武汉大学2013年数学分析考研试题参考解答

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一：

1：解：\[\because \underset{x\to 0}{\mathop{\lim }}\,\ln (1+x)=x\]

\[\therefore \underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt[n]{1+x}-1}{\ln (1+x)}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt[n]{1+x}-1}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{n}{{(1+x)}^{\frac{1}{n}-1}}=\frac{1}{n}\]

2：解：$\int{\frac{x\ln (x+\sqrt{1+{{x}^{2}}})}{{{(1+{{x}^{2}})}^{2}}}}dx=-\frac{1}{2}\int{\frac{-2x}{{{(1+{{x}^{2}})}^{2}}}\ln (x+\sqrt{1+{{x}^{2}}})}dx=$

$-\frac{1}{2}[\frac{1}{(1+{{x}^{2}})}\ln (x+\sqrt{1+{{x}^{2}}})-\int{\frac{1}{(1+{{x}^{2}})}\cdot \frac{\frac{x+\sqrt{1+{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}}}{x+\sqrt{1+{{x}^{2}}}}}dx=\int{\frac{dx}{{{(1+{{x}^{2}})}^{\frac{3}{2}}}}}$

而$\int{\frac{dx}{{{(1+{{x}^{2}})}^{\frac{3}{2}}}}}\overset{x=\tan \theta }{\mathop{=}}\,\int{\frac{d\tan \theta }{\frac{1}{{{(\cos \theta )}^{3}}}}}=\int{\cos \theta d\theta }=\sin \theta +C=\frac{x}{\sqrt{1+{{x}^{2}}}}+C$

于是$\int{\frac{x\ln (x+\sqrt{1+{{x}^{2}}})}{{{(1+{{x}^{2}})}^{2}}}}dx=-\frac{\ln (x+\sqrt{1+{{x}^{2}}})}{2(1+{{x}^{2}})}+\frac{x}{2\sqrt{1+{{x}^{2}}}}+C$（其中$C$为任意常数）

3：解：$\because \int_{0}^{\frac{\pi }{2}}{\sqrt{1-\sin 2x}}dx=\int_{0}^{\frac{\pi }{2}}{\sqrt{{{(\sin x-\cos x)}^{2}}}}dx=\int_{0}^{\frac{\pi }{2}}{\left| \sin x-\cos x \right|}dx$

\[=\int_{0}^{\frac{\pi }{4}}{(\cos x-\sin x)dx+}\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}{(sinx-\cos x)dx=2(\sqrt{2}}-1)\]

4：解：$\because y=\arcsin x=x+\frac{1}{2}\cdot \frac{{{x}^{3}}}{3}+\frac{1\times 3}{2\times 4}\cdot \frac{{{x}^{5}}}{5}+\cdots +\frac{(2n-1)!!}{(2n)!!}\cdot \frac{{{x}^{2n+1}}}{2n+1}+O({{x}^{2n+1}})$

于是当$n=2k$时，${{y}^{(n)}}(0)=0$

当$n=2k+1$时，${{y}^{(2k+1)}}(x)=\frac{(2k-1)!!}{(2k)!!}\cdot \frac{(2k+1)!}{2k+1}+O(1)$

则\[{{y}^{(2k+1)}}(0)=\frac{(2k-1)!!}{(2k)!!}\cdot \frac{(2k+1)!}{2k+1}={{[(2k-1)!!]}^{2}}={{[(n-2)!!]}^{2}}\]

5：解：$\because nx+k-1\sqrt{(nx+k)(nx+k-1)}nx+k$

\[\therefore \frac{1}{n}\sum\limits_{k=1}^{n}{(x+\frac{k-1}{n}}){{S}_{n}}=\frac{1}{{{n}^{2}}}\sum\limits_{k=1}^{n}{\sqrt{(nx+k)(nx+k-1)}}\frac{1}{n}\sum\limits_{k=1}^{n}{(x+\frac{k}{n}})\]

$\therefore \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{(x+\frac{k-1}{n}})\le \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}\le \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{(x+\frac{k}{n}})$

而不等式左边

$\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{(x+\frac{k-1}{n}})\overset{i=k-1}{\mathop{=}}\,\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{1=0}^{n-1}{(x+\frac{i}{n}})=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}[\sum\limits_{1=1}^{n}{(x+\frac{i}{n}})+(x+0)-(x+1)]=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{1=1}^{n}{(x+\frac{i}{n}})=$由迫敛性知：$\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{1=1}^{n}{(x+\frac{i}{n}})=\int_{0}^{1}{(x+t)dt=x+\frac{1}{2}}$

二：证明：由于${{a}_{2}}={{a}^{\frac{3}{4}}},{{a}_{3}}={{a}^{\frac{7}{8}}}$ $(a0)$

于是分三种情况：

1 当$a=1$时，此时${{x}_{n}}=1$，则$\{{{x}_{n}}\}$收敛且$\underset{n\to \infty }{\mathop{\lim }}\,{{x}_{n}}=1$

2 当$0a1$时，下证$a{{x}_{n}}\le \sqrt{a}$ （数学归纳法）

（1） 当$n=1$时，$a{{a}_{1}}=\sqrt{a}$成立；

（2） 设$n=k$时，$a{{x}_{k}}\le \sqrt{a}$，则$a{{x}_{n+1}}=\sqrt{a{{x}_{n}}}\le \sqrt{a\sqrt{a}}\le \sqrt{a}$

即当$n=k+1$时，也成立

于是对\[\forall n\in {{N}_{+}},a{{x}_{n}}\le \sqrt{a}\]

则$\frac{{{x}_{n+1}}}{{{x}_{n}}}=\frac{\sqrt{a}}{\sqrt{{{x}_{n}}}}1$

于是$\{{{x}_{n}}\}$单调递减且${{x}_{n}}a$

由单调有解原理知：$\{{{x}_{n}}\}$收敛，设$\underset{n\to \infty }{\mathop{\lim }}\,{{x}_{n}}=l$

由${{x}_{n+1}}=\sqrt{a{{x}_{n}}}$，两边取极限，于是有$\underset{n\to \infty }{\mathop{\lim }}\,{{x}_{n}}=a$

3 当$a1$时，同理可证$\sqrt{a}\le {{x}_{n}}a$，得$\{{{x}_{n}}\}$单调递增

同样由单调有解原理知：$\{{{x}_{n}}\}$收敛并且$\underset{n\to \infty }{\mathop{\lim }}\,{{x}_{n}}=a$

综上所述：对$\forall a0$，$\{{{x}_{n}}\}$收敛且$\underset{n\to \infty }{\mathop{\lim }}\,{{x}_{n}}=a$

三：证明：$\because \int_{0}^{+\infty }{\frac{dx}{{{(1+x)}^{2}}(1+{{x}^{\alpha }})}}=\int_{0}^{1}{\frac{dx}{{{(1+x)}^{2}}(1+{{x}^{\alpha }})}}+\int_{1}^{+\infty }{\frac{dx}{{{(1+x)}^{2}}(1+{{x}^{\alpha }})}}$

而\[\int_{1}^{+\infty }{\frac{dx}{{{(1+x)}^{2}}(1+{{x}^{\alpha }})}}\overset{t=\frac{1}{x}}{\mathop{=}}\,\int_{1}^{0}{\frac{-\frac{1}{{{x}^{2}}}dx}{{{(1+\frac{1}{x})}^{2}}(1+\frac{1}{{{x}^{\alpha }}})}=}\int_{0}^{1}{\frac{{{x}^{\alpha }}dx}{{{(1+x)}^{2}}(1+{{x}^{\alpha }})}}\]

于是

\[\int_{0}^{+\infty }{\frac{dx}{{{(1+x)}^{2}}(1+{{x}^{\alpha }})}}=\int_{0}^{1}{\frac{1+{{x}^{\alpha }}}{{{(1+x)}^{2}}(1+{{x}^{\alpha }})}dx=\int_{0}^{1}{\frac{dx}{{{(1+x)}^{2}}}}}=\frac{1}{2}(\]

反常积分\[\int_{0}^{+\infty }{\frac{dx}{{{(1+x)}^{2}}(1+{{x}^{\alpha }})}}\]与$\alpha $无关且值为$\frac{1}{2}$

四、证明：不妨设$F(x)=f(x+1)-f(x),x\in [0,1]$

于是$F(0)=f(1)-f(0)$

$F(1)=f(2)-f(1)=f(0)-f(1)$

于是$F(0)\cdot F(1)=-{{[f(0)-f(1)]}^{2}}\le 0$

（1）若$F(0)=0$或，可得$f(0)=f(1)=f(2)$

则只需取$\xi =0$或$1$，即证

（2）若$F(0)\cdot F(1)0$，于是$F(0)$与$F(1)$异号，于是由介值定理知：

$\exists \xi \in [0,1]$，使得$f(\xi )=f(\xi +1)$

五、证明：$\because u=xy,v=x-y$

$\therefore \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial x}=y\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}$

$\frac{{{\partial }^{2}}z}{\partial {{x}^{2}}}=y[\frac{\partial (\frac{\partial z}{\partial u})}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial (\frac{\partial z}{\partial u})}{\partial v}\cdot \frac{\partial v}{\partial x}]+[\frac{\partial (\frac{\partial z}{\partial v})}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial (\frac{\partial z}{\partial v})}{\partial v}\cdot \frac{\partial v}{\partial x}]$

$={{y}^{2}}\frac{{{\partial }^{2}}z}{\partial {{u}^{2}}}+2y\frac{{{\partial }^{2}}z}{\partial u\partial v}+\frac{{{\partial }^{2}}z}{\partial {{v}^{2}}}$

$\frac{{{\partial }^{2}}z}{\partial x\partial y}=xy\frac{{{\partial }^{2}}z}{\partial {{u}^{2}}}+(x-y)\frac{{{\partial }^{2}}z}{\partial x\partial y}-\frac{{{\partial }^{2}}z}{\partial {{v}^{2}}}+\frac{\partial z}{\partial u}$

同理可证：$\therefore \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial y}=x\frac{\partial z}{\partial u}-\frac{\partial z}{\partial v}$

$\frac{{{\partial }^{2}}z}{\partial {{x}^{2}}}=={{x}^{2}}\frac{{{\partial }^{2}}z}{\partial {{u}^{2}}}-2x\frac{{{\partial }^{2}}z}{\partial u\partial v}+\frac{{{\partial }^{2}}z}{\partial {{v}^{2}}}$

带入已知化简得：\[\frac{{{\partial }^{2}}z}{\partial {{u}^{2}}}+\frac{1}{{{(x+y)}^{2}}}\frac{\partial z}{\partial u}=0\]

而${{(x+y)}^{2}}={{(x-y)}^{2}}+4xy={{v}^{2}}+4u$

代入即得\[\frac{{{\partial }^{2}}z}{\partial {{u}^{2}}}+\frac{1}{{{v}^{2}}+4u}\frac{\partial z}{\partial u}=0\]

六：1：解：如图：

$A=\iint_{D}{\left| xy-\frac{1}{4} \right|}dxdy=A=\iint_{{{D}_{1}}}{(xy-\frac{1}{4})}dxdy+\iint_{{{D}_{2}}}{(-xy+\frac{1}{4})}dxdy$

$=\int_{\frac{1}{4}}^{1}{dx\int_{\frac{1}{4x}}^{1}{(xy-\frac{1}{4})dy+[}}\int_{0}^{\frac{1}{4}}{dx\int_{0}^{1}{(-xy+\frac{1}{4})dy+\int_{\frac{1}{4}}^{1}{dx\int_{0}^{\frac{1}{4x}}{(-xy+\frac{1}{4})dy]}}}}$

$=\frac{1}{8}\ln 2+\frac{3}{32}$

2：证明：$\because \left| \iint_{D}{(xy-\frac{1}{4})f(x,y)dxdy} \right|\le \iint_{D}{\left| xy-\frac{1}{4} \right|\left| f(x,y) \right|dxdy}$

由于$\left| xy-\frac{1}{4} \right|\ge 0,\left| f(x,y) \right|\ge 0$

于是$\exists ({{x}^{*}},{{y}^{*}})\in D$，使得

根据积分定理知：$\iint_{D}{\left| xy-\frac{1}{4} \right|\left| f(x,y) \right|dxdy}=\iint_{D}{\left| (xy-\frac{1}{4}) \right|dxdy}\cdot \left| f({{x}^{*}},{{y}^{*}}) \right|$

$=A\left| f({{x}^{*}},{{y}^{*}}) \right|$

而

\[\left| \iint_{D}{(xy-\frac{1}{4})f(x,y)dxdy} \right|=\left| \iint_{D}{xyf(x,y)dxdy-\frac{1}{4}\iint_{D}{f(x,y)dxdy}} \right|=1\]

于是$\exists ({{x}^{*}},{{y}^{*}})\in D$使得$\left| f({{x}^{*}},{{y}^{*}}) \right|\ge \frac{1}{A}$

七：解：设切点为$({{x}_{0}},{{y}_{0}},{{z}_{0}})$，设$f(x,y,z)=\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}+\frac{{{z}^{2}}}{{{c}^{2}}}$

从而${{f}_{x}}({{x}_{0}},{{y}_{0}},{{z}_{0}})=\frac{2{{x}_{0}}}{{{a}^{2}}},{{f}_{y}}({{x}_{0}},{{y}_{0}},{{z}_{0}})=\frac{2{{y}_{0}}}{{{b}^{2}}},{{f}_{z}}({{x}_{0}},{{y}_{0}},{{z}_{0}})=\frac{2{{z}_{0}}}{{{c}^{2}}}$

从而$\pi $的表达式为$\frac{2{{x}_{0}}}{{{a}^{2}}}(x-{{x}_{0}})+\frac{2{{y}_{0}}}{{{b}^{2}}}(y-{{y}_{0}})+\frac{2{{z}_{0}}}{{{c}^{2}}}(z-{{z}_{0}})=0$

且$\frac{x_{0}^{2}}{{{a}^{2}}}+\frac{y_{0}^{2}}{{{b}^{2}}}+\frac{z_{0}^{2}}{{{c}^{2}}}=1$，代入化简得：$\frac{{{x}_{0}}}{{{a}^{2}}}x+\frac{{{y}_{0}}}{{{b}^{2}}}y+\frac{{{z}_{0}}}{{{c}^{2}}}z=1$

于是$\pi $在第一象限的部分与三个坐标的坐标分别为

$(\frac{{{a}^{2}}}{{{x}_{0}}},0,0),(0,\frac{{{b}^{2}}}{{{y}_{0}}},0),(0,0,\frac{{{c}^{2}}}{{{z}_{0}}})$，可知${{x}_{0}},{{y}_{0}},{{z}_{0}}0$

于是$V=\frac{1}{6}\cdot \frac{{{a}^{2}}}{{{x}_{0}}}\cdot \frac{{{b}^{2}}}{{{y}_{0}}}\cdot \frac{{{c}^{2}}}{{{z}_{0}}}$，且$\frac{x_{0}^{2}}{{{a}^{2}}}+\frac{y_{0}^{2}}{{{b}^{2}}}+\frac{z_{0}^{2}}{{{c}^{2}}}=1$

由广义均值不等式知：\[\frac{x_{0}^{2}}{{{a}^{2}}}+\frac{y_{0}^{2}}{{{b}^{2}}}+\frac{z_{0}^{2}}{{{c}^{2}}}\ge 3\sqrt[3]{\frac{x_{0}^{2}}{{{a}^{2}}}\cdot \frac{y_{0}^{2}}{{{b}^{2}}}\cdot \frac{z_{0}^{2}}{{{c}^{2}}}}\]

当且仅当${{x}_{0}}=\frac{\sqrt{3}}{3}a,{{y}_{0}}=\frac{\sqrt{3}}{3}b,{{z}_{0}}=\frac{\sqrt{3}}{3}c$等号成立

于是当$\pi $的方程为$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=\sqrt{3}$时，${{V}_{\min }}=\frac{\sqrt{3}}{2}abc$

八、1：证明：$\because f(y)=\int_{0}^{+\infty }{x{{e}^{-{{x}^{2}}}}\cos xydx},-\infty y+\infty $

于是对$\forall {{y}_{0}}\in (-\infty ,+\infty )$，取$[a,b]\subset (-\infty ,+\infty )$，使得${{y}_{0}}\in [a,b]$

由于\[\left| x{{e}^{-{{x}^{2}}}}\cos x{{y}_{0}} \right|\le x{{e}^{-{{x}^{2}}}}\]，且\[\int_{0}^{+\infty }{x{{e}^{-{{x}^{2}}}}dx}=\frac{1}{2}\]收敛

从而$f(y)$在$[a,b]$上一致收敛，于是$f(y)$在$[a,b]$上连续

所以$f(y)$在点${{y}_{0}}$连续，由${{y}_{0}}$的任意性知，$f(y)$在点$(-\infty ,+\infty )$连续

记$F(x,y)=x{{e}^{-{{x}^{2}}}}\cos xy$，取$[c,d]\subset (-\infty ,+\infty )$，使得${{y}_{0}}\in [c,d]$

则${{F}_{y}}=-{{x}^{2}}{{e}^{-{{x}^{2}}}}\sin xy$和$F(x,y)$在$[0.+\infty )\times (-\infty ,+\infty )$上连续

对$\forall A0$，由于$\int_{0}^{A}{\sin xydx}=\frac{-\operatorname{cosAy}}{y}$，而$\left| \frac{-\operatorname{cosAy}}{y} \right|\le \frac{1}{y}\le \frac{1}{c}$

即$\int_{0}^{A}{\sin xydx}$对$y$在$[c,d]$上一致有界

而当$x1$时，${{x}^{2}}{{e}^{-{{x}^{2}}}}$是关于$x$的单调递减的函数，且$\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}{{e}^{-{{x}^{2}}}}=0$

从而对一切$x$，有${{x}^{2}}{{e}^{-{{x}^{2}}}}\to 0(x\to +\infty )$

从而由狄利克雷判别法知，$f(y)$有连续的导数

由于$F_{y}^{(2n)}={{(-1)}^{n}}{{x}^{2n+1}}{{e}^{-{{x}^{2}}}}\cos xy,F_{y}^{(2n+1)}={{(-1)}^{n+1}}{{x}^{2n+2}}{{e}^{-{{x}^{2}}}}\sin xy$

同理可证$f(y)$有$(2n)$和$(2n+1)$连续的导函数

于是$f(y)$有任意阶连续的导数

2：证明：由$F_{y}^{(2n)}={{(-1)}^{n}}{{x}^{2n+1}}{{e}^{-{{x}^{2}}}}\cos xy,F_{y}^{(2n+1)}={{(-1)}^{n+1}}{{x}^{2n+2}}{{e}^{-{{x}^{2}}}}\sin xy$

所以

$F_{y}^{(2n)}(x,0)={{(-1)}^{n}}{{x}^{2n+1}}{{e}^{-{{x}^{2}}}}\cos x0={{(-1)}^{n}}{{x}^{2n+1}}{{e}^{-{{x}^{2}}}},F_{y}^{(2n+1)}(x,0)={{(-)}^{n+1}}{{x}^{2n+2}}{{e}^{-{{x}^{2}}}}\sin x0=0$ 于是

\[{{f}^{(n)}}(0)=\int_{0}^{+\infty }{{{(-1)}^{n}}{{x}^{2n+1}}{{e}^{-{{x}^{2}}}}dx}={{(-1)}^{n}}\times (-\frac{1}{2})[{{e}^{-{{x}^{2}}}}\cdot {{x}^{2n}}|_{0}^{+\infty }-2n\cdot \int_{0}^{+\infty }{{{e}^{-{{x}^{2}}}}\cdot {{x}^{2n-1}}dx}]\] \[={{(-1)}^{n+1}}\times \frac{1}{2}\times (2n)\times \int_{0}^{+\infty }{{{e}^{-{{x}^{2}}}}\cdot {{x}^{2n-1}}dx}\]

$={{(-1)}^{n+2}}\times {{(\frac{1}{2})}^{2}}\times (2n)\times (2n-2)\times \int_{0}^{+\infty }{{{e}^{-{{x}^{2}}}}\cdot {{x}^{2n-3}}dx}$

$=\cdots $

$=-{{(\frac{1}{2})}^{n}}\times (2n)!!\times \int_{0}^{+\infty }{{{e}^{-{{x}^{2}}}}\cdot xdx={{(\frac{1}{2})}^{n+1}}\times (2n)!!}$

由泰勒展开式的定义知，$f(y)$的麦克劳林级数为

$f(y)=\sum\limits_{n=0}^{+\infty }{\frac{{{(\frac{1}{2})}^{n+1}}\times (2n)!!}{(2n)!}}{{y}^{2n}}=\sum\limits_{n=0}^{+\infty }{\frac{{{y}^{2n}}}{{{2}^{n+1}}(2n+1)!!}}$

九：法一：证明：由对称性知：$I=\iint\limits_{\sum }{({{x}^{2}}}+{{y}^{2}}+{{z}^{2}}{{)}^{-\frac{3}{2}}}{{(\frac{{{x}^{2}}}{{{a}^{4}}}+\frac{{{y}^{2}}}{{{b}^{4}}}+\frac{{{z}^{2}}}{{{c}^{4}}})}^{-\frac{1}{2}}}dS$

$=8I=\iint\limits_{{{S}_{1}}}{({{x}^{2}}}+{{y}^{2}}+{{z}^{2}}{{)}^{-\frac{3}{2}}}{{(\frac{{{x}^{2}}}{{{a}^{4}}}+\frac{{{y}^{2}}}{{{b}^{4}}}+\frac{{{z}^{2}}}{{{c}^{4}}})}^{-\frac{1}{2}}}dS$

其中${{S}_{1}}:z=c\sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}},(x,y)\in {{D}_{1}}=\{\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}\le 1,x0,y0\}$

同时$\sqrt{1+z_{x}^{2}+z_{y}^{2}}=\sqrt{1+{{[\frac{\frac{c}{{{a}^{2}}}x}{\sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}}}]}^{2}}+{{[\frac{\frac{c}{{{b}^{2}}}y}{\sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}}}]}^{2}}}$

$=\frac{c}{z}\sqrt{\frac{{{z}^{2}}}{{{c}^{2}}}+\frac{{{c}^{2}}{{x}^{2}}}{{{a}^{4}}}+\frac{{{c}^{2}}{{y}^{2}}}{{{b}^{4}}}}=\frac{{{c}^{2}}}{z}\sqrt{\frac{{{z}^{2}}}{{{c}^{4}}}+\frac{{{x}^{2}}}{{{a}^{4}}}+\frac{{{y}^{2}}}{{{b}^{4}}}}$

于是$I=8\iint\limits_{{{S}_{1}}}{({{x}^{2}}}+{{y}^{2}}+{{z}^{2}}{{)}^{-\frac{3}{2}}}\cdot {{(\frac{{{x}^{2}}}{{{a}^{4}}}+\frac{{{y}^{2}}}{{{b}^{4}}}+\frac{{{z}^{2}}}{{{c}^{4}}})}^{-\frac{1}{2}}}\cdot \frac{{{c}^{2}}}{z}\sqrt{\frac{{{z}^{2}}}{{{c}^{4}}}+\frac{{{x}^{2}}}{{{a}^{4}}}+\frac{{{y}^{2}}}{{{b}^{4}}}}dxdy$

$=8{{c}^{2}}\iint_{{{S}_{1}}}{\frac{dxdy}{z\cdot {{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{\frac{3}{2}}}}}$

$=8c\iint_{{{D}_{1}}}{\frac{dxdy}{\sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}}\cdot {{[{{x}^{2}}+{{y}^{2}}+(1-\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}})]}^{\frac{3}{2}}}}}$

其中 ${{D}_{1}}=\{\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}\le 1,x0,y0\}$

于是设$x=ra\cos \theta ,y=rb\sin \theta ,\theta \in [0,\frac{\pi }{2}],r\in [0,1]$

原式$=8ab\int_{0}^{\frac{\pi }{2}}{d\theta }\int_{0}^{1}{\frac{rdr}{\sqrt{1-{{r}^{2}}}{{[{{r}^{2}}{{a}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{b}^{2}}{{\sin }^{2}}\theta +{{c}^{2}}(1-{{r}^{2}})]}^{\frac{3}{2}}}}}$

$=4ab\int_{0}^{\frac{\pi }{2}}{d\theta }\int_{0}^{1}{\frac{d{{r}^{2}}}{\sqrt{1-{{r}^{2}}}{{[{{r}^{2}}{{a}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{b}^{2}}{{\sin }^{2}}\theta +{{c}^{2}}(1-{{r}^{2}})]}^{\frac{3}{2}}}}}$

\[\overset{t={{r}^{2}}}{\mathop{=}}\,4ab\int_{0}^{\frac{\pi }{2}}{d\theta }\int_{0}^{1}{\frac{dt}{\sqrt{1-t}{{[t{{a}^{2}}{{\cos }^{2}}\theta +t{{b}^{2}}{{\sin }^{2}}\theta +{{c}^{2}}(1-t)]}^{\frac{3}{2}}}}}\]

\[=4ab\int_{0}^{\frac{\pi }{2}}{d\theta }\int_{0}^{1}{\frac{dt}{\sqrt{1-t}{{[t[{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -{{c}^{2}}]+{{c}^{2}}]}^{\frac{3}{2}}}}}\]

\[=\frac{4ab}{{{c}^{3}}}\int_{0}^{\frac{\pi }{2}}{d\theta }\int_{0}^{1}{\frac{dt}{\sqrt{1-t}{{[t[{{(\frac{a}{c})}^{2}}{{\cos }^{2}}\theta +{{(\frac{b}{c})}^{2}}{{\sin }^{2}}\theta -1]+1]}^{\frac{3}{2}}}}}\]

为此，设$S(d)=\int_{0}^{1}{\frac{dt}{\sqrt{1-t}{{(1+dt)}^{\frac{3}{2}}}}}$，其中$d={{(\frac{a}{c})}^{2}}{{\cos }^{2}}\theta +{{(\frac{b}{c})}^{2}}{{\sin }^{2}}\theta -1$

令$h=\sqrt{1-t}$，则$S(d)=2\int_{0}^{1}{\frac{dh}{{{(1+d-d{{h}^{2}})}^{\frac{3}{2}}}}}$

易知：若$d=0$，则$S(0)=2$

若$d0$，则令\[h=\sqrt{\frac{1+d}{d}}\sin \varphi \]

于是\[S(d)=2\int_{0}^{\arcsin \sqrt{\frac{d}{1+d}}}{\frac{\sqrt{\frac{1+d}{d}}\cos \varphi }{{{[1+d-(1+d){{\sin }^{2}}\varphi ]}^{\frac{3}{2}}}}}d\varphi \]

=\[\frac{2}{\sqrt{d}(1+d)}\int_{0}^{\arcsin \sqrt{\frac{d}{1+d}}}{\frac{d\varphi }{{{\cos }^{2}}\varphi }}\]

=$\frac{2}{\sqrt{d}(1+d)}\tan (arcsin\sqrt{\frac{d}{1+d}})$

于是令

$m=\arcsin \sqrt{\frac{d}{1+d}}$，则$\operatorname{sinm}=\sqrt{\frac{d}{1+d}}$

从而

\[\tan (arcsin\sqrt{\frac{d}{1+d}})=\tan m=\sqrt{d}\]

即 $S(d)=\frac{2}{1+d}$

同理，若$d0$

$S(d)=2\int_{0}^{1}{\frac{dh}{{{(1+d+\left| d \right|{{h}^{2}})}^{\frac{3}{2}}}}}$

于是令$k=\sqrt{\frac{1+d}{\left| d \right|}}\tan \varphi $

则\[S(d)=2\int_{0}^{\arctan \sqrt{\frac{\left| d \right|}{1+d}}}{\frac{\sqrt{\frac{1+d}{\left| d \right|}}\frac{1}{{{\cos }^{2}}\varphi }}{{{[1+d+(1+d){{\tan }^{2}}\varphi ]}^{\frac{3}{2}}}}}d\varphi \]

=\[\frac{2}{\sqrt{\left| d \right|}(1+d)}\int_{0}^{\arctan \sqrt{\frac{\left| d \right|}{1+d}}}{\cos \varphi d\varphi }\]

$=\frac{2}{1+d}$

这里$d$不可能为$-1$，原因是$d={{(\frac{a}{c})}^{2}}{{\cos }^{2}}\theta +{{(\frac{b}{c})}^{2}}{{\sin }^{2}}\theta -1$

综上所述：$S(d)=\frac{2}{1+c}$

于是$S({{(\frac{a}{c})}^{2}}{{\cos }^{2}}\theta +{{(\frac{b}{c})}^{2}}{{\sin }^{2}}\theta -1)=\frac{2}{{{(\frac{a}{c})}^{2}}{{\cos }^{2}}\theta +{{(\frac{b}{c})}^{2}}{{\sin }^{2}}\theta }$

$=\frac{2{{c}^{2}}}{{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta }$

于是$I=\frac{8ab}{c}\int_{0}^{\frac{\pi }{2}}{\frac{d\theta }{{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta }}$

$=\frac{8ab}{c}\int_{0}^{\frac{\pi }{2}}{\frac{1+{{\tan }^{2}}\theta }{{{a}^{2}}+{{b}^{2}}{{\tan }^{2}}\theta }}d\theta $

于是令$v=\tan \theta $

则$I=\frac{8ab}{c}\int_{0}^{+\infty }{\frac{1+{{v}^{2}}}{{{a}^{2}}+{{b}^{2}}{{v}^{2}}}\cdot \frac{1}{1+{{v}^{2}}}}dv=\frac{4\pi }{c}$

法二：解：设

$\left\{\begin{array}{ll} x = a\sin \varphi \cos \theta , \\ y = b\sin \varphi \sin \theta ， \\ z = c\cos \varphi . \end{array} \right.$ $0 \le \varphi \le \pi ,0 \le \theta \le 2\pi$

则

${{x}_{\varphi }}=a\cos \varphi \cos \theta ,{{y}_{\varphi }}=b\cos \varphi \sin \theta ,{{z}_{\varphi }}=-c\sin \varphi $

${{x}_{\theta }}=-a\sin \varphi \sin \theta ,{{y}_{\theta }}=b\sin \varphi \cos \theta ,{{z}_{\theta }}=0$

从而

\[E={{\cos }^{2}}\varphi ({{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta )+{{c}^{2}}{{\sin }^{2}}\varphi \]

\[F={{\sin }^{2}}\varphi ({{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta )\]

\[G=({{b}^{2}}-{{a}^{2}})\sin \varphi \cos \varphi \sin \theta \cos \theta \]

$\sqrt{EF-{{G}^{2}}}=\sin \varphi \sqrt{{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\varphi +{{b}^{2}}{{c}^{2}}{{\sin }^{2}}\varphi {{\cos }^{2}}\theta +{{a}^{2}}{{c}^{2}}{{\sin }^{2}}\varphi {{\sin }^{2}}\theta }$

$=abc\sin \varphi \sqrt{\frac{{{x}^{2}}}{{{a}^{4}}}+\frac{{{y}^{2}}}{{{b}^{4}}}+\frac{{{z}^{2}}}{{{c}^{4}}}}$

于是

$I=\iint\limits_{0\le \varphi \le \pi ,0\le \theta \le 2\pi }{abc[{{a}^{2}}{{\sin }^{2}}\varphi {{\cos }^{2}}}\theta +{{b}^{2}}{{\sin }^{2}}\varphi {{\sin }^{2}}\theta +{{c}^{2}}{{\cos }^{2}}\varphi {{]}^{-\frac{3}{2}}}\sin \varphi d\varphi d\theta $

于是设

$J=\int_{0}^{\pi }{{{({{a}^{2}}{{\sin }^{2}}\varphi {{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\varphi {{\sin }^{2}}\theta +{{c}^{2}}{{\cos }^{2}}\varphi )}^{-\frac{3}{2}}}\sin \varphi d\varphi }$

\[\overset{t=\cos \varphi }{\mathop{=}}\,\int_{-1}^{1}{[({{a}^{2}}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta )(1-{{t}^{2}})+{{c}^{2}}{{t}^{2}}{{]}^{-\frac{3}{2}}}dt\]

$=\frac{2}{{{A}^{3}}}\int_{0}^{1}{[1-(\frac{{{A}^{2}}-{{c}^{2}}}{{{A}^{2}}}}){{t}^{2}}{{]}^{-\frac{3}{2}}}dt$

其中$A=\sqrt{{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta }$

不失一般性，不妨设$a\ge b\ge c$

于是

${{A}^{2}}={{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta ={{b}^{2}}+({{a}^{2}}-{{b}^{2}}){{\cos }^{2}}\theta \ge {{b}^{2}}\ge {{c}^{2}}$

当且仅当$a=b=c$取等号

（1）若$A=c$，则$a=b=c$，由此可得$I=4\pi $

（2）若$Ac$，则

$J=\frac{2}{{{A}^{2}}\sqrt{{{A}^{2}}-{{c}^{2}}}}\int_{0}^{\frac{\sqrt{{{A}^{2}}-{{c}^{2}}}}{A}}{{{(1-{{s}^{2}})}^{-\frac{3}{2}}}}ds$

$=\frac{2}{{{A}^{2}}\sqrt{{{A}^{2}}-{{c}^{2}}}}\frac{s}{\sqrt{1-{{s}^{2}}}}|_{0}^{\frac{\sqrt{{{A}^{2}}-{{c}^{2}}}}{A}}$

$=\frac{2}{{{A}^{2}}c}$

于是

$I=2ab\int_{0}^{2\pi }{\frac{1}{{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta }}d\theta $

$=8ab\int_{0}^{\frac{\pi }{2}}{\frac{1}{{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta }}d\theta $

$\overset{y=\tan \theta }{\mathop{=}}\,8ab\int_{0}^{+\infty }{\frac{dt}{({{a}^{2}}-{{b}^{2}})+{{b}^{2}}(1+{{t}^{2}})}}$

$=8ab\int_{0}^{+\infty }{\frac{dt}{{{a}^{2}}+{{b}^{2}}{{t}^{2}}}}$

$=4\pi $

由（1）（2）可知，$I=4\pi $