第七章 典型综合题分析

证法 1 将 (7.1) 式改写成 \[ \frac{2\left( {b - a}\right) }{a + b} < \ln \frac{b}{a} < \frac{b - a}{\sqrt{ab}}\;\left( {b > a > 0}\right) . \tag{7.2} \] 令 \(x = b/a\) ,则 (7.2) 式又可改写成 \[ \frac{2\left( {x - 1}\right) }{x + 1} < \ln x < \frac{x - 1}{\sqrt{x}}\;\left( {x > 1}\right) . \tag{7.3} \] 作辅助函数 \(F\left( x\right) \frac{\text{ 定义 }}{}\ln x - \frac{2\left( {x - 1}\right) }{x + 1}\) . 求 \({F}^{\prime }\left( x\right)\) 得 \[ {F}^{\prime }\left( x\right) = \frac{{\left( x - 1\right) }^{2}}{x{\left( x + 1\right) }^{2}} > 0\;\left( {x > 1}\right) . \] 所以 \(F\left( x\right)\) 在 \(x \geq 1\) 上严格上升,于是对 \(\forall x > 1\) ,有 \(F\left( x\right) > F\left( 1\right) =\) 0,即 \[ \frac{2\left( {x - 1}\right) }{x + 1} < \ln x\;\left( {x > 1}\right) . \] 这就是 (7.3) 式的第一个不等式. 为了证 (7.3) 式的第二个不等式,再作辅助函数 \(G\left( x\right) \frac{\text{ 定义 }}{\sqrt{x}}\frac{x - 1}{\sqrt{x}} - \ln x\) . 求 \({G}^{\prime }\left( x\right)\) 得 \[ {G}^{\prime }\left( x\right) = \frac{{\left( \sqrt{x} - 1\right) }^{2}}{{2x}\sqrt{x}} > 0\;\left( {x > 1}\right) . \] 所以 \(G\left( x\right)\) 在 \(x \geq 1\) 上严格上升,于是对 \(\forall x > 1\) ,有 \(G\left( x\right) > G\left( 1\right) = 0\) , 即 \[ \ln x < \frac{x - 1}{\sqrt{x}}\;\left( {x > 1}\right) . \] 证法 2 令 \(t = \ln b - \ln a\) ,则 (7.1) 式改写成 \[ {\mathrm{e}}^{\frac{t}{2}} < \frac{{\mathrm{e}}^{t} - 1}{t} < \frac{1 + {\mathrm{e}}^{t}}{2}\;\left( {t > 0}\right) . \tag{7.4} \] 利用 \({\mathrm{e}}^{t}\) 的泰勒级数,容易求得,对 \(\forall t > 0\) , \[ \frac{{\mathrm{e}}^{t} - 1}{t} = 1 + \frac{t}{2} + \mathop{\sum }\limits_{{n = 2}}^{\infty }\frac{{t}^{n}}{\left( {n + 1}\right) !}, \] \[ {\mathrm{e}}^{\frac{t}{2}} = 1 + \frac{t}{2} + \mathop{\sum }\limits_{{n = 2}}^{\infty }\frac{{t}^{n}}{n!{2}^{n}} \] \[ \frac{1 + {\mathrm{e}}^{t}}{2} = 1 + \frac{t}{2} + \mathop{\sum }\limits_{{n = 2}}^{\infty }\frac{{t}^{n}}{{2n}!} \] 比较这些级数的系数, 由于 \[ {2n}! < \left( {n + 1}\right) ! < n!{2}^{n}\;\left( {n \geq 2}\right) , \] 立得 (7.4)式. 证法 3 (7.4) 式乘以 \({\mathrm{e}}^{-\frac{t}{2}}\) ,并令 \(x = \frac{t}{2}\) ,则 (7.4) 式可改写成 \[ 1 < \frac{\operatorname{sh}x}{x} < \operatorname{ch}x\;\left( {x > 0}\right) . \tag{7.5} \] 根据微分中值定理,对 \(\forall x > 0,\exists \xi \in \left( {0,x}\right)\) ,使得 \[ \operatorname{sh}x = \operatorname{sh}x - \operatorname{sh}0 = \operatorname{ch}\xi \left( {x - 0}\right) = x\operatorname{ch}\xi . \] 又因为 \[ 1 < \operatorname{ch}\xi < \operatorname{ch}x\;\left( {0 < \xi < x}\right) , \] 即得 (7.5)式 \[ 1 < \operatorname{ch}\xi = \frac{\operatorname{sh}x}{x} < \operatorname{ch}x. \] 证法 4 (7.5) 式可改写成 \[ \operatorname{th}x = \frac{\operatorname{sh}x}{\operatorname{ch}x} < x < \operatorname{sh}x\;\left( {x > 0}\right) , \tag{7.6} \] 容易验证直线 \(y = x\) 是曲线 \(y = \operatorname{sh}x\) 与曲线 \(y = \operatorname{th}x\) 在 \(x = 0\) 点的公共切线,以及当 \(x > 0\) 时, \(y = \operatorname{sh}x\) 为下凸曲线 (因 \({y}^{\prime \prime } = \operatorname{sh}x > 0\) ),所以曲线在其切线的上方; \(y = \operatorname{th}x\) 为上凸曲线 (因 \({y}^{\prime \prime } = - 2\mathrm{{sh}}x/{\mathrm{{ch}}}^{3}x <\) 0 ),所以曲线在其切线的下方. 因此 (7.6) 式成立. 证法 5 将 (7.4) 式改写成 \[ {\mathrm{e}}^{t/2} < \frac{1}{t}{\int }_{0}^{t}{\mathrm{e}}^{x}\mathrm{\;d}x < \frac{1 + {\mathrm{e}}^{t}}{2}. \tag{7.7} \] 因为 \(y = {\mathrm{e}}^{x}\) 是凹函数,所以 \[ {\mathrm{e}}^{x} = {\mathrm{e}}^{\frac{x}{t} \cdot t + \left( {1 - \frac{x}{t}}\right) \cdot 0} < \frac{x}{t}{\mathrm{e}}^{t} + \left( {1 - \frac{x}{t}}\right) \;\left( {0 < x < t}\right) . \] 上式自 0 至 \(t\) 积分,即得 \[ \frac{1}{t}{\int }_{0}^{t}{\mathrm{e}}^{x}\mathrm{\;d}x < \frac{1 + {\mathrm{e}}^{t}}{2}. \] 另一方面,还因为 \(y = {\mathrm{e}}^{x}\) 是凹函数,所以 \[ \frac{1}{t}{\int }_{0}^{t}{\mathrm{e}}^{x}\mathrm{\;d}x = \frac{1}{t}\left\lbrack {{\int }_{0}^{\frac{t}{2}}{\mathrm{e}}^{x}\mathrm{\;d}x + {\int }_{\frac{t}{2}}^{t}{\mathrm{e}}^{x}\mathrm{\;d}x}\right\rbrack \] \[ = \frac{1}{t}{\int }_{0}^{\frac{t}{2}}\left( {{\mathrm{e}}^{\frac{t}{2} - \xi } + {\mathrm{e}}^{\frac{t}{2} + \xi }}\right) \mathrm{d}\xi > \frac{1}{t}{\int }_{0}^{\frac{t}{2}}2{\mathrm{e}}^{\frac{t}{2}}\mathrm{\;d}\xi = {\mathrm{e}}^{\frac{t}{2}}. \] 证法 6 将 (7.1) 式改写成 \[ \frac{2}{a + b} < \frac{\ln b - \ln a}{b - a} < \frac{1}{\sqrt{ab}}\;\left( {b > a > 0}\right) . \tag{7.8} \] 因为 \(\displaystyle{\int }_{a}^{\sqrt{ab}}\frac{1}{x}\mathrm{\;d}x = \ln \sqrt{\frac{b}{a}} = {\int }_{\sqrt{ab}}^{b}\frac{1}{x}\mathrm{\;d}x}\) ,所以 \[ \frac{\ln b - \ln a}{b - a} = \frac{1}{b - a}{\int }_{a}^{b}\frac{1}{x}\mathrm{\;d}x = \frac{2}{b - a}{\int }_{a}^{\sqrt{ab}}\frac{1}{x}\mathrm{\;d}x. \tag{7.9} \] 又因为 \(y = 1/x\) 是凹函数,所以 \[ {\int }_{a}^{\sqrt{ab}}\frac{1}{x}\mathrm{\;d}x < \frac{1}{2}\left( {\frac{1}{a} + \frac{1}{\sqrt{ab}}}\right) \left( {\sqrt{ab} - a}\right) = \frac{b - a}{2\sqrt{ab}}. \tag{7.10} \] 联立 (7.9) 和 (7.10) 式, 即得 \[ \frac{\ln b - \ln a}{b - a} < \frac{1}{\sqrt{ab}}. \] 另一方面,还因为 \(y = 1/x\) 是凹函数,所以曲线 \(y = 1/x\) 在横坐标为 \(\frac{a + b}{2}\) 的点处的切线: \[ T\left( x\right) = \frac{2}{a + b} - \frac{4}{{\left( a + b\right) }^{2}}\left( {x - \frac{a + b}{2}}\right) \] 在曲线 \(y = \frac{1}{x}\) 的下方. 于是 \[ \frac{\ln b - \ln a}{b - a} = \frac{1}{b - a}{\int }_{a}^{b}\frac{1}{x}\mathrm{\;d}x > \frac{1}{b - a}{\int }_{a}^{b}T\left( x\right) \mathrm{d}x \] \[ = \frac{1}{b - a} \cdot \frac{1}{2}\left( {b - a}\right) \left( {T\left( a\right) + T\left( b\right) }\right) = \frac{2}{a + b}. \] 证法 7 注意 \[ \bar{x}\frac{\text{ 定义 }}{}\frac{b - a}{\ln b - \ln a} = \frac{{\int }_{a}^{b}x \cdot \frac{1}{x}\mathrm{\;d}x}{{\int }_{a}^{b}\frac{1}{x}\mathrm{\;d}x} \] 表示曲线 \(y = \frac{1}{x}\) 在 \(\left\lbrack {a,b}\right\rbrack\) 上形成的曲边梯形薄片 (记为 \(A\) ) 的重心横坐标 (薄片的面密度为 1 ). 于是 (7.1) 式可改写成 \[ \sqrt{ab} < \bar{x} = \frac{{\int }_{a}^{b}x \cdot \frac{1}{x}\mathrm{\;d}x}{{\int }_{a}^{b}\frac{1}{x}\mathrm{\;d}x} < \frac{a + b}{2}. \] 再将上式改写成两个不等式, 其中一个为 \[ {\int }_{a}^{\frac{a + b}{2}}\left( {\frac{a + b}{2} - x}\right) \frac{1}{x}\mathrm{\;d}x > {\int }_{\frac{a + b}{2}}^{b}\left( {x - \frac{a + b}{2}}\right) \frac{1}{x}\mathrm{\;d}x. \tag{7.11} \] 其物理意义是: 曲边梯形 \(A\) 在直线 \(x = \frac{a + b}{2}\) 的左边部分比右边部分对直线 \(x = \frac{a + b}{2}\) 轴的静力矩来得大. 另一个不等式为 \[ {\int }_{\sqrt{ab}}^{b}\left( {x - \sqrt{ab}}\right) \frac{1}{x}\mathrm{\;d}x > {\int }_{a}^{\sqrt{ab}}\left( {\sqrt{ab} - x}\right) \frac{1}{x}\mathrm{\;d}x, \tag{7. 12} \] 其物理意义是: 曲边梯形 \(A\) 在直线 \(x = \sqrt{ab}\) 的右边部分比左边部分对直线 \(x = \sqrt{ab}\) 轴的静力矩来得大. 先证 (7.11) 式. 对下面积分分别作 \(\frac{a + b}{2} - x = u\) 和 \(x - \frac{a + b}{2} = u\) 的变换: \[ {\int }_{a}^{\frac{a + b}{2}}\left( {\frac{a + b}{2} - x}\right) \frac{1}{x}\mathrm{\;d}x - {\int }_{\frac{a + b}{2}}^{b}\left( {x - \frac{a + b}{2}}\right) \frac{1}{x}\mathrm{\;d}x \] \[ = {\int }_{0}^{\frac{b - a}{2}}\frac{u\mathrm{\;d}u}{\frac{a + b}{2} - u} - {\int }_{0}^{\frac{b - a}{2}}\frac{u\mathrm{\;d}u}{u + \frac{a + b}{2}} \] \[ = {\int }_{0}^{\frac{b - a}{2}}\frac{2{u}^{2}\mathrm{\;d}u}{{\left( \frac{a + b}{2}\right) }^{2} - {u}^{2}} > 0. \] 再证 (7.12) 式. 对下面积分分别作 \(\frac{x}{\sqrt{ab}} = t\) 和 \(\frac{\sqrt{ab}}{x} = t\) 的变换: \[ {\int }_{\sqrt{ab}}^{b}\left( {x - \sqrt{ab}}\right) \frac{1}{x}\mathrm{\;d}x - {\int }_{a}^{\sqrt{ab}}\left( {\sqrt{ab} - x}\right) \frac{1}{x}\mathrm{\;d}x \] \[ = \sqrt{ab}{\int }_{1}^{\sqrt{b/a}}\frac{t - 1}{t}\mathrm{\;d}t - \sqrt{ab}{\int }_{1}^{\sqrt{b/a}}\frac{t - 1}{{t}^{2}}\mathrm{\;d}t \] \[ = \sqrt{ab}{\int }_{1}^{\sqrt{b/a}}{\left( \frac{t - 1}{t}\right) }^{2}\mathrm{\;d}t > 0. \] 联立 (7.11) 和 (7.12) 式, 即得 (7.1) 式成立. 证法 8 设 \(f\left( x\right) \overset{\text{ 定义 }}{ = }\ln \frac{b + x}{a + x}\) ,则 \[ \ln b - \ln a = \ln \frac{b}{a} = f\left( 0\right) - f\left( {+\infty }\right) = - {\int }_{0}^{+\infty }{f}^{\prime }\left( x\right) \mathrm{d}x \] \[ = \left( {b - a}\right) {\int }_{0}^{+\infty }\frac{\mathrm{d}x}{\left( {x + a}\right) \left( {b + x}\right) }. \tag{7.13} \] 又 \[ {\left( x + \sqrt{ab}\right) }^{2} < \left( {x + a}\right) \left( {x + b}\right) < {\left( x + \frac{a + b}{2}\right) }^{2}\;\left( {x > 0}\right) . \] (7.14) 联立 (7.13) 和 (7.14) 式,一方面, \[ \frac{\ln b - \ln a}{b - a} = {\int }_{0}^{+\infty }\frac{\mathrm{d}x}{\left( {x + a}\right) \left( {x + b}\right) } > {\int }_{0}^{+\infty }\frac{\mathrm{d}x}{{\left( x + \frac{a + b}{2}\right) }^{2}} = \frac{2}{a + b}; \] 另一方面, \[ \frac{\ln b - \ln a}{b - a} = {\int }_{0}^{+\infty }\frac{\mathrm{d}x}{\left( {x + a}\right) \left( {x + b}\right) } < {\int }_{0}^{+\infty }\frac{\mathrm{d}x}{{\left( x + \sqrt{ab}\right) }^{2}} = \frac{1}{\sqrt{ab}}. \]

例 2 设 \(f\left( x\right) = {\mathrm{e}}^{{x}^{2}}{\int }_{x}^{+\infty }{\mathrm{e}}^{-{t}^{2}}\mathrm{\;d}t\) ,求证: \(f\left( x\right) \leq \sqrt{\pi }/2\left( {x \geq 0}\right)\) .

例 3 求证: \(f\left( x\right) = x{\mathrm{e}}^{-{x}^{2}}{\int }_{0}^{x}{\mathrm{e}}^{{t}^{2}}\mathrm{\;d}t\) 在 \(\left( {-\infty ,\infty }\right)\) 上有界.

例 4 设 \(f\left( x\right)\) 在 \(\mathbf{R}\) 上连续可导,且 \(\mathop{\sup }\limits_{{x \in \mathbf{R}}}\left| {{\mathrm{e}}^{-{x}^{2}}{f}^{\prime }\left( x\right) }\right| < + \infty\) ,求证: \[ \mathop{\sup }\limits_{{x \in \mathbb{R}}}\left| {x{\mathrm{e}}^{-{x}^{2}}f\left( x\right) }\right| < + \infty . \]

例 5 设 \({f}_{n}\left( x\right) = \cos x + {\cos }^{2}x + \cdots + {\cos }^{n}x\) . 求证: (1)对任意自然数 \(n\) ,方程 \({f}_{n}\left( x\right) = 1\) 在 \(\lbrack 0,\pi /3)\) 内有且仅有一个根; (2)设 \({x}_{n} \in \left\lbrack {0,\frac{\pi }{3}}\right)\) 是 \({f}_{n}\left( x\right) = 1\) 的根,则 \(\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \frac{\pi }{3}}\) .

例 6 设 \(f\left( x\right) = \frac{1}{1 - x - {x}^{2}}\) ,求证 \(\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{n!}{{f}^{\left( n\right) }\left( 0\right) }\) 收敛.

例 7 设 \(f\left( x\right)\) 在 \(\left( {-\infty ,\infty }\right)\) 连续, \[ {\int }_{-\infty }^{+\infty }\left| {f\left( x\right) }\right| \mathrm{d}x < + \infty ,\;{\int }_{-\infty }^{+\infty }{\left| f\left( x\right) \right| }^{2}\mathrm{\;d}x < + \infty . \] 定义 \[ \psi \left( x\right) = {\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{\mathrm{e}}^{-\left( {\left| {x - \xi }\right| + \left| {x - \eta }\right| }\right) }\left| {f\left( \xi \right) }\right| \left| {f\left( \eta \right) }\right| \mathrm{d}\xi \mathrm{d}\eta . \] 求证: \[ {\int }_{-\infty }^{+\infty }\psi \left( x\right) \mathrm{d}x \leq 4{\int }_{-\infty }^{+\infty }{\left| f\left( x\right) \right| }^{2}\mathrm{\;d}x. \]

例 8 设 \(f\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }n{\mathrm{e}}^{-n}\cos {nx}\) ,求证: (1) \(\mathop{\max }\limits_{{0 \leq x \leq {2\pi }}}\left| {f\left( x\right) }\right| \geq \frac{2}{\mathrm{e}}\) ； (2) \({f}^{\prime }\left( x\right)\) 存在; (3) \(\mathop{\max }\limits_{{0 \leq x \leq {2\pi }}}\left| {{f}^{\prime }\left( x\right) }\right| \geq \frac{2}{\pi \mathrm{e}}\) .

例 9 求证: \(t\) 的方程 \[ \frac{{x}^{2}}{a - t} + \frac{{y}^{2}}{b - t} + \frac{{z}^{2}}{c - t} = 1\;\left( {a > b > c}\right) \] ( 1 )有三个不同实根 \({t}_{1},{t}_{2},{t}_{3}\) ,分别属于区间 \(\displaystyle{- \infty < {t}_{1} < c,c < {t}_{2} <}\) \(b,b < {t}_{3} < a\) ,其中(x, y, z)不在坐标平面上; (2)过任意点的三个曲面 \({t}_{i}\left( {x,y,z}\right) =\) 常数 \(\left( {i = 1,2,3}\right)\) 互相正交.

例 10 求证: \(\displaystyle{\int }_{0}^{1}\frac{\ln \left( {1 + x}\right) }{1 + {x}^{2}}\mathrm{\;d}x = \frac{\pi }{8}\ln 2\) .

例 11 设 \(f\left( x\right)\) 在 \(\left( {-\infty , + \infty }\right)\) 上二次连续可微, \[ \left| {f\left( x\right) }\right| \leq 1,\;{\left| f\left( 0\right) \right| }^{2} + {\left| {f}^{\prime }\left( 0\right) \right| }^{2} = 4. \] 求证: \(\exists \xi \in \mathbf{R}\) ,使得 \(f\left( \xi \right) + {f}^{\prime \prime }\left( \xi \right) = 0\) .

例 12 设 \(f\left( x\right)\) 无穷次可微,且 \[ \left| {{f}^{\left( n\right) }\left( x\right) }\right| \leq M,\;f\left( {1/n}\right) = 0\;\left( {n = 1,2,\cdots }\right) . \] 求证: \(f\left( x\right) \equiv 0\) .

例 13 设 \(f\left( x\right)\) 在 \(\left( {-\infty , + \infty }\right)\) 可微, \(f\left( 0\right) = 0\) ,且处处有 \(\left| {{f}^{\prime }\left( x\right) }\right|\) \(\leq \left| {f\left( x\right) }\right|\) . 求证: \(f\left( x\right) \equiv 0\left( {\forall x \in \mathbf{R}}\right)\) .

例 14 设 \(f\left( x\right) \in C\left\lbrack {0,1}\right\rbrack\) ,且 \(\displaystyle{\int }_{0}^{1}f\left( x\right) \mathrm{d}x = 0,{\int }_{0}^{1}{xf}\left( x\right) \mathrm{d}x = 1\) . 求证: \[ \mathop{\max }\limits_{{0 \leq x \leq 1}}\left| {f\left( x\right) }\right| > 4 \]

例 15 设 \({f}^{\prime \prime }\left( x\right) \geq 0\left( {-\infty < x < + \infty }\right) ,u\left( x\right) \in C\left( {-\infty , + \infty }\right)\) . 求证: 对 \(\forall a > 0\) ,有 \[ \frac{1}{a}{\int }_{0}^{a}f\left\lbrack {u\left( x\right) }\right\rbrack \mathrm{d}x \geq f\left( {\frac{1}{a}{\int }_{0}^{a}u\left( x\right) \mathrm{d}x}\right) . \]

例 16 设 \(f\left( x\right)\) 在 \(\left\lbrack {a,b}\right\rbrack\) 上可微, \({f}^{\prime }\left( x\right)\) 单调上升并满足 \(\left| {{f}^{\prime }\left( x\right) }\right|\) \(\geq m > 0\) . 求证: \[ \left| {{\int }_{a}^{b}\cos f\left( x\right) \mathrm{d}x}\right| \leq \frac{2}{m}. \]

例 17 设 \(f\left( x\right)\) 在 \(\left\lbrack {a,b}\right\rbrack\) 上可微,且 \(\left| {{f}^{\prime }\left( x\right) }\right| \leq M\) . 求证: \[ \left| {\frac{1}{b - a}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x - \frac{f\left( a\right) + f\left( b\right) }{2}}\right| \leq \frac{M\left( {b - a}\right) }{4}\left( {1 - {\theta }^{2}}\right) , \] 其中 \(\theta \overset{\text{ 定义 }}{ = }\frac{f\left( b\right) - f\left( a\right) }{M\left( {b - a}\right) }\) .

例 18 设 \(f\left( x\right)\) 在 \(\left( {-\infty , + \infty }\right)\) 上二次连续可微,并满足: (1) \(f\left( x\right) \leq {f}^{\prime \prime }\left( x\right)\) ; (2) \(\mathop{\lim }\limits_{{x \rightarrow \pm \infty }}{\mathrm{e}}^{-\left| x\right| }f\left( x\right) = 0\) . 求证: \(f\left( x\right) \leq 0\left( {\forall x \in \left( {-\infty , + \infty }\right) }\right)\) .

例 19 设 \(f\left( x\right) \in {C}^{1}\lbrack 0,\infty ),x{f}^{\prime }\left( x\right)\) 在 \(\lbrack 0,\infty )\) 上有界,并且 \[ \frac{1}{x}{\int }_{x}^{2x}\left| {f\left( t\right) }\right| \mathrm{d}t \rightarrow 0\;\left( {x \rightarrow + \infty }\right) . \] 求证: \(f\left( x\right) \rightarrow 0\left( {x \rightarrow + \infty }\right)\) .

例 20 假设函数 \(f\left( x\right)\) 在闭区间 \(\left\lbrack {0,b}\right\rbrack\) 上单调增加,函数 \(g\left( x\right)\) 使得广义积分 \(\displaystyle{\int }_{0}^{+\infty }\frac{g\left( x\right) }{x}\mathrm{\;d}x\) 收敛. 求证: \[ \mathop{\lim }\limits_{{p \rightarrow + \infty }}{\int }_{0}^{b}f\left( x\right) \frac{g\left( {px}\right) }{x}\mathrm{\;d}x = f\left( {0}^{ + }\right) {\int }_{0}^{+\infty }\frac{g\left( x\right) }{x}\mathrm{\;d}x. \]

例 21 设 \(F\left( {x,y,z}\right)\) 在 \({\mathbf{R}}^{3}\) 中有连续的一阶偏导数,并满足 \[ y\frac{\partial F}{\partial x} - x\frac{\partial F}{\partial y} + \frac{\partial F}{\partial z} \geq \alpha > 0\;\left( {\alpha \text{ 为常数 }}\right) . \] 求证: 当点(x, y, z)沿着曲线 \(\Gamma : x = - \cos t,y = \sin t,z = t\left( {t \geq 0}\right)\) 趋向无穷时, \(F\left( {x,y,z}\right) \rightarrow + \infty\) .

例 22 设 \(u = u\left( {x,y}\right)\) 在平面区域 \(D\) 上二阶连续可微. 求证: \[ \frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} \geq 0\;\left( {\forall \left( {x,y}\right) \in D}\right) \] 成立的充要条件为 \[ u\left( {{x}_{0},{y}_{0}}\right) \leq \frac{1}{2\pi }{\int }_{0}^{2\pi }u\left( {{x}_{0} + r\cos \theta ,{y}_{0} + r\sin \theta }\right) \mathrm{d}\theta \;\left( {\forall \left( {{x}_{0},{y}_{0}}\right) \in D}\right) , \] 其中 \(0 \leq r < d\left( {{x}_{0},{y}_{0}}\right) ,d\) 是点 \(\left( {{x}_{0},{y}_{0}}\right)\) 到 \(D\) 的边界 \(\partial D\) 的距离.

例 23 设 \(U\left( {{x}_{0},{\delta }_{0}}\right) \subset {\mathbf{R}}^{n},f \in {C}^{2}\left( {U\left( {{x}_{0},{\delta }_{0}}\right) }\right) ,\nabla f\left( {x}_{0}\right) = 0\) . 又设对任意单位向量 \(\mathbf{\alpha } \in {\mathbf{R}}^{n},{\left( \mathbf{\alpha } \cdot \nabla \right) }^{2}f\left( {\mathbf{x}}_{0}\right) > 0\) . 求证: \(\exists 0 < \delta < {\delta }_{0}\) ,使 \[ \left( {x - {x}_{0}}\right) \cdot \nabla f\left( x\right) > 0.\;\left( {\forall x \in U\left( {{x}_{0},\delta }\right) \smallsetminus \left\{ {x}_{0}\right\} }\right) . \]

例 24 设 \(U\left( {{x}_{0},\delta }\right) \subset {\mathbf{R}}^{n},f\) 在 \(U\left( {{x}_{0},\delta }\right)\) 内连续,在 \(U\left( {{x}_{0},\delta }\right) \smallsetminus \left\{ {x}_{0}\right\}\) 内可微. 求证: (1)如果 \(\left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) \nabla f\left( \mathbf{x}\right) < 0\left( {\forall \mathbf{x} \in U\left( {{\mathbf{x}}_{0},\delta }\right) \smallsetminus \left\{ {\mathbf{x}}_{0}\right\} }\right)\) ,那么 \({\mathbf{x}}_{0}\) 是 \(f\) 的一个极大值点; (2)如果 \(\left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) \cdot \nabla f\left( \mathbf{x}\right) > 0\left( {\forall \mathbf{x} \in U\left( {{\mathbf{x}}_{0},\delta }\right) \smallsetminus \left\{ {\mathbf{x}}_{0}\right\} }\right)\) ,那么 \({\mathbf{x}}_{0}\) 是 \(f\) 的一个极小值点; (3) 如果 \(f\left( {x,y}\right) = {x}^{2} + {2xy} + 3{y}^{2} + {2x} + {10y} + 9\) ,则(1, - 2)是 \(f\left( {x,y}\right)\) 的一个极小值点.

例 25 设 \(f\left( x\right)\) 在 \(\mathbf{R}\) 上有二阶连续导数,且有 \[ {M}_{0} = {\int }_{-\infty }^{+\infty }\left| {f\left( x\right) }\right| \mathrm{d}x < + \infty ,\;{M}_{2} = {\int }_{-\infty }^{+\infty }\left| {{f}^{\prime \prime }\left( x\right) }\right| \mathrm{d}x < + \infty . \] 求证: (1) \({M}_{1} = {\int }_{-\infty }^{\infty }\left| {{f}^{\prime }\left( x\right) }\right| \mathrm{d}x < + \infty\) ; (2) \({M}_{1}^{2} \leq 4{M}_{0}{M}_{2}\) .

例 26 设 \(D\) 为 \({\mathbf{R}}^{n}\) 中的有界闭集,映射 \(f : D \rightarrow D\) 满足: \(\forall x,y \in\) \(D,\mathbf{x} \neq \mathbf{y}\) ,有 \(\left| {\mathbf{f}\left( \mathbf{x}\right) - \mathbf{f}\left( \mathbf{y}\right) }\right| < \left| {\mathbf{x} - \mathbf{y}}\right|\) . 证明: 映射 \(\mathbf{f}\) 有惟一的不动点 \({\mathbf{x}}^{ * } \in D\) ,即有惟一点 \({\mathbf{x}}^{ * } \in D\) ,使 \(\mathbf{f}\left( {\mathbf{x}}^{ * }\right) = {\mathbf{x}}^{ * }\) .

例 27 设 \(D : {x}^{2} + {y}^{2} < 1.f\left( {x,y}\right)\) 为有界正值函数,在 \(D\) 上有二阶连续偏导数, 且满足 \(\Delta \ln f\left( {x,y}\right) \geq {f}^{2}\left( {x,y}\right) \;\left( {\Delta \text{ 为拉普拉斯算符,即 }\Delta = \frac{{\partial }^{2}}{\partial {x}^{2}} + \frac{{\partial }^{2}}{\partial {y}^{2}}}\right) .\)

例 28 设 \(f\left( {x,y}\right)\) 在 \({x}^{2} + {y}^{2} < 1\) 上二次连续可微,且满足 \[ \frac{{\partial }^{2}f}{\partial {x}^{2}} + \frac{{\partial }^{2}f}{\partial {y}^{2}} = {\mathrm{e}}^{-\left( {{x}^{2} + {y}^{2}}\right) }. \]

例 29 设 \(\Omega\) 为空间第一卦限区域,函数 \(f\left( {x,y,z}\right)\) 在 \(\Omega\) 上有连续一阶偏导数. \(S\) 为 \(\mathbf{\Omega }\) 中任一光滑闭曲面,试给出第二型曲面积分 \[ {\iint }_{S}f\left( {x,y,z}\right) \left( {x\mathrm{\;d}y\mathrm{\;d}z + y\mathrm{\;d}z\mathrm{\;d}x + z\mathrm{\;d}x\mathrm{\;d}y}\right) = 0 \] 的充要条件, 并证明之.

例 30 设 \(A > 0,{AC} - {B}^{2} > 0\) . 求平面曲线 \(A{x}^{2} + {2Bxy} + C{y}^{2} = 1\) 所围的图形面积.

例 31 利用级数收敛性证明无穷积分 \(\displaystyle{\int }_{0}^{+\infty }{\left( -1\right) }^{\left\lbrack {x}^{2}\right\rbrack }\) 收敛,其中 \(\left\lbrack {x}^{2}\right\rbrack\) 表示不超过 \({x}^{2}\) 的最大整数.

证 对 \(\forall n,k \in N\) ,注意到 \[ \mathop{\sum }\limits_{{n = 2}}^{k}\frac{{a}_{n}}{{n}^{2} + {m}^{2}} \leq {\int }_{1}^{k}\frac{{a}_{n}}{{n}^{2} + {x}^{2}}\mathrm{\;d}x \] \[ = \frac{{a}_{n}}{n}\left( {\arctan \frac{k}{n} - \arctan \frac{1}{n}}\right) < \frac{\pi }{2} \cdot \frac{{a}_{n}}{n}, \] 于是有 \[ {c}_{n}\overset{\text{ 定义 }}{ = }\mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{n}}{{n}^{2} + {m}^{2}} = \frac{{a}_{n}}{{n}^{2} + 1} + \mathop{\sum }\limits_{{m = 2}}^{\infty }\frac{{a}_{n}}{{n}^{2} + {m}^{2}} \] \[ \leq \frac{1}{2}\left( {\pi + 1}\right) \frac{{a}_{n}}{n}, \] 于是 \(\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n}}\) 收敛,即 \(\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{n}}{{n}^{2} + {m}^{2}}}\) 收敛.

例 33 设有一座小山,取它的底面所在的平面为 \({Oxy}\) 坐标平面, 其底部所占的区域为 \[ D = \left\{ {\left( {x,y}\right) \mid {x}^{2} + {y}^{2} - {xy} \leq {75}}\right\} . \] 小山的高度函数为 \(h\left( {x,y}\right) = {75} - {x}^{2} - {y}^{2} + {xy}\) . (1)设 \(M\left( {{x}_{0},{y}_{0}}\right)\) 为区域 \(D\) 上一点,问 \(h\left( {x,y}\right)\) 在该点沿平面上什么方向的方向导数最大? 若记此方向导数的最大值为 \(g\left( {{x}_{0},{y}_{0}}\right)\) , 试写出 \(g\left( {{x}_{0},{y}_{0}}\right)\) 的表达式. (2)现欲利用此小山开展攀岩活动,为此需要在山脚寻找一上山坡度最大的点作为攀登的起点. 也就是说,要在 \(D\) 的边界线 \({x}^{2} +\) \({y}^{2} - {xy} = {75}\) 上找出使 (1) 中的 \(g\left( {x,y}\right)\) 达到最大值的点. 试确定攀登起点的位置.