来源 [尊重原有作者劳动成果]
一：
1：解：[ecause underset{x o 0}{mathop{lim }}\,ln (1+x)=x]
[ herefore underset{x o 0}{mathop{lim }}\,frac{sqrt[n]{1+x}-1}{ln (1+x)}=underset{x o 0}{mathop{lim }}\,frac{sqrt[n]{1+x}-1}{x}=underset{x o 0}{mathop{lim }}\,frac{1}{n}{{(1+x)}^{frac{1}{n}-1}}=frac{1}{n}]
2：解：$int{frac{xln (x+sqrt{1+{{x}^{2}}})}{{{(1+{{x}^{2}})}^{2}}}}dx=-frac{1}{2}int{frac{-2x}{{{(1+{{x}^{2}})}^{2}}}ln (x+sqrt{1+{{x}^{2}}})}dx=$
$-frac{1}{2}[frac{1}{(1+{{x}^{2}})}ln (x+sqrt{1+{{x}^{2}}})-int{frac{1}{(1+{{x}^{2}})}cdot frac{frac{x+sqrt{1+{{x}^{2}}}}{sqrt{1+{{x}^{2}}}}}{x+sqrt{1+{{x}^{2}}}}}dx=int{frac{dx}{{{(1+{{x}^{2}})}^{frac{3}{2}}}}}$
而$int{frac{dx}{{{(1+{{x}^{2}})}^{frac{3}{2}}}}}overset{x= an heta }{mathop{=}}\,int{frac{d an heta }{frac{1}{{{(cos heta )}^{3}}}}}=int{cos heta d heta }=sin heta +C=frac{x}{sqrt{1+{{x}^{2}}}}+C$
于是$int{frac{xln (x+sqrt{1+{{x}^{2}}})}{{{(1+{{x}^{2}})}^{2}}}}dx=-frac{ln (x+sqrt{1+{{x}^{2}}})}{2(1+{{x}^{2}})}+frac{x}{2sqrt{1+{{x}^{2}}}}+C$（其中$C$为任意常数）
3：解：$ecause int_{0}^{frac{pi }{2}}{sqrt{1-sin 2x}}dx=int_{0}^{frac{pi }{2}}{sqrt{{{(sin x-cos x)}^{2}}}}dx=int_{0}^{frac{pi }{2}}{left| sin x-cos x ight|}dx$
[=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：$ecause y=arcsin x=x+frac{1}{2}cdot frac{{{x}^{3}}}{3}+frac{1 imes 3}{2 imes 4}cdot frac{{{x}^{5}}}{5}+cdots +frac{(2n-1)!!}{(2n)!!}cdot frac{{{x}^{2n+1}}}{2n+1}+O({{x}^{2n+1}})$
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：$ecause nx+k-1sqrt{(nx+k)(nx+k-1)}nx+k$
[ herefore frac{1}{n}sumlimits_{k=1}^{n}{(x+frac{k-1}{n}}){{S}_{n}}=frac{1}{{{n}^{2}}}sumlimits_{k=1}^{n}{sqrt{(nx+k)(nx+k-1)}}frac{1}{n}sumlimits_{k=1}^{n}{(x+frac{k}{n}})]
$ herefore underset{n o infty }{mathop{lim }}\,frac{1}{n}sumlimits_{k=1}^{n}{(x+frac{k-1}{n}})le underset{n o infty }{mathop{lim }}\,{{S}_{n}}le underset{n o infty }{mathop{lim }}\,frac{1}{n}sumlimits_{k=1}^{n}{(x+frac{k}{n}})$
而不等式左边
$underset{n o infty }{mathop{lim }}\,frac{1}{n}sumlimits_{k=1}^{n}{(x+frac{k-1}{n}})overset{i=k-1}{mathop{=}}\,underset{n o infty }{mathop{lim }}\,frac{1}{n}sumlimits_{1=0}^{n-1}{(x+frac{i}{n}})=underset{n o infty }{mathop{lim }}\,frac{1}{n}[sumlimits_{1=1}^{n}{(x+frac{i}{n}})+(x+0)-(x+1)]=underset{n o infty }{mathop{lim }}\,frac{1}{n}sumlimits_{1=1}^{n}{(x+frac{i}{n}})=$由迫敛性知：$underset{n o infty }{mathop{lim }}\,{{S}_{n}}=underset{n o infty }{mathop{lim }}\,frac{1}{n}sumlimits_{1=1}^{n}{(x+frac{i}{n}})=int_{0}^{1}{(x+t)dt=x+frac{1}{2}}$
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$underset{n o infty }{mathop{lim }}\,{{x}_{n}}=1$
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$a{{x}_{n+1}}=sqrt{a{{x}_{n}}}le sqrt{asqrt{a}}le sqrt{a}$
即当$n=k+1$时，也成立
于是对[forall nin {{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$underset{n o infty }{mathop{lim }}\,{{x}_{n}}=l$
由${{x}_{n+1}}=sqrt{a{{x}_{n}}}$，两边取极限，于是有$underset{n o infty }{mathop{lim }}\,{{x}_{n}}=a$
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$underset{n o infty }{mathop{lim }}\,{{x}_{n}}=a$
综上所述：对$forall a0$，${{{x}_{n}}}$收敛且$underset{n o infty }{mathop{lim }}\,{{x}_{n}}=a$
三：证明：$ecause int_{0}^{+infty }{frac{dx}{{{(1+x)}^{2}}(1+{{x}^{alpha }})}}=int_{0}^{1}{frac{dx}{{{(1+x)}^{2}}(1+{{x}^{alpha }})}}+int_{1}^{+infty }{frac{dx}{{{(1+x)}^{2}}(1+{{x}^{alpha }})}}$
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$F(x)=f(x+1)-f(x),xin [0,1]$
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：$ecause u=xy,v=x-y$
$ herefore 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}=yfrac{partial z}{partial u}+frac{partial z}{partial v}$
$frac{{{partial }^{2}}z}{partial {{x}^{2}}}=y[frac{partial (frac{partial z}{partial u})}{partial u}cdot frac{partial u}{partial x}+frac{partial (frac{partial z}{partial u})}{partial v}cdot frac{partial v}{partial x}]+[frac{partial (frac{partial z}{partial v})}{partial u}cdot frac{partial u}{partial x}+frac{partial (frac{partial z}{partial v})}{partial v}cdot frac{partial v}{partial x}]$
$={{y}^{2}}frac{{{partial }^{2}}z}{partial {{u}^{2}}}+2yfrac{{{partial }^{2}}z}{partial upartial v}+frac{{{partial }^{2}}z}{partial {{v}^{2}}}$
$frac{{{partial }^{2}}z}{partial xpartial y}=xyfrac{{{partial }^{2}}z}{partial {{u}^{2}}}+(x-y)frac{{{partial }^{2}}z}{partial xpartial y}-frac{{{partial }^{2}}z}{partial {{v}^{2}}}+frac{partial z}{partial u}$
同理可证：$ herefore 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}=xfrac{partial z}{partial u}-frac{partial z}{partial v}$
$frac{{{partial }^{2}}z}{partial {{x}^{2}}}=={{x}^{2}}frac{{{partial }^{2}}z}{partial {{u}^{2}}}-2xfrac{{{partial }^{2}}z}{partial upartial v}+frac{{{partial }^{2}}z}{partial {{v}^{2}}}$
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：
$A=iint_{D}{left| xy-frac{1}{4} ight|}dxdy=A=iint_{{{D}_{1}}}{(xy-frac{1}{4})}dxdy+iint_{{{D}_{2}}}{(-xy+frac{1}{4})}dxdy$
$=int_{frac{1}{4}}^{1}{dxint_{frac{1}{4x}}^{1}{(xy-frac{1}{4})dy+[}}int_{0}^{frac{1}{4}}{dxint_{0}^{1}{(-xy+frac{1}{4})dy+int_{frac{1}{4}}^{1}{dxint_{0}^{frac{1}{4x}}{(-xy+frac{1}{4})dy]}}}}$
$=frac{1}{8}ln 2+frac{3}{32}$
2：证明：$ecause left| iint_{D}{(xy-frac{1}{4})f(x,y)dxdy} ight|le iint_{D}{left| xy-frac{1}{4} ight|left| f(x,y) ight|dxdy}$
由于$left| xy-frac{1}{4} ight|ge 0,left| f(x,y) ight|ge 0$
于是$exists ({{x}^{*}},{{y}^{*}})in D$，使得
根据积分定理知：$iint_{D}{left| xy-frac{1}{4} ight|left| f(x,y) ight|dxdy}=iint_{D}{left| (xy-frac{1}{4}) ight|dxdy}cdot left| f({{x}^{*}},{{y}^{*}}) ight|$
$=Aleft| f({{x}^{*}},{{y}^{*}}) ight|$
而
[left| iint_{D}{(xy-frac{1}{4})f(x,y)dxdy} ight|=left| iint_{D}{xyf(x,y)dxdy-frac{1}{4}iint_{D}{f(x,y)dxdy}} ight|=1]
于是$exists ({{x}^{*}},{{y}^{*}})in D$使得$left| f({{x}^{*}},{{y}^{*}}) ight|ge frac{1}{A}$
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 3sqrt[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：$ecause 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}} ight|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$[0.+infty ) imes (-infty ,+infty )$上连续
对$forall A0$，由于$int_{0}^{A}{sin xydx}=frac{-operatorname{cosAy}}{y}$，而$left| frac{-operatorname{cosAy}}{y} ight|le frac{1}{y}le frac{1}{c}$
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$underset{x o infty }{mathop{lim }}\,{{x}^{2}}{{e}^{-{{x}^{2}}}}=0$
从而对一切$x$，有${{x}^{2}}{{e}^{-{{x}^{2}}}} o 0(x o +infty )$
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}} imes (-frac{1}{2})[{{e}^{-{{x}^{2}}}}cdot {{x}^{2n}}|_{0}^{+infty }-2ncdot int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}cdot {{x}^{2n-1}}dx}]] [={{(-1)}^{n+1}} imes frac{1}{2} imes (2n) imes int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}cdot {{x}^{2n-1}}dx}]
$={{(-1)}^{n+2}} imes {{(frac{1}{2})}^{2}} imes (2n) imes (2n-2) imes int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}cdot {{x}^{2n-3}}dx}$
$=cdots $
$=-{{(frac{1}{2})}^{n}} imes (2n)!! imes int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}cdot xdx={{(frac{1}{2})}^{n+1}} imes (2n)!!}$
由泰勒展开式的定义知，$f(y)$的麦克劳林级数为
$f(y)=sumlimits_{n=0}^{+infty }{frac{{{(frac{1}{2})}^{n+1}} imes (2n)!!}{(2n)!}}{{y}^{2n}}=sumlimits_{n=0}^{+infty }{frac{{{y}^{2n}}}{{{2}^{n+1}}(2n+1)!!}}$
九：法一：证明：由对称性知：$I=iintlimits_{sum }{({{x}^{2}}}+{{y}^{2}}+{{z}^{2}}{{)}^{-frac{3}{2}}}{{(frac{{{x}^{2}}}{{{a}^{4}}}+frac{{{y}^{2}}}{{{b}^{4}}}+frac{{{z}^{2}}}{{{c}^{4}}})}^{-frac{1}{2}}}dS$
$=8I=iintlimits_{{{S}_{1}}}{({{x}^{2}}}+{{y}^{2}}+{{z}^{2}}{{)}^{-frac{3}{2}}}{{(frac{{{x}^{2}}}{{{a}^{4}}}+frac{{{y}^{2}}}{{{b}^{4}}}+frac{{{z}^{2}}}{{{c}^{4}}})}^{-frac{1}{2}}}dS$
其中${{S}_{1}}:z=csqrt{1-frac{{{x}^{2}}}{{{a}^{2}}}-frac{{{y}^{2}}}{{{b}^{2}}}},(x,y)in {{D}_{1}}={frac{{{x}^{2}}}{{{a}^{2}}}+frac{{{y}^{2}}}{{{b}^{2}}}le 1,x0,y0}$
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$I=8iintlimits_{{{S}_{1}}}{({{x}^{2}}}+{{y}^{2}}+{{z}^{2}}{{)}^{-frac{3}{2}}}cdot {{(frac{{{x}^{2}}}{{{a}^{4}}}+frac{{{y}^{2}}}{{{b}^{4}}}+frac{{{z}^{2}}}{{{c}^{4}}})}^{-frac{1}{2}}}cdot frac{{{c}^{2}}}{z}sqrt{frac{{{z}^{2}}}{{{c}^{4}}}+frac{{{x}^{2}}}{{{a}^{4}}}+frac{{{y}^{2}}}{{{b}^{4}}}}dxdy$
$=8{{c}^{2}}iint_{{{S}_{1}}}{frac{dxdy}{zcdot {{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{frac{3}{2}}}}}$
$=8ciint_{{{D}_{1}}}{frac{dxdy}{sqrt{1-frac{{{x}^{2}}}{{{a}^{2}}}-frac{{{y}^{2}}}{{{b}^{2}}}}cdot {{[{{x}^{2}}+{{y}^{2}}+(1-frac{{{x}^{2}}}{{{a}^{2}}}-frac{{{y}^{2}}}{{{b}^{2}}})]}^{frac{3}{2}}}}}$
其中 ${{D}_{1}}={frac{{{x}^{2}}}{{{a}^{2}}}+frac{{{y}^{2}}}{{{b}^{2}}}le 1,x0,y0}$
于是设$x=racos heta ,y=rbsin heta , heta in [0,frac{pi }{2}],rin [0,1]$
原式$=8abint_{0}^{frac{pi }{2}}{d heta }int_{0}^{1}{frac{rdr}{sqrt{1-{{r}^{2}}}{{[{{r}^{2}}{{a}^{2}}{{cos }^{2}} heta +{{r}^{2}}{{b}^{2}}{{sin }^{2}} heta +{{c}^{2}}(1-{{r}^{2}})]}^{frac{3}{2}}}}}$
$=4abint_{0}^{frac{pi }{2}}{d heta }int_{0}^{1}{frac{d{{r}^{2}}}{sqrt{1-{{r}^{2}}}{{[{{r}^{2}}{{a}^{2}}{{cos }^{2}} heta +{{r}^{2}}{{b}^{2}}{{sin }^{2}} heta +{{c}^{2}}(1-{{r}^{2}})]}^{frac{3}{2}}}}}$
[overset{t={{r}^{2}}}{mathop{=}}\,4abint_{0}^{frac{pi }{2}}{d heta }int_{0}^{1}{frac{dt}{sqrt{1-t}{{[t{{a}^{2}}{{cos }^{2}} heta +t{{b}^{2}}{{sin }^{2}} heta +{{c}^{2}}(1-t)]}^{frac{3}{2}}}}}]
[=4abint_{0}^{frac{pi }{2}}{d heta }int_{0}^{1}{frac{dt}{sqrt{1-t}{{[t[{{a}^{2}}{{cos }^{2}} heta +{{b}^{2}}{{sin }^{2}} heta -{{c}^{2}}]+{{c}^{2}}]}^{frac{3}{2}}}}}]
[=frac{4ab}{{{c}^{3}}}int_{0}^{frac{pi }{2}}{d heta }int_{0}^{1}{frac{dt}{sqrt{1-t}{{[t[{{(frac{a}{c})}^{2}}{{cos }^{2}} heta +{{(frac{b}{c})}^{2}}{{sin }^{2}} heta -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$d={{(frac{a}{c})}^{2}}{{cos }^{2}} heta +{{(frac{b}{c})}^{2}}{{sin }^{2}} heta -1$
令$h=sqrt{1-t}$，则$S(d)=2int_{0}^{1}{frac{dh}{{{(1+d-d{{h}^{2}})}^{frac{3}{2}}}}}$
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)=2int_{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}}}}}dvarphi ]
=[frac{2}{sqrt{d}(1+d)}int_{0}^{arcsin sqrt{frac{d}{1+d}}}{frac{dvarphi }{{{cos }^{2}}varphi }}]
=$frac{2}{sqrt{d}(1+d)} an (arcsinsqrt{frac{d}{1+d}})$
于是令
$m=arcsin sqrt{frac{d}{1+d}}$，则$operatorname{sinm}=sqrt{frac{d}{1+d}}$
从而
[ an (arcsinsqrt{frac{d}{1+d}})= an m=sqrt{d}]
即 $S(d)=frac{2}{1+d}$
同理，若$d0$
$S(d)=2int_{0}^{1}{frac{dh}{{{(1+d+left| d ight|{{h}^{2}})}^{frac{3}{2}}}}}$
于是令$k=sqrt{frac{1+d}{left| d ight|}} an varphi $
则[S(d)=2int_{0}^{arctan sqrt{frac{left| d ight|}{1+d}}}{frac{sqrt{frac{1+d}{left| d ight|}}frac{1}{{{cos }^{2}}varphi }}{{{[1+d+(1+d){{ an }^{2}}varphi ]}^{frac{3}{2}}}}}dvarphi ]
=[frac{2}{sqrt{left| d ight|}(1+d)}int_{0}^{arctan sqrt{frac{left| d ight|}{1+d}}}{cos varphi dvarphi }]
$=frac{2}{1+d}$
这里$d$不可能为$-1$，原因是$d={{(frac{a}{c})}^{2}}{{cos }^{2}} heta +{{(frac{b}{c})}^{2}}{{sin }^{2}} heta -1$
综上所述：$S(d)=frac{2}{1+c}$
于是$S({{(frac{a}{c})}^{2}}{{cos }^{2}} heta +{{(frac{b}{c})}^{2}}{{sin }^{2}} heta -1)=frac{2}{{{(frac{a}{c})}^{2}}{{cos }^{2}} heta +{{(frac{b}{c})}^{2}}{{sin }^{2}} heta }$
$=frac{2{{c}^{2}}}{{{a}^{2}}{{cos }^{2}} heta +{{b}^{2}}{{sin }^{2}} heta }$
于是$I=frac{8ab}{c}int_{0}^{frac{pi }{2}}{frac{d heta }{{{a}^{2}}{{cos }^{2}} heta +{{b}^{2}}{{sin }^{2}} heta }}$
$=frac{8ab}{c}int_{0}^{frac{pi }{2}}{frac{1+{{ an }^{2}} heta }{{{a}^{2}}+{{b}^{2}}{{ an }^{2}} heta }}d heta $
于是令$v= an heta $
则$I=frac{8ab}{c}int_{0}^{+infty }{frac{1+{{v}^{2}}}{{{a}^{2}}+{{b}^{2}}{{v}^{2}}}cdot frac{1}{1+{{v}^{2}}}}dv=frac{4pi }{c}$
法二：解：设
$left{egin{array}{ll} x = asin varphi cos heta , \ y = bsin varphi sin heta ， \ z = ccos varphi . end{array} ight.$ $0 le varphi le pi ,0 le heta le 2pi$
则
${{x}_{varphi }}=acos varphi cos heta ,{{y}_{varphi }}=bcos varphi sin heta ,{{z}_{varphi }}=-csin varphi $
${{x}_{ heta }}=-asin varphi sin heta ,{{y}_{ heta }}=bsin varphi cos heta ,{{z}_{ heta }}=0$
从而
[E={{cos }^{2}}varphi ({{a}^{2}}{{cos }^{2}} heta +{{b}^{2}}{{sin }^{2}} heta )+{{c}^{2}}{{sin }^{2}}varphi ]
[F={{sin }^{2}}varphi ({{a}^{2}}{{cos }^{2}} heta +{{b}^{2}}{{sin }^{2}} heta )]
[G=({{b}^{2}}-{{a}^{2}})sin varphi cos varphi sin heta cos heta ]
$sqrt{EF-{{G}^{2}}}=sin varphi sqrt{{{a}^{2}}{{b}^{2}}{{cos }^{2}}varphi +{{b}^{2}}{{c}^{2}}{{sin }^{2}}varphi {{cos }^{2}} heta +{{a}^{2}}{{c}^{2}}{{sin }^{2}}varphi {{sin }^{2}} heta }$
$=abcsin varphi sqrt{frac{{{x}^{2}}}{{{a}^{4}}}+frac{{{y}^{2}}}{{{b}^{4}}}+frac{{{z}^{2}}}{{{c}^{4}}}}$
于是
$I=iintlimits_{0le varphi le pi ,0le heta le 2pi }{abc[{{a}^{2}}{{sin }^{2}}varphi {{cos }^{2}}} heta +{{b}^{2}}{{sin }^{2}}varphi {{sin }^{2}} heta +{{c}^{2}}{{cos }^{2}}varphi {{]}^{-frac{3}{2}}}sin varphi dvarphi d heta $
于是设
$J=int_{0}^{pi }{{{({{a}^{2}}{{sin }^{2}}varphi {{cos }^{2}} heta +{{b}^{2}}{{sin }^{2}}varphi {{sin }^{2}} heta +{{c}^{2}}{{cos }^{2}}varphi )}^{-frac{3}{2}}}sin varphi dvarphi }$
[overset{t=cos varphi }{mathop{=}}\,int_{-1}^{1}{[({{a}^{2}}}{{cos }^{2}} heta +{{b}^{2}}{{sin }^{2}} heta )(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$A=sqrt{{{a}^{2}}{{cos }^{2}} heta +{{b}^{2}}{{sin }^{2}} heta }$
不失一般性，不妨设$age bge c$
于是
${{A}^{2}}={{a}^{2}}{{cos }^{2}} heta +{{b}^{2}}{{sin }^{2}} heta ={{b}^{2}}+({{a}^{2}}-{{b}^{2}}){{cos }^{2}} heta ge {{b}^{2}}ge {{c}^{2}}$
当且仅当$a=b=c$取等号
（1）若$A=c$，则$a=b=c$，由此可得$I=4pi $
（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=2abint_{0}^{2pi }{frac{1}{{{a}^{2}}{{cos }^{2}} heta +{{b}^{2}}{{sin }^{2}} heta }}d heta $
$=8abint_{0}^{frac{pi }{2}}{frac{1}{{{a}^{2}}{{cos }^{2}} heta +{{b}^{2}}{{sin }^{2}} heta }}d heta $
$overset{y= an heta }{mathop{=}}\,8abint_{0}^{+infty }{frac{dt}{({{a}^{2}}-{{b}^{2}})+{{b}^{2}}(1+{{t}^{2}})}}$
$=8abint_{0}^{+infty }{frac{dt}{{{a}^{2}}+{{b}^{2}}{{t}^{2}}}}$
$=4pi $
由（1）（2）可知，$I=4pi $