武汉大学2010年数学分析试题解答
一、 1、解 $underset{x o 0}{mathop{lim }}\,frac{ln {{left( 1+x ight)}^{frac{1}{x}}}-1}{x}$$=underset{x o 0}{mathop{lim }}\,frac{frac{ln left( 1+x ight)}{x}-1}{x}$$=underset{x o 0}{mathop{lim }}\,frac{ln left( 1+x ight)-x}{{{x}^{2}}}$$=underset{x o 0}{mathop{lim }}\,frac{frac{1}{1+x}-1}{2x}$$=-frac{1}{2}underset{x o 0}{mathop{lim }}\,frac{1}{1+x}=-frac{1}{2}$.
2、 解 $sumlimits_{k=1}^{n}{frac{{{2}^{{}^{k}/{}_{n}}}}{n+1}}le sumlimits_{k=1}^{n}{frac{{{2}^{{}^{k}/{}_{n}}}}{n+frac{1}{k}}}le sumlimits_{k=1}^{n}{frac{{{2}^{{}^{k}/{}_{n}}}}{n+frac{1}{n}}}$，
$underset{n o infty }{mathop{lim }}\,sumlimits_{k=1}^{n}{frac{{{2}^{{}^{k}/{}_{n}}}}{n+1}}=underset{n o infty }{mathop{lim }}\,frac{n}{n+1}sumlimits_{k=1}^{n}{frac{1}{n}{{2}^{{}^{k}/{}_{n}}}}=int_{0}^{1}{{{2}^{x}}dx}$$=left. left( frac{1}{ln 2}{{2}^{x}} ight) ight|_{0}^{1}=frac{1}{ln 2}$，
$underset{n o infty }{mathop{lim }}\,sumlimits_{k=1}^{n}{frac{{{2}^{{}^{k}/{}_{n}}}}{n+frac{1}{n}}}=underset{n o infty }{mathop{lim }}\,frac{n}{n+frac{1}{n}}sumlimits_{k=1}^{n}{frac{1}{n}{{2}^{{}^{k}/{}_{n}}}}=int_{0}^{1}{{{2}^{x}}dx}=frac{1}{ln 2}$，所以$underset{n o infty }{mathop{lim }}\,left( frac{{{2}^{{}^{1}/{}_{n}}}}{n+1}+frac{{{2}^{{}^{2}/{}_{n}}}}{n+frac{1}{2}}+cdots +frac{{{2}^{{}^{n}/{}_{n}}}}{n+frac{1}{n}} ight)=frac{1}{ln 2}$.
3、 解：
$int{frac{dx}{1+ an x}}=int{frac{cos x}{cos x+sin x}}dx=frac{1}{2}int{frac{left( cos x+sin x ight)+left( cos x-sin x ight)}{cos x+sin x}}dx$
$=frac{1}{2}x+frac{1}{2}ln left| sin x+cos x ight|+C$.
4、 解 ：
设$G(x,alpha )=intlimits_{x-2alpha }^{x+3alpha }{cos ({{x}^{2}}}+{{y}^{2}}+{{alpha }^{2}})dy$，则
$F(alpha )=intlimits_{0}^{{{e}^{alpha }}}{dxintlimits_{x-2alpha }^{x+3alpha }{cos ({{x}^{2}}}}+{{y}^{2}}+{{alpha }^{2}})dy=intlimits_{0}^{{{e}^{alpha }}}{G(x,alpha )dx}$
于是
$F'(alpha )=intlimits_{0}^{{{e}^{alpha }}}{{{G}_{alpha }}(x,alpha )dx+{{e}^{alpha }}}G({{e}^{alpha }},alpha )$
而
${{G}_{alpha }}(x,alpha )=-2alpha intlimits_{x-2alpha }^{x+3alpha }{sin ({{x}^{2}}}+{{y}^{2}}+{{alpha }^{2}})dy+3sin (2{{x}^{2}}+9{{alpha }^{2}}+6ax)+2sin (2{{x}^{2}}+4{{alpha }^{2}}-4alpha x)$于是
$F'(alpha )=intlimits_{0}^{{{e}^{alpha }}}{[-2alpha intlimits_{x-2alpha }^{x+3alpha }{sin ({{x}^{2}}}+{{y}^{2}}+{{alpha }^{2}})dy+3sin (2{{x}^{2}}+9{{alpha }^{2}}+6ax)+2sin (2{{x}^{2}}+4{{alpha }^{2}}-4alpha x)]dx}$$+{{e}^{alpha }}intlimits_{{{e}^{alpha }}-2alpha }^{{{e}^{alpha }}+3alpha }{cos ({{e}^{2alpha }}}+{{y}^{2}}+{{alpha }^{2}})dy$
5、 解 $iiintlimits_{V}{{{e}^{x}}{{y}^{2}}{{z}^{3}}dxdydz}$$=int_{0}^{1}{{{e}^{x}}dxint_{0}^{x}{{{y}^{2}}dyint_{0}^{xy}{{{z}^{3}}dz}}}$$=int_{0}^{1}{{{e}^{x}}dxint_{0}^{x}{{{y}^{2}}frac{1}{4}{{x}^{4}}{{y}^{4}}dy}}$
$=frac{1}{4}frac{1}{7}int_{0}^{1}{{{x}^{11}}{{e}^{x}}dx}=frac{1}{28}int_{0}^{1}{{{x}^{11}}{{e}^{x}}dx}$ $=1425600-frac{7342285e}{14}$，（多次分部积）。
$I(m)=int_{0}^{1}{{{x}^{m}}{{e}^{x}}dx}=e-mI(m-1) $ 。
二、证明
方法一（1）当 $a>frac{1}{4}$时，${{x}_{1}}ge 0$，${{x}_{n+1}}=sqrt{a+{{x}_{n}}}$，
因为$|{{x}_{n+1}}-{{x}_{n}}|=|sqrt{a+{{x}_{n}}}-sqrt{a+{{x}_{n-1}}}|$
$=frac{1}{sqrt{a+{{x}_{n}}}+sqrt{a+{{x}_{n-1}}}}|{{x}_{n}}-{{x}_{n-1}}|$le frac{1}{2sqrt{a}}|{{x}_{n}}-{{x}_{n-1}}|$，$n=2,3,cdots$，
于是得压缩序列${{{x}_{n}}}$是收敛的，设$underset{n o infty }{mathop{lim }}\,{{x}_{n}}=A$，显然$Age sqrt{a}$；
在${{x}_{n+1}}=sqrt{a+{{x}_{n}}}$两边令$n o infty $取极限得到$A=sqrt{a+A}$，
从而${{A}^{2}}-A-a=0$，解得$A=frac{1pm sqrt{1+4a}}{2}$，因为$Age sqrt{a}$，故$A=frac{1+sqrt{1+4a}}{2}$ .
$underset{n o infty }{mathop{lim }}\,{{x}_{n}}=A=frac{1+sqrt{1+4a}}{2}$ ；
（2）当 $0<ale frac{1}{4}$时${{x}_{1}}=sqrt{a}$，${{x}_{n+1}}=sqrt{a+{{x}_{n}}}$，
得${{x}_{n}}ge sqrt{a}$，${{x}_{2}}ge {{x}_{1}}$，${{x}_{n+1}}=sqrt{a+{{x}_{n}}}$，由此推出${{{x}_{n}}}$单调递增，
$f(x)=sqrt{a+x}$单调递增，令$A=frac{1+sqrt{1+4a}}{2}$，则有$f(A)=A$，
${{x}_{1}}<A=frac{1+sqrt{1+4a}}{2}$，由此得${{x}_{n}}<A=frac{1+sqrt{1+4a}}{2}$，
${{{x}_{n}}}$单调递增有界，设$underset{n o infty }{mathop{lim }}\,{{x}_{n}}=A$，显然$Age sqrt{a}$；
在${{x}_{n+1}}=sqrt{a+{{x}_{n}}}$两边令$n o infty$取极限得到$A=sqrt{a+A}$，
从而${{A}^{2}}-A-a=0$，解得$A=frac{1pm sqrt{1+4a}}{2}$，因为$Age sqrt{a}$，故[A=frac{1+sqrt{1+4a}}{2}$ .
$underset{n o infty }{mathop{lim }}\,{{x}_{n}}=A=frac{1+sqrt{1+4a}}{2}$
方法二 令$A=frac{1+sqrt{1+4a}}{2}$，则有$A=frac{1+sqrt{1+4a}}{2}>1$，${{x}_{n+1}}^{2}=a+{{x}_{n}}$，${{A}^{2}}=a+A$，
从而$|{{x}_{n+1}}-A|=frac{|{{x}_{n}}-A|}{{{x}_{n}}+A}le frac{1}{A}|{{x}_{n}}-A|le cdots le frac{1}{{{A}^{n}}}|{{x}_{1}}-A|$，
于是$underset{n o infty }{mathop{lim }}\,{{x}_{n}}=A=frac{1+sqrt{1+4a}}{2}$。
三、 证明 令$F(x)=xf(x) $，由积分中值定理，存在${{xi }_{1}}in left( 0,frac{1}{2} ight) $，使得
$f(2)=int_{0}^{frac{1}{2}}{xfleft( x ight)dx}=frac{1}{2}{{xi }_{1}}f({{xi }_{1}})$，于是有$F(2)=2f(2)={{xi }_{1}}f({{xi }_{1}})=F({{xi }_{1}})$，
由罗尔中值定理，得存在$xi in left( 0,2 ight) $，使得${F}'(xi )=0$，即 $fleft( xi ight)+xi {f}'left( xi ight)=0$ 。
四、 ${{u}_{x}}=v+x{{v}_{x}}+y{varphi }'left( v ight){{v}_{x}}+{psi }'left( v ight){{v}_{x}}=v$，${{u}_{xx}}={{v}_{x}}$，${{u}_{xy}}={{v}_{y}}$，
${{u}_{y}}=x{{v}_{y}}+varphi left( v ight)+y{varphi }'left( v ight){{v}_{y}}+{psi }'left( v ight){{v}_{y}}=varphi left( v ight) $，${{u}_{yy}}={varphi }'left( v ight){{v}_{y}}$，${{u}_{yx}}={varphi }'left( v ight){{v}_{x}}$；
于是${{u}_{xx}}={{v}_{x}}$，${{u}_{yy}}={varphi }'left( v ight){{v}_{y}}$，${{u}_{xy}}={{v}_{y}}$，${{u}_{yx}}={varphi }'left( v ight){{v}_{x}}$，
$frac{{{partial }^{2}}u}{partial {{x}^{2}}}cdot frac{{{partial }^{2}}u}{partial {{y}^{2}}}-{{left( frac{{{partial }^{2}}u}{partial xpartial y} ight)}^{2}}={{v}_{x}}{varphi }'(v){{v}_{y}}-{{u}_{xy}}cdot {{u}_{yx}}$ $={{v}_{x}}{varphi }'(v){{v}_{y}}-{{v}_{y}}{varphi }'(v){{v}_{x}}=0$。
五、 解 两曲面的交线为${{x}^{2}}+{{y}^{2}}={{a}^{2}},z=a$，$D={(x,y):{{x}^{2}}+{{y}^{2}}le {{a}^{2}}}$，
${{S}_{1}}:z=2a-sqrt{{{x}^{2}}+{{y}^{2}}},(x,y)in D$；${{S}_{2}}:z=frac{1}{a}({{x}^{2}}+{{y}^{2}}),(x,y)in D$，
$dsigma =sqrt{1+{{(frac{partial z}{partial x})}^{2}}+{{(frac{partial z}{partial y})}^{2}}}dxdy$，曲面的面积
${{S}_{1}}=iintlimits_{D}{sqrt{1+{{(frac{partial z}{partial x})}^{2}}+{{(frac{partial z}{partial y})}^{2}}}}dxdy=iintlimits_{D}{sqrt{2}}dxdy=sqrt{2}pi {{a}^{2}}$，
${{S}_{2}}=iintlimits_{D}{sqrt{1+{{(frac{partial z}{partial x})}^{2}}+{{(frac{partial z}{partial y})}^{2}}}}dxdy$$=iintlimits_{D}{frac{sqrt{{{a}^{2}}+4{{x}^{2}}+4{{y}^{2}}}}{a}}dxdy$$=int_{0}^{2pi }{d} heta int_{0}^{a}{frac{sqrt{{{a}^{2}}+4{{r}^{2}}}}{a}}cdot rdr$
$=2pi frac{1}{a}frac{1}{3}frac{1}{4}{{({{a}^{2}}+4{{r}^{2}})}^{{}^{3}/{}_{2}}}|_{0}^{a}$$=frac{pi (5sqrt{5}-1){{a}^{2}}}{6}$ 。
则所求曲面的面积为$S={{S}_{1}}+{{S}_{2}}=sqrt{2}pi {{a}^{2}}+frac{pi (5sqrt{5}-1){{a}^{2}}}{6}$。
六、证明 对任意$a>0$，当$xge 1+a$时，设${{u}_{n}}left( x ight)=frac{ln left( 1+nx ight)}{n{{x}^{n}}}$，
$0<{{u}_{n}}left( x ight)le frac{nx}{n{{x}^{n}}}=frac{1}{{{x}^{n-1}}}le frac{1}{{{left( 1+a ight)}^{n-1}}}$，而$sumlimits_{n=1}^{infty }{frac{1}{{{left( 1+a ight)}^{n-1}}}}$收敛，所以$sumlimits_{n=1}^{infty }{{{u}_{n}}left( x ight)} $在$xin left[ 1+a,+infty ight) $上一致收敛；由${{u}_{n}}left( x ight) $在$left[ 1+a,+infty ight) $上连续，所以$fleft( x ight)=sumlimits_{n=1}^{infty }{frac{ln left( 1+nx ight)}{n{{x}^{n}}}}$在$left[ 1+a,+infty ight) $上连续，由$a>0$的任意性，知$fleft( x ight) $在$left( 1,+infty ight) $上连续；由${{u}_{n}}left( 1 ight)=frac{ln left( 1+n ight)}{n}>frac{1}{n},(n>2) $，显然$sumlimits_{n=1}^{infty }{{{u}_{n}}left( 1 ight)} $发散，所以$sumlimits_{n=1}^{infty }{{{u}_{n}}left( x ight)} $在$left( 1,+infty ight) $不一致收敛.
七、设$phi (u)=int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}}cos uxdx$.证明（1） 设$f(x,u)={{e}^{-{{x}^{2}}}}cos ux$，则有$f(x,u) $在$ [0,+infty ) imes (-infty ,+infty ) $上连续，且有$|f(x,u)|le {{e}^{-{{x}^{2}}}}$，而$int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}}dx$收敛，根据魏尔斯特拉斯判别法，积分$int_{0}^{+infty }{f(x,u)}dx$$ (-infty ,+infty ) $上一致收敛，所以$phi (u)=int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}}cos uxdx$的定义域为$ (-infty ,+infty ) $；
（2） $f(x,u) $，$frac{{{partial }^{k}}}{partial {{u}^{k}}}f(x,u) $在$ [0,+infty ) imes (-infty ,+infty ) $上连续，且有$|f(x,u)|le {{e}^{-{{x}^{2}}}}$，$|frac{{{partial }^{k}}}{partial {{u}^{k}}}f(x,u)|le {{x}^{k}}{{e}^{-{{x}^{2}}}}$，而$int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}}dx$，$int_{0}^{+infty }{{{x}^{k}}{{e}^{-{{x}^{2}}}}}dx$收敛，根据魏尔斯特拉斯判别法，积分$int_{0}^{+infty }{f(x,u)}dx$，$int_{0}^{+infty }{frac{{{partial }^{k}}}{partial {{u}^{k}}}f(x,u)}dx$均在$ (-infty ,+infty ) $上一致收敛，$k=1,2,cdots $
于是$phi (u)=int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}}cos uxdx$在$ (-infty ,+infty ) $上连续可微，且$phi (u)=int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}}cos uxdx$在$ (-infty ,+infty ) $上具任意阶的连续导数；${{varphi }^{(k)}}(u)=int_{0}^{+infty }{frac{{{partial }^{k}}}{partial {{u}^{k}}}f(x,u)}dx$，$k=1,2,cdots $；
（3） ${varphi }'(u)=int_{0}^{+infty }{frac{partial }{partial u}({{e}^{-{{x}^{2}}}}}cos ux)dx=int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}}(-xsin ux)dx$
$=int_{0}^{+infty }{sin ux}d(frac{1}{2}{{e}^{-{{x}^{2}}}})$$=-int_{0}^{+infty }{frac{1}{2}{{e}^{-{{x}^{2}}}}(}ucos ux)dx$$=-frac{u}{2}phi (u) $，由此得$ [ln phi (u){]}'=-frac{u}{2}$，积分得$ln phi (u)=-frac{{{u}^{2}}}{4}+{{C}_{1}}$从而有$phi (u)=C{{e}^{-frac{{{u}^{2}}}{4}}}$，[C=phi (0)=int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}}dx=frac{sqrt{pi }}{2}$，
故$phi (u)=int_{0}^{+infty }{{{e}^{-{{x}^{2}}}}}cos uxdx$$=frac{sqrt{pi }}{2}{{e}^{-frac{{{u}^{2}}}{4}}}$ .
八、证明 设$D={(x,y,z):frac{{{left( x-3 ight)}^{2}}}{16}+frac{{{left( y-2 ight)}^{2}}}{9}le 1,z=0}$，$n=(0,0,1) $$D$与$Sigma$所围的区域为$V$，显然点$ (0,0,0) $在$V$的外部，曲面$sum$没有罩着点$ (0,0,0) $。$div(frac{1}{{{left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} ight)}^{{scriptstyle{}^{3}/{}_{2}}}}}(x,y,z))=0$，利用高斯公式，得
$iintlimits_{sum }{frac{xdydz+ydzdx+zdxdy}{{{left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} ight)}^{{scriptstyle{}^{3}/{}_{2}}}}}}=iintlimits_{D}{frac{xdydz+ydzdx+zdxdy}{{{left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} ight)}^{{scriptstyle{}^{3}/{}_{2}}}}}}=0$.此题出的错误。
应把曲面方程改为使$D$与$Sigma $所围的区域$V$含点$ (0,0,0) $。曲面$sum$应罩着点$ (0,0,0) $。改为：$sum $为$1-frac{z}{5}=frac{{{left( x-2 ight)}^{2}}}{16}+frac{{{left( y-2 ight)}^{2}}}{9}$（$zge 0$）的上侧.求证$iintlimits_{sum }{frac{xdydz+ydzdx+zdxdy}{{{left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} ight)}^{{scriptstyle{}^{3}/{}_{2}}}}}}=2pi $.
证明 取$varepsilon >0$充分小，${{S}_{varepsilon }}:{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{varepsilon }^{2}}(zge 0) $，${{D}_{varepsilon }}=(x,y,z):{{x}^{2}}+{{y}^{2}}ge {{varepsilon }^{2}},frac{{{left( x-2 ight)}^{2}}}{16}+frac{{{left( y-2 ight)}^{2}}}{9}le 1,z=0}$，$div(frac{1}{{{left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} ight)}^{{scriptstyle{}^{3}/{}_{2}}}}}(x,y,z))=0$
${{D}_{varepsilon }}$，${{S}_{varepsilon }}$与$Sigma $所围的区域为${{V}_{varepsilon }}$，利用高斯公式，得
$iintlimits_{sum }{frac{xdydz+ydzdx+zdxdy}{{{left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} ight)}^{{scriptstyle{}^{3}/{}_{2}}}}}}=iintlimits_{{{S}_{varepsilon }}+{{D}_{varepsilon }}}{frac{xdydz+ydzdx+zdxdy}{{{left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} ight)}^{{scriptstyle{}^{3}/{}_{2}}}}}}$
$=iintlimits_{{{S}_{varepsilon }}}{frac{xdydz+ydzdx+zdxdy}{{{left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} ight)}^{{scriptstyle{}^{3}/{}_{2}}}}}}+iintlimits_{{{D}_{varepsilon }}}{frac{xdydz+ydzdx+zdxdy}{{{left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} ight)}^{{scriptstyle{}^{3}/{}_{2}}}}}}$
$=frac{1}{{{varepsilon }^{3}}}iintlimits_{{{S}_{varepsilon }}}{xdydz+ydzdx+zdxdy}+0$
$=frac{1}{{{varepsilon }^{3}}}iintlimits_{{{S}_{varepsilon }}}{frac{1}{varepsilon }}({{x}^{2}}+{{y}^{2}}+{{z}^{2}})dS$
$=frac{1}{{{varepsilon }^{2}}}iintlimits_{{{S}_{varepsilon }}}{dS}=frac{1}{{{varepsilon }^{2}}}frac{1}{2}4pi {{varepsilon }^{2}}=2pi $ 。
九、证明 （1）对$forall {{x}_{1}},{{x}_{2}}in [0,+infty ) $，成立$|sqrt{{{x}_{2}}}-sqrt{{{x}_{1}}}|le sqrt{|{{x}_{2}}-{{x}_{1}}|}$，$left| fleft( {{x}_{2}} ight)-fleft( {{x}_{1}} ight) ight|le sqrt{|{{x}_{2}}-{{x}_{1}}|}$，由此知$fleft( x ight)=sqrt{x}$在$left[ 0,+infty ight) $上是一致连续的；
（2） 因为$fleft( x ight)=sqrt{x}$在$left( 0,+infty ight) $内可导，导函数${f}'left( x ight)=frac{1}{2sqrt{x}}$在$left( 0,+infty ight) $内无界，
所以$fleft( x ight)=sqrt{x}$在$left[ 0,+infty ight) $上不是$Lipschitz$连续的。 注：设$pge 1$，则对$forall {{x}_{1}},{{x}_{2}}in [0,+infty ) $，
成立$|{{x}_{2}}^{frac{1}{p}}-{{x}_{1}}^{frac{1}{p}}|le |{{x}_{2}}-{{x}_{1}}{{|}^{frac{1}{p}}}$。（这个结果，用简单初等方法就能证出。）