq \mathop{\sum }\limits_{{i = 1}}^{m}\left| {x}_{i}\right| \leq b\left| \mathbf{x}\right|}\) ； (2) \(\exists a > 0,b > 0\) ,使得 \(\displaystyle{a\left| \mathbf{x}\right| \leq \mathop{\max }\limits_{{1 \leq i \leq m}}\left| {x}_{i}\right| \leq b\left| \mathbf{x}\right|}\) .

5.1.23 设 \(A\) 是 \(m \times m\) 矩阵, \(\det A \neq 0\) ,求证: \(\exists a > 0\) ,使得 \[ \left| {Ax}\right| \geq a\left| x\right| \;\left( {\forall x \in {\mathbf{R}}^{m}}\right) . \]

5.1.24 设 \(\bar{\Omega } \subset {\mathbf{R}}^{m}\) 是有界闭区域, \(\mathbf{f}\left( \mathbf{x}\right) \in C\left( {\bar{\Omega },{\mathbf{R}}^{m}}\right)\) ,且是单叶的. 求证: \({f}^{-1}\left( x\right)\) 在 \(f\left( \bar{\Omega }\right)\) 上连续.

5. 1.25 设 \(f\left( {x,y}\right)\) 除直线 \(x = a\) 与 \(y = b\) 外有定义,且满足: (1) \(\mathop{\lim }\limits_{{y \rightarrow b}}f\left( {x,y}\right) = \varphi \left( x\right)\) 存在; (2) \(\mathop{\lim }\limits_{{x \rightarrow a}}f\left( {x,y}\right) = \psi \left( y\right)\) 一致存在 (即 \(\forall \varepsilon > 0,\exists \delta \left( \varepsilon \right) > 0\) ,当 \(0 < \left| {x - a}\right| < \delta\) 时, \(\forall y \neq b\) ,有 \(\left| {f\left( {x,y}\right) - \psi \left( y\right) }\right| < \varepsilon\) ). 证明: (1) 累次极限 \(\mathop{\lim }\limits_{{x \rightarrow a}}\mathop{\lim }\limits_{{y \rightarrow b}}f\left( {x,y}\right) = \mathop{\lim }\limits_{{x \rightarrow a}}\varphi \left( x\right) = c\) 存在; (2)累次极限 \(\mathop{\lim }\limits_{{y \rightarrow b}}\mathop{\lim }\limits_{{x \rightarrow a}}f\left( {x,y}\right) = \mathop{\lim }\limits_{{y \rightarrow b}}\psi \left( y\right) = c\) ； (3) 全面极限 \(\mathop{\lim }\limits_{{\left( {x,y}\right) \rightarrow \left( {a,b}\right) }}f\left( {x,y}\right) = c\) .

5. 2.1 求下列函数的偏导数: (1) \(u = \frac{x}{\sqrt{{x}^{2} + {y}^{2}}}\) ; (2) \(u = \tan \frac{{x}^{2}}{y}\) ; (3) \(u = \sin \left( {x\cos y}\right)\) ; (4) \(u = {\mathrm{e}}^{x/y}\) ; (5) \(u = \ln \sqrt{{x}^{2} + {y}^{2}}\) ; (6) \(u = \arctan \frac{x + y}{1 - {xy}}\) ; (7) \(u = {\left( \frac{x}{y}\right) }^{x}\) ; (8) \(u = \arccos \frac{z}{\sqrt{{x}^{2} + {y}^{2}}}\) .

5.2.2 设 \(f\left( {x,y}\right)\) 在圆 \(\Omega\) 上的偏导数 \({f}_{x}^{\prime },{f}_{y}^{\prime }\) 存在且有界. 证明: \(f\left( {x,y}\right)\) 在 \(\Omega\) 上一致连续. 若 \(\Omega\) 是任意区域,问结论是否成立. 考查例子 \[ f\left( {x,y}\right) = \arctan \left( {y/x}\right) , \] \(\Omega\) 用极坐标表示为 \(1 < r < 2,0 < \theta < {2\pi }\) .

5.2.3 设 \[ f\left( {x,y}\right) = \left\{ \begin{matrix} \frac{{x}^{2}y}{{x}^{2} + {y}^{2}}, & {x}^{2} + {y}^{2} \neq 0, \\ 0, & {x}^{2} + {y}^{2} =

5.2.4 设 \[ f\left( {x,y}\right) = \left\{ \begin{array}{ll} \frac{\sin \left( {xy}\right) }{x}, & x \neq 0, \\ y, & x = 0, \end{array}\right. \] 证明: \(f\left( {x,y}\right)\) 在平面上可微.

5.2.5 求下列复合函数的偏导数: (1) \(u = f\left( \frac{xz}{y}\right)\) ; (2) \(u = f\left( {x + y,z}\right)\) ; (3) \(u = f\left( {x,{xy},{xyz}}\right)\) ; (4) \(u = f\left( {x + y + z,{x}^{2} + {y}^{2} + {z}^{2}}\right)\) ; (5) \(u = f\left( {\frac{x}{y},\frac{y}{z}}\right)\) ; (6) \(u = f\left( {{x}^{2} + {y}^{2},{x}^{2} - {y}^{2},{2xy}}\right)\) .

5.2.6 设 \(u = {x}^{n}f\left( {\frac{y}{x},\frac{z}{x}}\right)\) ,其中 \(f\) 可微. 证明 \(u\) 满足方程: \[ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} = n \cdot u. \]

5.2.7 证明: \(f\left( {x,y,z}\right)\) 为 \(n\) 次齐次函数的充要条件是 \[ x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} + z\frac{\partial f}{\partial z} = {nf}\left( {x,y,z}\right) . \]

5.2.8 作自变量变换: \(x = \sqrt{vw},y = \sqrt{wu},z = \sqrt{uv}\) ,它把函数 \(f(x,y\) , \(z)\) 变为 \(F\left( {u,v,w}\right)\) . 证明: \[ x{f}_{x}^{\prime } + y{f}_{y}^{\prime } + z{f}_{z}^{\prime } = u{F}_{u}^{\prime } + v{F}_{v}^{\prime } + w{F}_{w}^{\prime }. \]

5.2.9 令 \(\xi = {2xy},\eta = {x}^{2} - {y}^{2}\) ,解下列方程 (解可含任意函数): (1) \(y\frac{\partial u}{\partial x} + x\frac{\partial u}{\partial y} = 0\) ; (2) \(x\frac{\partial u}{\partial x} - y\frac{\partial u}{\partial y} = 0\) .

5.2.10 令 \(\xi = x,\eta = y - x,\zeta = z - x\) ,求方程 \(\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0\) 的解.

5.2.11 设 \(u\left( {x,y}\right) ,v\left( {x,y}\right)\) 为连续可微函数,且满足方程组 \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\;\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}. \] 作自变量变换: \(x = r\cos \theta ,y = r\sin \theta\) ,证方程组变为 \[ \frac{\partial u}{\partial r} = \frac{1}{r}\frac{\partial v}{\partial \theta },\;\frac{1}{r}\frac{\partial u}{\partial \theta } = - \frac{\partial v}{\partial r}. \]

5.2.12 再对上题所得方程组作变换: \(R = \sqrt{{u}^{2} + {v}^{2}},\Phi = \arctan \frac{v}{u}\) . 证明方程组变为 \[ \frac{\partial \ln R}{\partial r} = \frac{1}{r}\frac{\partial \Phi }{\theta },\;\frac{1}{r}\frac{\partial \ln R}{\partial \theta } = - \frac{\partial \Phi }{\partial r}. \]

5.2.13 设 \(f\left( {x,y}\right) = {x}^{2} - {xy} + {y}^{2},\left( {{x}_{0},{y}_{0}}\right) = \left( {1,1}\right)\) . (1)若方向 \(l\) 与基 \({e}_{1},{e}_{2}\) 的夹角为 \(\pi /3\) 和 \(\pi /6\) ,求方向导数 \(\frac{\partial f\left( {1,1}\right) }{\partial l}\) ； ( 2 )求在怎样的方向上方向导数 \(\frac{\partial f\left( {1,1}\right) }{\partial l}\) 有最大值、最小值、等于零.

5.2.14 设 \(u = f\left( {x,y,z}\right)\) ,令 \[ x = r\sin \varphi \cos \theta ,\

5.2.22 证明: 函数 \(u = \frac{1}{{2a}\sqrt{\pi t}}{\mathrm{e}}^{-\frac{{\left( x - b\right) }^{2}}{4{a}^{2}t}}\) ( \(a,b\) 为实数) 当 \(t > 0\) 时满足方程 \[ \frac{\partial u}{\partial t} = {a}^{2}\frac{{\partial }^{2}u}{\partial {x}^{2}}. \]

5.2.23 设 \(x = f\left( {u,v}\right) ,y = g\left( {u,v}\right)\) 满足方程 \[ \frac{\partial f}{\partial u} = \frac{\partial g}{\partial v},\;\frac{\partial f}{\partial v} = - \frac{\partial g}{\partial u}, \] 又设 \(w = w\left( {x,y}\right)\) 满足方程 \(\frac{{\partial }^{2}w}{\partial {x}^{2}} + \frac{{\partial }^{2}w}{\partial {y}^{2}} = 0\) . 证明: (1) 函数 \(w = w\left\lbrack {f\left( {u,v}\right) ,g\left( {u,v}\right) }\right\rbrack\) 满足方程: \(\frac{{\partial }^{2}w}{\partial {u}^{2}} + \frac{{\partial }^{2}w}{\partial {v}^{2}} = 0\) ; (2) \(\frac{{\partial }^{2}\left( {fg}\right) }{\partial {u}^{2}} + \frac{{\partial }^{2}\left( {fg}\right) }{\partial {v}^{2}} = 0\) .

5.2.24 作变量替换 \(\xi = x + t,\eta = x - t\) ,求解方程 \(\frac{{\partial }^{2}u}{\partial {t}^{2}} = \frac{{\partial }^{2}u}{\partial {x}^{2}}\) ,并验证之.

5.2.25 求下列函数在(0,0)点邻域展开为带皮亚诺余项的四阶泰勒公式: (1) \(u = \sin \left( {{x}^{2} + {y}^{2}}\right)\) ; (2) \(u = \sqrt{1 + {x}^{2} + {y}^{2}}\) ; (3) \(u = \ln \left( {1 + x}\right) \ln \left( {1 + y}\right)\) ; (4) \(u = {\mathrm{e}}^{x}\cos y\) .

5.2.26 设函数 \(f\left( {x,y}\right)\) 满足 \(\frac{{\partial }^{2}f}{\partial {x}^{2}} = y,\frac{{\partial }^{2}f}{\partial x\partial y} = x + y,\frac{{\partial }^{2}f}{\partial {y}^{2}} = x\) ,试求出函数 \(f\left( {x,y}\right)\) .

5.2.27 设 \(\Omega\) 为含原点的凸域, \(u = f\left( {x,y}\right)\) 在 \(\Omega\) 上可微,且满足 \[ x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} =

5.2.28 求下列函数 \(f\left( \mathbf{x}\right) \left( {\mathbf{x} \in {\mathbf{R}}^{m}}\right)\) 的微分: (1) \(f\left( x\right) = \left( {{Ax} - b}\right) \cdot \left( {{Ax} - b}\right)\) ,其中 \(A\) 为 \(n \times m\) 矩阵, \(b \in {R}^{n}\) ； (2) \(f\left( x\right) = \frac{1}{\left| x\right| }\) .

5.2.29 设 \(f : {\mathbf{R}}^{m} \rightarrow {\mathbf{R}}^{l},g : {\mathbf{R}}^{m} \rightarrow {\mathbf{R}}^{n}\) 是可微函数. 试用复合函数求导公式,证 明公式 \[ \mathrm{D}f\left( \mathbf{x}\right) g\left( \mathbf{x}\right) = f\left( \mathbf{x}\right) \mathrm{D}g\left( \mathbf{x}\right) + g\left( \mathbf{x}\right) \mathrm{D}f\left( \mathbf{x}\right) . \]

5.2.30 设 \(f\left( x\right) = \frac{x}{\left| x\right| },x \in {R}^{m}\) . (1) 求 \(\mathrm{D}f\left( \mathbf{x}\right)\) ; (2)取方向 \(l = \frac{x}{\left| x\right| }\) ,求方向导数 \(\frac{\partial f}{\partial l}\) ； (3)取方向 \(l\) 满足 \(l \cdot x = 0\) ,求方向导数 \(\frac{\partial f}{\partial l}\) ； (4)求导数的范数 \(\parallel \mathrm{D}f\left( \mathbf{x}\right) \parallel\) .

5.2.31 求下列变换的雅可比行列式: (1) \({x}_{1} = r\cos \theta ,{x}_{2} = r\sin \theta\) ,求 \(\frac{\partial \left( {{x}_{1},{x}_{2}}\right) }{\partial \left( {r,\theta }\right) }\) ; (2) \({x}_{1} = r\cos {\theta }_{1},{x}_{2} = r\sin {\theta }_{1}\cos {\theta }_{2},{x}_{3} = r\sin {\theta }_{1}\sin {\theta }_{2}\) ,求 \(\frac{\partial \left( {{x}_{1},{x}_{2},{x}_{3}}\right) }{\partial \left( {r,{\theta }_{1},{\theta }_{2}}\right) }\) ; (3) \(\left\{ \begin{array}{l} {x}_{1} = r\cos {\theta }_{1}, \\ {x}_{2} = r\sin {\theta }_{1}\cos {\theta }_{2}, \\ {x}_{3} = r\sin {\theta }_{1}\sin {\theta }_{2}\cos {\theta }_{3}, \\ \;\vdots \\ {x}_{m - 1} = r\sin {\theta }_{1}\sin {\theta }_{2}\sin {\theta }_{3}\cdots \sin {\theta }_{m - 2} < \pi , \\ {x}_{m} = r\sin {\theta }_{1}\sin {\theta }_{2}\sin {\theta }_{3}\cdots \sin {\theta }_{m - 2}\cos {\theta }_{m - 1}, \\ {x}_{m} = r\sin {\theta }_{1}\sin {\theta }_{2}\sin {\theta }_{3}\cdots \sin {\theta }_{m - 2}\sin {\theta }_{m - 1}, \end{array}\right.\) 试用数学归纳法求 \(\frac{\partial \left( {{x}_{1},{x}_{2},\cdots ,{x}_{m}}\right) }{\partial \left( {r,{\theta }_{1},\cdots ,{\theta }_{m - 1}}\right) }\) .

5.2.32 设 \(\Omega\) 为 \({\mathbf{R}}^{m}\) 中凸区域, \(f\left( \mathbf{x}\right) \in {C}^{2}\left( {\Omega ,\mathbf{R}}\right)\) . 若 \(f\) 的海色矩阵 \({H}_{f}\left( \mathbf{x}\right)\) 是半正定的. 证明: \(f\left( \mathbf{x}\right)\) 是 \(\mathbf{\Omega }\) 上的凹函数.

5. 3.1 求由下列方程所定义的函数 \(y\) 的一阶、二阶导数: (1) \(\ln \sqrt{{x}^{2} + {y}^{2}} = \arctan \frac{y}{x}\) ; (2) \({xy} + {2}^{y} = 0\) .

5.3.2 对下列方程所确定的 \(z = z\left( {x,y}\right)\) ,求一阶偏导数: (1) \({x}^{n} + {y}^{n} + {z}^{n} = {a}^{n}\) ; (2) \(x + y + z = {\mathrm{e}}^{x + y + z}\) .

5.3.3 对下列方程所确定的 \(z = z\left( {x,y}\right)\) ,求二阶偏导数: (1) \({xy} + {yz} + {zx} = 1\) ; (2) \(\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = 1\) .

5.3.4 求由下列方程所确定的 \(z = z\left( {x,y}\right)\) 的微分: (1) \(f\left( {{xy},z - y}\right) = 0\) ; (2) \(f\left( {x,x + y,x + y + z}\right) = 0\) .

5.3.5 设 \(z = z\left( {x,y}\right)\) 由方程 \({x}^{2} + {y}^{2} + {z}^{2} = {yf}\left( \frac{z}{y}\right)\) 所确定,证明: \[ \left( {{x}^{2} - {y}^{2} - {z}^{2}}\right) \frac{\partial z}{\partial x} + {2xy}\frac{\partial z}{\partial y} = {2xz}. \]

5.3.6 设 \(z = z\left( {x,y}\right)\) 由方程 \(F\left( {x + \frac{z}{y},y + \frac{z}{x}}\right) = 0\) 所确定,证明: \[ x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = z - {xy}. \]

5.3.7 设 \(u = u\left( {x,y,z}\right)\) 由方程 \(F\left( {{u}^{2} - {x}^{2},{u}^{2} - {y}^{2},{u}^{2} - {z}^{2}}\right) = 0\) 所确定,证明: \[ \frac{{u}_{x}^{\prime }}{x} + \frac{{u}_{y}^{\prime }}{y} + \frac{{u}_{z}^{\prime }}{z} = \frac{1}{u}. \]

5.3.8 设 \(u = u\left( {x,y,z}\right)\) 由方程 \(\frac{{x}^{2}}{{a}^{2} + u} + \frac{{y}^{2}}{{b}^{2} + u} + \frac{{z}^{2}}{{c}^{2} + u} = 1\) 所确定,证明: \[ {\left( \frac{\partial u}{\partial x}\right) }^{2} + {\left( \frac{\partial u}{\partial y}\right) }^{2} + {\left( \frac{\partial u}{\partial z}\right) }^{2} = {2x}\frac{\partial u}{\partial x} + {2y}\frac{\partial u}{\partial y} + {2z}\frac{\partial u}{\partial z}. \]

5.3.9 求函数 \(z = f\left( {x + y,z + y}\right)\) 的二阶偏导数.

5.3.10 证明: 由方程组 \[ \left\{ \begin{array}{l} z = {ax} + {y\varphi }\left( a\right) + \psi \left( a\right) , \\ 0 = x + {y\varphi }\left( a\right) + {\psi }^{\prime }\left( a\right) \end{array}\right. \] 所确定的函数 \(z = z\left( {x,y}\right)\) 满足方程 \[ \frac{{\partial }^{2}z}{\partial {x}^{2}} \cdot \frac{{\partial }^{2}z}{\partial {y}^{2}} - {\left( \frac{{\partial }^{2}z}{\partial x\partial y}\right) }^{2} =

0. \] 5.3.11 若 \(z = z\left( {x,y}\right)\) 满足方程 \(\frac{{\partial }^{2}z}{\partial {x}^{2}} \cdot \frac{{\partial }^{2}z}{\partial {y}^{2}} - {\left( \frac{{\partial }^{2}z}{\partial x\partial y}\right) }^{2} = 0\) . 证明: 若把 \(z =\) \(z\left( {x,y}\right)\) 中的 \(y\) 看成 \(x,z\) 的函数,则它满足同样形状的方程: \[ \frac{{\partial }^{2}y}{\partial {x}^{2}} \cdot \frac{{\partial }^{2}y}{\partial {z}^{2}} - {\left( \frac{{\partial }^{2}y}{\partial x\partial z}\right) }^{2} =

5.3.12 设 \(u = {\mathrm{e}}^{x}\cos y,v = {\mathrm{e}}^{x}\sin y\) . 求证: (1) 当 \(\left( {x,y}\right) \in {\mathbf{R}}^{2}\) 时, \(\frac{\partial \left( {u,v}\right) }{\partial \left( {x,y}\right) } \neq 0\) ,但变换不是一一的; (2)记 \(\Omega = \{ \left( {x,y}\right) \mid 0 < y < {2\pi }, - \infty < x < + \infty \}\) ,这时变换在 \(\Omega\) 上是一一的, 并求出逆变换.

5.3.13 求下列变换的雅可比行列式 \(\frac{\partial \left( {x,y}\right) }{\partial \left( {u,v}\right) },\frac{\partial \left( {u,v}\right) }{\partial \left( {x,y}\right) }\) : (1) \(\left\{ \begin{array}{l} u = {xy}, \\ v = \frac{x}{y}; \end{array}\right.\) (2) \(\left\{ \begin{array}{l} u = {x}^{2} + {y}^{2}, \\ v = {2xy}. \end{array}\right.\)

5.3.14 由下列方程组求 \(\frac{\mathrm{d}y}{\mathrm{\;d}x},\frac{\mathrm{d}z}{\mathrm{\;d}x},\frac{{\mathrm{d}}^{2}y}{\mathrm{\;d}{x}^{2}},\frac{{\mathrm{d}}^{2}z}{\mathrm{\;d}{x}^{2}}\) : (1) \(\left\{ \begin{array}{l} x + y + z = 0, \\ {x}^{2} + {y}^{2} + {z}^{2} = 6; \end{array}\right.\) (2) \(\left\{ \begin{array}{l} z = {x}^{2} + {y}^{2}, \\ 2{x}^{2} + 2{y}^{2} - {z}^{2} =

5.3.15 求由方程组 \(x = u\cos v,y = u\sin v,z = v\) 所确定的函数 \(z = z\left( {x,y}\right)\) 的一阶、二阶偏导数.

5.3.16 设 \(u = f\left( {x - {ut},y - {ut},z - {ut}}\right) ,g\left( {x,y,z}\right) = 0\) . 求 \(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}\) ,并问这时 \(t\) 是自变量还是因变量?

5.3.17 设 \(z = z\left( {x,y}\right)\) 满足方程组 \(f\left( {x,y,z,t}\right) = 0,g\left( {x,y,z,t}\right) = 0\) ,求 \(\mathrm{d}z\) .

5.3.18 设 \(\Omega \subset {\mathbf{R}}^{m}\) 是凸区域, \(\mathbf{f}\left( \mathbf{x}\right) \in {C}^{1}\left( {\Omega ,{\mathbf{R}}^{m}}\right) ,\mathrm{D}\mathbf{f}\left( \mathbf{x}\right)\) 在 \(\Omega\) 上是正定矩阵. 求证: \(f\left( x\right)\) 是 \(\Omega\) 上的单叶函数.

5.3.19 设 \(x \in {\mathbf{R}}^{m},f\left( \mathbf{x}\right) \in {C}^{2}\left( {U\left( {\mathbf{x}}_{0}\right) ,\mathbf{R}}\right) ,\mathrm{D}f\left( {\mathbf{x}}_{0}\right) = 0,\det {H}_{f}\left( {\mathbf{x}}_{0}\right) \neq 0\) . 求证: \(\exists \delta > 0\) ,当 \(x \in U\left( {{x}_{0};\delta }\right) \smallsetminus \left\{ {x}_{0}\right\}\) 时, \(\mathrm{D}f\left( x\right) \neq 0\) .

5. 3.20 假设 \(\mathbf{f}\left( \mathbf{x}\right) \in {C}^{1}\left( {{\mathbf{R}}^{m},{\mathbf{R}}^{m}}\right)\) ,并且在 \({\mathbf{R}}^{m}\) 上 \(\det D\mathbf{f}\left( \mathbf{x}\right) \neq 0\) . 又当 \(\left| \mathbf{x}\right| \rightarrow\) \(\displaystyle{+ \infty}\) 时, \(\left| {\mathbf{f}\left( \mathbf{x}\right) }\right| \rightarrow + \infty\) . 证明: \(\mathbf{f}\left( {\mathbf{R}}^{m}\right) = {\mathbf{R}}^{m}\) .

5. 4.1 求曲线 \(z = {x}^{2} + {y}^{2},2{x}^{2} + 2{y}^{2} - {z}^{2} = 0\) 在(1,1,2)点的切线方程.

5.4.2 在曲线 \(y = {x}^{2},z = {x}^{3}\) 上求一点,使该点的切线平行于平面 \[ x + {2y} + z =

5.4.3 求下列曲面在指定点的切平面和法线方程: (1) \({x}^{2} + {y}^{2} + {z}^{2} = {169},\;M\left( {3,4,{12}}\right)\) ; (2) \(z = \arctan \left( {x/y}\right) ,\;M\left( {1,1,\pi /4}\right)\) ; (3) \(3{x}^{2} + 2{y}^{2} = {2x} + 1\) , (4) \(z = y + \ln \left( {x/z}\right)\) ,

5.4.4 求曲面 \(x = u\cos v,y = u\sin v,z = v\) 在 \(\left( {\sqrt{2},\sqrt{2},\pi /4}\right)\) 点的切平面方程.

5.4.5 求曲面 \({x}^{2} + 2{y}^{2} + 3{z}^{2} = {21}\) 的平行于平面 \(x + {4y} + {6z} = 0\) 的各切平面.

5.4.6 求曲面 \({x}^{2} + {y}^{2} + {z}^{2} = x\) 的切平面,使其垂直于平面 \[ x - y - z = 2\text{ 和 }x - y - z/2 =

5.4.7 试确定正数 \(\lambda\) ,使曲面 \({xyz} = \lambda\) 与椭球面 \(\frac{{x}^{2}}{{a}^{2}} + \frac{{y}^{2}}{{b}^{2}} + \frac{{z}^{2}}{{c}^{2}} = 1\) 的某点相切.

5.4.8 证明: 曲面 \(\sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{a}\) 的切平面在坐标轴上割下的诸线段之和为常量.

5.4.9 证明: 曲面 \(F\left( {x - {az},y - {bz}}\right) = 0\) 的切平面与某一定直线平行,其中 \(a,b\) 为常数.

5.4.10 证明: 曲面 \({ax} + {by} + {cz} = \Phi \left( {{x}^{2} + {y}^{2} + {z}^{2}}\right)\) 在 \(M\left( {{x}_{0},{y}_{0},{z}_{0}}\right)\) 点的法向量与向量 \(\left( {{x}_{0},{y}_{0},{z}_{0}}\right)\) 及(a, b, c)共面.

5.4.11 求下列函数的极值: (1) \(f\left( {x,y}\right) = {x}^{2} - {xy} + {y}^{2} - {2x} + y\) ; (2) \(f\left( {x,y}\right) = \sin x\sin y\sin \left( {x + y}\right) \left( {0 \leq x,y \leq \pi }\right)\) ； (3) \(f\left( {x,y,z}\right) = \left( {x + y + z}\right) {\mathrm{e}}^{-\left( {{x}^{2} + {y}^{2} + {z}^{2}}\right) }\) .

5.4.12 确立最小正数 \(A\) 和最大负数 \(B\) ,使不等式 \[ \frac{B}{xy} \leq \ln \left( {{x}^{2} + {y}^{2}}\right) \leq A\left( {{x}^{2} + {y}^{2}}\right) \] 在第一象限内成立.

5.4.13 求函数 \(f\left( {a,b}\right) = {\int }_{0}^{1}{\left\lbrack {x}^{2} - a - bx\right\rbrack }^{2}\mathrm{\;d}x\) 的最小值.

5.4.14 作容积为 \(V\) 的闭口长方形容器,问长、宽、高成何比例时用料最省?

5.4.15 有一块铁片,宽 \(b = {24}\mathrm{\;{cm}}\) ,要把它的两边折起做成一个槽,使得容积最大,求每边的倾角 \(\alpha\) 和折起的宽度 \(x\) .

5.4.16 在椭球面 \(\frac{{x}^{2}}{{a}^{2}} + \frac{{y}^{2}}{{b}^{2}} + \frac{{z}^{2}}{{c}^{2}} = 1\) 内接长方体中,求体积为最大的那个长方体.

5.4.17 给定曲面 \(z = 1 - {x}^{2} - {z}^{2}\) ,求过第一象限中曲面的切平面,使它与第一象限坐标面所围的四面体体积最小.

5.4.18 设 \(u\left( {x,y}\right)\) 在 \({x}^{2} + {y}^{2} \leq 1\) 上连续,在 \({x}^{2} + {y}^{2} < 1\) 上满足 \[ \frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} = u \] 且在 \({x}^{2} + {y}^{2} = 1\) 上 \(u\left( {x,y}\right) > 0\) . 证明: (1)当 \({x}^{2} + {y}^{2} \leq 1\) 时, \(u\left( {x,y}\right) \geq 0\) ； (2)当 \({x}^{2} + {y}^{2} \leq 1\) 时, \(u\left( {x,y}\right) > 0\) .

5.4.19 设 \(a > 0,c > 0,{ac} - {b}^{2} > 0\) ,则方程 \(a{x}^{2} + {2bxy} + c{y}^{2} = 1\) 表示椭圆. 试证该椭圆的面积为 \(\frac{\pi }{\sqrt{{ac} - {b}^{2}}}\) .

5.4.20 (1) 在 \({x}^{2} + {y}^{2} = 1\) 的条件下,求 \(f\left( {x,y}\right) = a{x}^{2} + {2bxy} + c{y}^{2}\) 的最大值与最小值; (2)利用(1)证明:当 \(a > 0,{ac} - {b}^{2} > 0\) 时,二次型 \(f\left( {x,y}\right)\) 是正定的；当 \(a <\) \(0,{ac} - {b}^{2} > 0\) 时,二次型 \(f\left( {x,y}\right)\) 是负定的.

5.4.21 求圆的内接 \(n\) 边形中面积最大者.

5.4.22 求圆的外切 \(n\) 边形中面积最小者.

5.4.23 证明: 椭圆的内接三角形中, 面积最大的三角形的顶点处的椭圆法线必与三角形的该顶点的对边垂直；由此求出面积最大的内接三角形.

5.4.24 给定椭球 \(\frac{{x}^{2}}{{a}^{2}} + \frac{{y}^{2}}{{b}^{2}} + \frac{{z}^{2}}{{c}^{2}} = 1\) . (1)求第一象限中椭球的切平面,使它与坐标平面围成的四面体体积最小; (2)证明体积最小的椭球外切八面体体积 \(\leq 4\sqrt{3}{abc}\) .

5.4.25 设凸四边形各边长分别为 \(a,b,c,d\) . 求证: 凸四边形对角和为 \(\pi\) 时面积最大.

5.4.26 长为 \(a\) 的铁丝切成两段,一段围成一个正方形,另一段围成一个圆. 这两段的长各为多少时, 由它们所围正方形面积和圆面积之和最大?

5.4.27 要制作一个中间是圆柱,两端为相同的正圆锥的空浮标,它的体积是一定的, 要使所用材料最省, 圆柱和圆锥的尺寸应成何比例?

5.4.28 求抛物线 \(y = {x}^{2}\) 和直线 \(x - y = 1\) 间的最短距离.

5.4.29 在 \({\mathbf{R}}^{m}\) 中给定超平面 \(\mathbf{x} \cdot \mathbf{a} - c = 0,\mathbf{x},\mathbf{a} \in {\mathbf{R}}^{m},c\) 为实数.

5. 5.1 求下列极限:} (1) \(\displaystyle{\mathop{\lim }\limits_{{x \rightarrow 0}}{\int }_{-1}^{1}\sqrt{{x}^{2} + {y}^{2}}\mathrm{\;d}y}\) ; (2) \(\displaystyle{\mathop{\lim }\limits_{{x \rightarrow 0}}{\int }_{0}^{2}{y}^{2}\cos {xy}\mathrm{\;d}y}\) ; (3) \(\displaystyle{\mathop{\lim }\limits_{{a \rightarrow 0}}{\int }_{a}^{1 + a}\frac{\mathrm{d}x}{1 + {x}^{2} + {a}^{2}}}\) .

5.5.2 设 \(f\left( x\right)\) 连续, \(F\left( x\right) = {\int }_{0}^{x}f\left( t\right) {\left( x - t\right) }^{n - 1}\mathrm{\;d}t\) ,求 \({F}^{\left( n\right) }\left( x\right)\) .

5.5.3 设 \(f\left( x\right) \in {C}^{2}\left( {-\infty ,\infty }\right) ,F\left( x\right) \in {C}^{1}\left( {-\infty ,\infty }\right)\) , \[ u = \frac{1}{2}\left\lbrack {f\left( {x + {at}}\right) + f\left( {x - {at}}\right) }\right\rbrack + \frac{1}{2n}{\int }_{x - {at}}^{x + {at}}F\left( y\right) \mathrm{d}y. \] 求证: 当 \(\displaystyle{- \infty < x < \infty ,t \geq 0}\) 时, \(u\left( {x,t}\right) ,\frac{{\partial }^{2}u}{\partial {t}^{2}},\frac{{\partial }^{2}u}{\partial {x}^{2}}\) 连续,且满足 \[ \frac{{\partial }^{2}u}{\partial {t}^{2}} = {a}^{2}\frac{{\partial }^{2}v}{\partial {x}^{2}},\;u\left( {x,0}\right) = f\left( x\right) ,\;\frac{\partial u\left( {x,0}\right) }{\partial t} = F\left( x\right) . \]

5.5.4 求 \({F}^{\prime }\left( x\right)\) : (1) \(F\left( x\right) = {\int }_{\sin x}^{\cos x}{\mathrm{e}}^{x\sqrt{1 - {y}^{2}}}\mathrm{\;d}y\) ; (2) \(F\left( x\right) = {\int }_{a + x}^{b + x}\frac{\sin {xy}}{y}\mathrm{\;d}y\) ; (3) \(F\left( x\right) = {\int }_{0}^{x}{\int }_{{t}^{2}}^{{x}^{2}}f\left( {t,s}\right) \mathrm{d}s\mathrm{\;d}t\) .

5.5.5 设 \(f\left( x\right) \in R\left\lbrack {-\pi ,\pi }\right\rbrack\) ,求函数 \[ F\left( {{\alpha }_{0},{\alpha }_{1},\cdots ,{\alpha }_{n},{\beta }_{1},\cdots ,{\beta }_{n}}\right) \] \[ = \frac{1}{\pi }{\int }_{-\pi }^{\pi }{\left\lbrack f\left( x\right) - \frac{{\alpha }_{0}}{2} - \mathop{\sum }\limits_{{k = 1}}^{n}\left( {\alpha }_{k}\cos kx + {\beta }_{k}\sin kx\right) \right\rbrack }^{2}\mathrm{\;d}x \] 的最小值.

5.5.6 设 \(f\left( x\right)\) 是周期为 \({2\pi }\) 的连续函数. \({a}_{n},{b}_{n}\) 为其傅氏系数, \({A}_{n},{B}_{n}\) 是卷积函数 \[ F\left( x\right) = \frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left( t\right) f\left( {x - t}\right) \mathrm{d}t \] 的傅氏系数. 求证: (1) \({A}_{0} = {a}_{0}^{2},{A}_{n} = {a}_{n}^{2} + {b}_{n}^{2},{B}_{n} = 0\left( {n = 1,2,\cdots }\right)\) ; (2) \(\frac{1}{\pi }{\int }_{-\pi }^{\pi }{f}^{2}\left( t\right) \mathrm{d}t = \frac{{a}_{0}^{2}}{2} + \mathop{\sum }\limits_{{n = 1}}^{\infty }\left( {{a}_{n}^{2} + {b}_{n}^{2}}\right)\) .

5.5.7 设 \(F\left( x\right) = {\int }_{0}^{2\pi }{\mathrm{e}}^{x\cos \theta }\cos \left( {x\sin \theta }\right) \mathrm{d}\theta\) ,求证: \(F\left( x\right) \equiv {2\pi }\) .

5.5.8 \(\ln \frac{3}{2}\) .

5. 6.1 证明下列积分在所给区间上一致收敛: (1) \(\displaystyle{\int }_{0}^{+\infty }\frac{\cos \left( {xy}\right) }{{x}^{2} + {y}^{2}}\mathrm{\;d}y\left( {x \geq a > 0}\right)\) ; (2) \(\displaystyle{\int }_{1}^{+\infty }{x}^{a}{\mathrm{e}}^{-x}\mathrm{\;d}x\left( {a \leq \alpha \leq b}\right)\) ； (3) \(\displaystyle{\int }_{0}^{+\infty }\frac{\sin {x}^{2}}{1 + {x}^{p}}\mathrm{\;d}x\left( {p \geq 0}\right)\) ; (4) \(\displaystyle{\int }_{0}^{+\infty }\frac{\sin x}{x}{\mathrm{e}}^{-{ax}}\mathrm{\;d}x\left( {a \geq 0}\right)\) .

5.6.2 设 \(f\left( x\right) \in C\lbrack 0, + \infty )\) 且有界, \(f\left( 0\right) > 0\) . 讨论函数 \[ F\left( y\right) = {\int }_{0}^{+\infty }\frac{{yf}\left( x\right) }{{x}^{2} + {y}^{2}}\mathrm{\;d}x \] 的连续性.

5.6.3 计算积分 \(I = {\int }_{0}^{+\infty }\frac{\arctan {bx} - \arctan {ax}}{x}\mathrm{\;d}x\left( {b > a > 0}\right)\) .

5.6.4 通过引入参数,计算积分 \[ {\int }_{0}^{+\infty }\frac{\ln \left( {1 + {x}^{2}}\right) }{1 + {x}^{2}}\mathrm{\;d}x. \]

5.6.5 通过引入收敛因子 \({\mathrm{e}}^{-{ax}}\) 的方法,计算积分 \[ I = {\int }_{0}^{\infty }\frac{\cos {bx} - \cos {ax}}{x}\mathrm{\;d}x\;\left( {b > a > 0}\right) . \]

5.6.6 利用已知积分求下列积分 \(\left( {b > a > 0}\right)\) : \customfootnote{ ① 黎曼引理 设 \(f\left( x\right)\) 在 \(\left\lbrack {a,b}\right\rbrack\) 上可积或绝对可积,则 \[ \mathop{\lim }\limits_{{\lambda \rightarrow + \infty }}{\int }_{a}^{b}f\left( x\right) \cos {\lambda x}\mathrm{\;d}x = 0,\;\mathop{\lim }\limits_{{\lambda \rightarrow + \infty }}{\int }_{a}^{b}f\left( x\right) \sin {\lambda x}\mathrm{\;d}x =

### 5.6.7 利用已知积分求下列积分 **(1)** \[ \int_{0}^{+\infty} \left( \frac{\sin x}{x} \right)^2 dx \] **解答** 已知 \[ \int_{0}^{+\infty} \frac{\sin x}{x} dx = \frac{\pi}{2} \] 利用分部积分法： 令 \( u = \sin^2 x, dv = \frac{dx}{x^2} \)，则 \[ du = 2\sin x \cos x dx = \sin 2x dx, \quad v = -\frac{1}{x} \] 于是 \[ \int_{0}^{+\infty} \frac{\sin^2 x}{x^2} dx = \left[-\frac{\sin^2 x}{x}\right]_{0}^{+\infty} + \int_{0}^{+\infty} \frac{\sin 2x}{x} dx \] 第一项在 \(x\to 0\) 时极限为 0，在 \(\displaystyle{x\to\infty}\) 时也为 0。 第二项令 \(t=2x\)，得 \[ \int_{0}^{+\infty} \frac{\sin 2x}{x} dx = \int_{0}^{+\infty} \frac{\sin t}{t} dt = \frac{\pi}{2} \] 所以 \[ \boxed{\frac{\pi}{2}} \] --- **(2)** \[ \int_{0}^{+\infty} \frac{\sin^4 x}{x^2} dx \] **解答** 利用三角恒等式 \[ \sin^4 x = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x \] 则 \[ \int_{0}^{+\infty} \frac{\sin^4 x}{x^2} dx = \frac{3}{8}\int_{0}^{+\infty}\frac{dx}{x^2} - \frac12 \int_{0}^{+\infty} \frac{\cos 2x}{x^2} dx + \frac18 \int_{0}^{+\infty} \frac{\cos 4x}{x^2} dx \] 但第一项发散，因此需用分部积分法处理。 令 \[ I = \int_{0}^{+\infty} \frac{\sin^4 x}{x^2} dx \] 分部积分： \(u = \sin^4 x, dv = x^{-2} dx\)，则 \[ du = 4\sin^3 x \cos x dx = 2\sin^2 x \sin 2x dx, \quad v = -\frac1x \] 于是 \[ I = \left[-\frac{\sin^4 x}{x}\right]_{0}^{+\infty} + \int_{0}^{+\infty} \frac{2\sin^2 x \sin 2x}{x} dx \] 第一项为0，第二项用 \(\sin^2 x = \frac{1-\cos 2x}{2}\)： \[ I = \int_{0}^{+\infty} \frac{(1-\cos 2x)\sin 2x}{x} dx = \int_{0}^{+\infty} \frac{\sin 2x}{x} dx - \int_{0}^{+\infty} \frac{\cos 2x \sin 2x}{x} dx \] 第一个积分 \(=\frac{\pi}{2}\)，第二个积分中 \(\cos 2x \sin 2x = \frac12 \sin 4x\)，所以 \[ \int_{0}^{+\infty} \frac{\frac12 \sin 4x}{x} dx = \frac12 \cdot \frac{\pi}{2} = \frac{\pi}{4} \] 因此 \[ I = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \] \[ \boxed{\frac{\pi}{4}} \] --- **(3)** \[ \int_{-\infty}^{+\infty} e^{-(a x^2 + b x + c)} dx \quad (a>0) \] **解答** 配方： \[ a x^2 + b x + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a} \] 则 \[ \int_{-\infty}^{+\infty} e^{-(a x^2+bx+c)} dx = e^{-c + \frac{b^2}{4a}} \int_{-\infty}^{+\infty} e^{-a (x+\frac{b}{2a})^2} dx \] 令 \(t = \sqrt{a}(x+\frac{b}{2a})\)，则 \[ \int_{-\infty}^{+\infty} e^{-a (x+\frac{b}{2a})^2} dx = \frac{1}{\sqrt{a}} \int_{-\infty}^{+\infty} e^{-t^2} dt = \sqrt{\frac{\pi}{a}} \] 所以 \[ \boxed{\sqrt{\frac{\pi}{a}} \, e^{\frac{b^2}{4a} - c}} \] --- **(4)** \[ \int_{-\infty}^{+\infty} e^{-(x^2 + \frac{a^2}{x^2})} dx \quad (a>0) \] **解答** 利用对称性，原积分 \(= 2\int_{0}^{+\infty} e^{-(x^2 + a^2/x^2)} dx\)。 令 \(t = x - \frac{a}{x}\)，则 \[ dt = \left(1 + \frac{a}{x^2}\right) dx \] 但更好的方法是利用公式： \[ \int_{0}^{+\infty} e^{-(x^2 + \frac{a^2}{x^2})} dx = \frac{\sqrt{\pi}}{2} e^{-2a} \] （这是已知结果，可通过变量代换 \(u = x - a/x\) 证明） 因此原积分 \[ = 2 \cdot \frac{\sqrt{\pi}}{2} e^{-2a} = \sqrt{\pi} e^{-2a} \] \[ \boxed{\sqrt{\pi} e^{-2a}} \] --- **(5)** \[ \int_{0}^{+\infty} \frac{e^{-x^2} - \cos x}{x^2} dx \] **解答** 分部积分：令 \(u = e^{-x^2} - \cos x, dv = x^{-2} dx\)，则 \[ du = (-2x e^{-x^2} + \sin x) dx, \quad v = -\frac1x \] 于是 \[ I = \left[-\frac{e^{-x^2} - \cos x}{x}\right]_{0}^{+\infty} + \int_{0}^{+\infty} \frac{-2x e^{-x^2} + \sin x}{x} dx \] 第一项在 \(x\to 0\) 时，\(e^{-x^2} - \cos x \sim (1-x^2) - (1 - x^2/2) = -x^2/2\)，除以 \(x\) 趋于0，所以第一项为0。 第二项 \[ I = \int_{0}^{+\infty} (-2 e^{-x^2}) dx + \int_{0}^{+\infty} \frac{\sin x}{x} dx = -2\cdot\frac{\sqrt{\pi}}{2} + \frac{\pi}{2} = -\sqrt{\pi} + \frac{\pi}{2} \] \[ \boxed{\frac{\pi}{2} - \sqrt{\pi}} \] --- **(6)** \[ \int_{0}^{+\infty} \frac{\sin(\alpha x) \cos(\beta x)}{x} dx \] **解答** 利用积化和差： \[ \sin(\alpha x)\cos(\beta x) = \frac12 [\sin((\alpha+\beta)x) + \sin((\alpha-\beta)x)] \] 则 \[ I = \frac12 \int_{0}^{+\infty} \frac{\sin((\alpha+\beta)x)}{x} dx + \frac12 \int_{0}^{+\infty} \frac{\sin((\alpha-\beta)x)}{x} dx \] 已知 \[ \int_{0}^{+\infty} \frac{\sin(kx)}{x} dx = \frac{\pi}{2} \operatorname{sgn}(k) \] 所以 \[ I = \frac{\pi}{4} \left[ \operatorname{sgn}(\alpha+\beta) + \operatorname{sgn}(\alpha-\beta) \right] \] 例如当 \(\alpha > \beta > 0\) 时，两项均为正，得 \(\pi/2\)；若 \(0<\alpha<\beta\)，则第二项为负，结果为0。 \[ \boxed{\frac{\pi}{

### 5.6.7 利用已知积分求下列积分 **(1)** \[ \int_{0}^{+\infty} \left( \frac{\sin x}{x} \right)^2 dx \] **解答** 已知 \[ \int_{0}^{+\infty} \frac{\sin x}{x} dx = \frac{\pi}{2} \] 利用分部积分法： 令 \( u = \sin^2 x, dv = \frac{dx}{x^2} \)，则 \[ du = 2\sin x \cos x dx = \sin 2x dx, \quad v = -\frac{1}{x} \] 于是 \[ \int_{0}^{+\infty} \frac{\sin^2 x}{x^2} dx = \left[-\frac{\sin^2 x}{x}\right]_{0}^{+\infty} + \int_{0}^{+\infty} \frac{\sin 2x}{x} dx \] 第一项在 \(x\to 0\) 时极限为 0，在 \(\displaystyle{x\to\infty}\) 时也为 0。 第二项令 \(t=2x\)，得 \[ \int_{0}^{+\infty} \frac{\sin 2x}{x} dx = \int_{0}^{+\infty} \frac{\sin t}{t} dt = \frac{\pi}{2} \] 所以 \[ \boxed{\frac{\pi}{2}} \] --- **(2)** \[ \int_{0}^{+\infty} \frac{\sin^4 x}{x^2} dx \] **解答** 利用三角恒等式 \[ \sin^4 x = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x \] 则 \[ \int_{0}^{+\infty} \frac{\sin^4 x}{x^2} dx = \frac{3}{8}\int_{0}^{+\infty}\frac{dx}{x^2} - \frac12 \int_{0}^{+\infty} \frac{\cos 2x}{x^2} dx + \frac18 \int_{0}^{+\infty} \frac{\cos 4x}{x^2} dx \] 但第一项发散，因此需用分部积分法处理。 令 \[ I = \int_{0}^{+\infty} \frac{\sin^4 x}{x^2} dx \] 分部积分： \(u = \sin^4 x, dv = x^{-2} dx\)，则 \[ du = 4\sin^3 x \cos x dx = 2\sin^2 x \sin 2x dx, \quad v = -\frac1x \] 于是 \[ I = \left[-\frac{\sin^4 x}{x}\right]_{0}^{+\infty} + \int_{0}^{+\infty} \frac{2\sin^2 x \sin 2x}{x} dx \] 第一项为0，第二项用 \(\sin^2 x = \frac{1-\cos 2x}{2}\)： \[ I = \int_{0}^{+\infty} \frac{(1-\cos 2x)\sin 2x}{x} dx = \int_{0}^{+\infty} \frac{\sin 2x}{x} dx - \int_{0}^{+\infty} \frac{\cos 2x \sin 2x}{x} dx \] 第一个积分 \(=\frac{\pi}{2}\)，第二个积分中 \(\cos 2x \sin 2x = \frac12 \sin 4x\)，所以 \[ \int_{0}^{+\infty} \frac{\frac12 \sin 4x}{x} dx = \frac12 \cdot \frac{\pi}{2} = \frac{\pi}{4} \] 因此 \[ I = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \] \[ \boxed{\frac{\pi}{4}} \] --- **(3)** \[ \int_{-\infty}^{+\infty} e^{-(a x^2 + b x + c)} dx \quad (a>0) \] **解答** 配方： \[ a x^2 + b x + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a} \] 则 \[ \int_{-\infty}^{+\infty} e^{-(a x^2+bx+c)} dx = e^{-c + \frac{b^2}{4a}} \int_{-\infty}^{+\infty} e^{-a (x+\frac{b}{2a})^2} dx \] 令 \(t = \sqrt{a}(x+\frac{b}{2a})\)，则 \[ \int_{-\infty}^{+\infty} e^{-a (x+\frac{b}{2a})^2} dx = \frac{1}{\sqrt{a}} \int_{-\infty}^{+\infty} e^{-t^2} dt = \sqrt{\frac{\pi}{a}} \] 所以 \[ \boxed{\sqrt{\frac{\pi}{a}} \, e^{\frac{b^2}{4a} - c}} \] --- **(4)** \[ \int_{-\infty}^{+\infty} e^{-(x^2 + \frac{a^2}{x^2})} dx \quad (a>0) \] **解答** 利用对称性，原积分 \(= 2\int_{0}^{+\infty} e^{-(x^2 + a^2/x^2)} dx\)。 令 \(t = x - \frac{a}{x}\)，则 \[ dt = \left(1 + \frac{a}{x^2}\right) dx \] 但更好的方法是利用公式： \[ \int_{0}^{+\infty} e^{-(x^2 + \frac{a^2}{x^2})} dx = \frac{\sqrt{\pi}}{2} e^{-2a} \] （这是已知结果，可通过变量代换 \(u = x - a/x\) 证明） 因此原积分 \[ = 2 \cdot \frac{\sqrt{\pi}}{2} e^{-2a} = \sqrt{\pi} e^{-2a} \] \[ \boxed{\sqrt{\pi} e^{-2a}} \] --- **(5)** \[ \int_{0}^{+\infty} \frac{e^{-x^2} - \cos x}{x^2} dx \] **解答** 分部积分：令 \(u = e^{-x^2} - \cos x, dv = x^{-2} dx\)，则 \[ du = (-2x e^{-x^2} + \sin x) dx, \quad v = -\frac1x \] 于是 \[ I = \left[-\frac{e^{-x^2} - \cos x}{x}\right]_{0}^{+\infty} + \int_{0}^{+\infty} \frac{-2x e^{-x^2} + \sin x}{x} dx \] 第一项在 \(x\to 0\) 时，\(e^{-x^2} - \cos x \sim (1-x^2) - (1 - x^2/2) = -x^2/2\)，除以 \(x\) 趋于0，所以第一项为0。 第二项 \[ I = \int_{0}^{+\infty} (-2 e^{-x^2}) dx + \int_{0}^{+\infty} \frac{\sin x}{x} dx = -2\cdot\frac{\sqrt{\pi}}{2} + \frac{\pi}{2} = -\sqrt{\pi} + \frac{\pi}{2} \] \[ \boxed{\frac{\pi}{2} - \sqrt{\pi}} \] --- **(6)** \[ \int_{0}^{+\infty} \frac{\sin(\alpha x) \cos(\beta x)}{x} dx \] **解答** 利用积化和差： \[ \sin(\alpha x)\cos(\beta x) = \frac12 [\sin((\alpha+\beta)x) + \sin((\alpha-\beta)x)] \] 则 \[ I = \frac12 \int_{0}^{+\infty} \frac{\sin((\alpha+\beta)x)}{x} dx + \frac12 \int_{0}^{+\infty} \frac{\sin((\alpha-\beta)x)}{x} dx \] 已知 \[ \int_{0}^{+\infty} \frac{\sin(kx)}{x} dx = \frac{\pi}{2} \operatorname{sgn}(k) \] 所以 \[ I = \frac{\pi}{4} \left[ \operatorname{sgn}(\alpha+\beta) + \operatorname{sgn}(\alpha-\beta) \right] \] 例如当 \(\alpha > \beta > 0\) 时，两项均为正，得 \(\pi/2\)；若 \(0<\alpha<\beta\)，则第二项为负，结果为0。 \[ \boxed{\frac{\pi}{

### 5.6.7 利用已知积分求下列积分 **(1)** \[ \int_{0}^{+\infty} \left( \frac{\sin x}{x} \right)^2 dx \] **解答** 已知 \[ \int_{0}^{+\infty} \frac{\sin x}{x} dx = \frac{\pi}{2} \] 利用分部积分法： 令 \( u = \sin^2 x, dv = \frac{dx}{x^2} \)，则 \[ du = 2\sin x \cos x dx = \sin 2x dx, \quad v = -\frac{1}{x} \] 于是 \[ \int_{0}^{+\infty} \frac{\sin^2 x}{x^2} dx = \left[-\frac{\sin^2 x}{x}\right]_{0}^{+\infty} + \int_{0}^{+\infty} \frac{\sin 2x}{x} dx \] 第一项在 \(x\to 0\) 时极限为 0，在 \(\displaystyle{x\to\infty}\) 时也为 0。 第二项令 \(t=2x\)，得 \[ \int_{0}^{+\infty} \frac{\sin 2x}{x} dx = \int_{0}^{+\infty} \frac{\sin t}{t} dt = \frac{\pi}{2} \] 所以 \[ \boxed{\frac{\pi}{2}} \] --- **(2)** \[ \int_{0}^{+\infty} \frac{\sin^4 x}{x^2} dx \] **解答** 利用三角恒等式 \[ \sin^4 x = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x \] 则 \[ \int_{0}^{+\infty} \frac{\sin^4 x}{x^2} dx = \frac{3}{8}\int_{0}^{+\infty}\frac{dx}{x^2} - \frac12 \int_{0}^{+\infty} \frac{\cos 2x}{x^2} dx + \frac18 \int_{0}^{+\infty} \frac{\cos 4x}{x^2} dx \] 但第一项发散，因此需用分部积分法处理。 令 \[ I = \int_{0}^{+\infty} \frac{\sin^4 x}{x^2} dx \] 分部积分： \(u = \sin^4 x, dv = x^{-2} dx\)，则 \[ du = 4\sin^3 x \cos x dx = 2\sin^2 x \sin 2x dx, \quad v = -\frac1x \] 于是 \[ I = \left[-\frac{\sin^4 x}{x}\right]_{0}^{+\infty} + \int_{0}^{+\infty} \frac{2\sin^2 x \sin 2x}{x} dx \] 第一项为0，第二项用 \(\sin^2 x = \frac{1-\cos 2x}{2}\)： \[ I = \int_{0}^{+\infty} \frac{(1-\cos 2x)\sin 2x}{x} dx = \int_{0}^{+\infty} \frac{\sin 2x}{x} dx - \int_{0}^{+\infty} \frac{\cos 2x \sin 2x}{x} dx \] 第一个积分 \(=\frac{\pi}{2}\)，第二个积分中 \(\cos 2x \sin 2x = \frac12 \sin 4x\)，所以 \[ \int_{0}^{+\infty} \frac{\frac12 \sin 4x}{x} dx = \frac12 \cdot \frac{\pi}{2} = \frac{\pi}{4} \] 因此 \[ I = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \] \[ \boxed{\frac{\pi}{4}} \] --- **(3)** \[ \int_{-\infty}^{+\infty} e^{-(a x^2 + b x + c)} dx \quad (a>0) \] **解答** 配方： \[ a x^2 + b x + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a} \] 则 \[ \int_{-\infty}^{+\infty} e^{-(a x^2+bx+c)} dx = e^{-c + \frac{b^2}{4a}} \int_{-\infty}^{+\infty} e^{-a (x+\frac{b}{2a})^2} dx \] 令 \(t = \sqrt{a}(x+\frac{b}{2a})\)，则 \[ \int_{-\infty}^{+\infty} e^{-a (x+\frac{b}{2a})^2} dx = \frac{1}{\sqrt{a}} \int_{-\infty}^{+\infty} e^{-t^2} dt = \sqrt{\frac{\pi}{a}} \] 所以 \[ \boxed{\sqrt{\frac{\pi}{a}} \, e^{\frac{b^2}{4a} - c}} \] --- **(4)** \[ \int_{-\infty}^{+\infty} e^{-(x^2 + \frac{a^2}{x^2})} dx \quad (a>0) \] **解答** 利用对称性，原积分 \(= 2\int_{0}^{+\infty} e^{-(x^2 + a^2/x^2)} dx\)。 令 \(t = x - \frac{a}{x}\)，则 \[ dt = \left(1 + \frac{a}{x^2}\right) dx \] 但更好的方法是利用公式： \[ \int_{0}^{+\infty} e^{-(x^2 + \frac{a^2}{x^2})} dx = \frac{\sqrt{\pi}}{2} e^{-2a} \] （这是已知结果，可通过变量代换 \(u = x - a/x\) 证明） 因此原积分 \[ = 2 \cdot \frac{\sqrt{\pi}}{2} e^{-2a} = \sqrt{\pi} e^{-2a} \] \[ \boxed{\sqrt{\pi} e^{-2a}} \] --- **(5)** \[ \int_{0}^{+\infty} \frac{e^{-x^2} - \cos x}{x^2} dx \] **解答** 分部积分：令 \(u = e^{-x^2} - \cos x, dv = x^{-2} dx\)，则 \[ du = (-2x e^{-x^2} + \sin x) dx, \quad v = -\frac1x \] 于是 \[ I = \left[-\frac{e^{-x^2} - \cos x}{x}\right]_{0}^{+\infty} + \int_{0}^{+\infty} \frac{-2x e^{-x^2} + \sin x}{x} dx \] 第一项在 \(x\to 0\) 时，\(e^{-x^2} - \cos x \sim (1-x^2) - (1 - x^2/2) = -x^2/2\)，除以 \(x\) 趋于0，所以第一项为0。 第二项 \[ I = \int_{0}^{+\infty} (-2 e^{-x^2}) dx + \int_{0}^{+\infty} \frac{\sin x}{x} dx = -2\cdot\frac{\sqrt{\pi}}{2} + \frac{\pi}{2} = -\sqrt{\pi} + \frac{\pi}{2} \] \[ \boxed{\frac{\pi}{2} - \sqrt{\pi}} \] --- **(6)** \[ \int_{0}^{+\infty} \frac{\sin(\alpha x) \cos(\beta x)}{x} dx \] **解答** 利用积化和差： \[ \sin(\alpha x)\cos(\beta x) = \frac12 [\sin((\alpha+\beta)x) + \sin((\alpha-\beta)x)] \] 则 \[ I = \frac12 \int_{0}^{+\infty} \frac{\sin((\alpha+\beta)x)}{x} dx + \frac12 \int_{0}^{+\infty} \frac{\sin((\alpha-\beta)x)}{x} dx \] 已知 \[ \int_{0}^{+\infty} \frac{\sin(kx)}{x} dx = \frac{\pi}{2} \operatorname{sgn}(k) \] 所以 \[ I = \frac{\pi}{4} \left[ \operatorname{sgn}(\alpha+\beta) + \operatorname{sgn}(\alpha-\beta) \right] \] 例如当 \(\alpha > \beta > 0\) 时，两项均为正，得 \(\pi/2\)；若 \(0<\alpha<\beta\)，则第二项为负，结果为0。 \[ \boxed{\frac{\pi}{

5.6.10 证明: \[ {\int }_{-1}^{1}{\left( 1 + x\right) }^{p}{\left( 1 - x\right) }^{q}\mathrm{\;d}x = {2}^{p + q + 1}\mathrm{\;B}\left( {p + 1,q + 1}\right) \] \[ \left( {p > - 1,q > - 1}\right) \text{ . } \]

5.6.11 证明: \[ {\int }_{0}^{\frac{\pi }{2}}{\sin }^{a}x\mathrm{\;d}x = {\int }_{0}^{\frac{\pi }{2}}{\cos }^{a}x\mathrm{\;d}x = \frac{1}{2}\mathrm{\;B}\left( {\frac{1}{2},\frac{a + 1}{2}}\right) . \]

5.6.12 证明: (1) \(\displaystyle{\int }_{0}^{+\infty }{\mathrm{e}}^{-{x}^{n}}\mathrm{\;d}x = \frac{1}{n}\Gamma \left( \frac{1}{n}\right) \left( {n > 0}\right)\) ； (2) \(\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{\int }_{0}^{\infty }{\mathrm{e}}^{-{x}^{n}}\mathrm{\;d}x = 1}\) .

5.6.13 设 \(\displaystyle{\int }_{-\infty }^{+\infty }\left| {f\left( x\right) }\right| \mathrm{d}x\) 存在,求证: (1) \(F\left( u\right) = {\int }_{-\infty }^{+\infty }f\left( x\right) \cos \left( {ux}\right) \mathrm{d}x\) 在 \(\left( {-\infty , + \infty }\right)\) 上连续; (2) \(F\left( u\right)\) 在 \(\left( {-\infty , + \infty }\right)\) 上一致连续.

5.6.14 设 \(f\left( x\right)\) 在 \(\lbrack 0, + \infty )\) 上可积, \(\forall A > 0,f\left( x\right) \in R\left\lbrack {0,A}\right\rbrack\) . 求证: \[ \mathop{\lim }\limits_{{a \rightarrow 0}}{\int }_{0}^{+\infty }{\mathrm{e}}^{-{ax}}f\left( x\right) \mathrm{d}x = {\int }_{0}^{+\infty }f\left( x\right) \mathrm{d}x. \]