武汉大学2014数学分析考研真题解答

一、（1）求积分[intlimits_{0}^{1}{{{(xInx)}^{n}}dx}]

解：不妨设$t=ln xRightarrow x={{e}^{t}},-infty <t<0$

于是[intlimits_{0}^{1}{{{(xInx)}^{n}}dx}=intlimits_{-infty }^{0}{{{({{e}^{t}}cdot t)}^{n}}d{{e}^{t}}}=intlimits_{-infty }^{0}{{{e}^{(n+1)t}}cdot {{t}^{n}}dt}]

[=frac{1}{n+1}{{e}^{(n+1)t}}{{t}^{n}}|_{-infty }^{0}-frac{n}{n+1}intlimits_{-infty }^{0}{{{e}^{(n+1)t}}cdot {{t}^{n-1}}dt}=-frac{n}{n+1}intlimits_{-infty }^{0}{{{e}^{(n+1)t}}cdot {{t}^{n-1}}dt}]

而[intlimits_{-infty }^{0}{{{e}^{(n+1)t}}cdot {{t}^{n-1}}dt}=frac{1}{n+1}{{e}^{(n+1)t}}{{t}^{n-2}}|_{-infty }^{0}-frac{n-1}{n+1}intlimits_{-infty }^{0}{{{e}^{(n+1)t}}cdot {{t}^{n-2}}dt}=-frac{n-1}{n+1}intlimits_{-infty }^{0}{{{e}^{(n+1)t}}cdot {{t}^{n-2}}dt}]

于是[intlimits_{0}^{1}{{{(xInx)}^{n}}dx}=frac{n(n-1)}{{{(n+1)}^{2}}}intlimits_{-infty }^{0}{{{e}^{(n+1)t}}cdot {{t}^{n-2}}dt}]

以此类推，既得[intlimits_{0}^{1}{{{(xInx)}^{n}}dx}=frac{n!}{{{(n+1)}^{n}}}intlimits_{-infty }^{0}{{{e}^{(n+1)t}}dt}=frac{{{(-1)}^{n}}n!}{{{(n+1)}^{n+1}}}{{e}^{(n+1)t}}|_{-infty }^{0}=frac{{{(-1)}^{n}}n!}{{{(n+1)}^{n+1}}}]

（2）求极限$underset{xto {{1}^{-}}}{mathop{lim }},{{(1-x)}^{3}}sumlimits_{n=1}^{+infty }{{{n}^{2}}{{x}^{n}}}$

解：不妨设$S(x)=sumlimits_{n=1}^{+infty }{{{n}^{2}}}{{x}^{n}}$

考虑到$underset{nto +infty }{mathop{lim }},frac{{{n}^{2}}}{{{(n+1)}^{2}}}=1$，且$sumlimits_{n=1}^{+infty }{{{n}^{2}}}$和$sumlimits_{n=1}^{+infty }{{{(-1)}^{n}}}{{n}^{2}}$都发散

于是$S(x)=sumlimits_{n=1}^{+infty }{{{n}^{2}}}{{x}^{n}},xin (-1,1)$

则$frac{S(x)}{x}=sumlimits_{n=1}^{+infty }{{{n}^{2}}}{{x}^{n-1}}Rightarrow int_{0}^{x}{frac{S(t)}{t}}dt=sumlimits_{n=1}^{+infty }{n}{{x}^{n}}$

设$g(x)=int_{0}^{x}{frac{S(t)}{t}dtRightarrow frac{g(x)}{x}=sumlimits_{n=1}^{+infty }{n{{x}^{n-1}}}}Rightarrow int_{0}^{x}{frac{g(t)}{t}}dt=sumlimits_{n=1}^{+infty }{{{x}^{n}}}=frac{x}{1-x}$

于是$frac{g(x)}{x}=[frac{x}{1-x}]'=frac{1}{{{(1-x)}^{2}}}Rightarrow g(x)=frac{x}{{{(1-x)}^{2}}}Rightarrow $

$frac{S(x)}{x}=[frac{x}{{{(1-x)}^{2}}}]'=frac{1+x}{{{(1-x)}^{3}}}Rightarrow S(x)=frac{x(1+x)}{{{(1-x)}^{3}}},xin (-1,1)$

于是$underset{xto {{1}^{-}}}{mathop{lim }},{{(1-x)}^{3}}sumlimits_{n=1}^{+infty }{{{n}^{2}}{{x}^{n}}}=underset{xto {{1}^{-}}}{mathop{lim }},x(1+x)=2$

（3）[underset{nto +infty }{mathop{lim }},n[{{(1+frac{1}{n})}^{n}}-e]]

解：先求[underset{xto +infty }{mathop{lim }},x[{{(1+frac{1}{x})}^{x}}-e]]

由于[underset{xto +infty }{mathop{lim }},x[{{(1+frac{1}{x})}^{x}}-e]overset{t=frac{1}{x}}{mathop{=}},underset{tto {{0}^{+}}}{mathop{lim }},frac{{{(1+t)}^{frac{1}{t}}}-e}{t}=underset{tto {{0}^{+}}}{mathop{lim }},[{{(1+t)}^{frac{1}{t}}}]']

设

[f(t)={{(1+t)}^{frac{1}{t}}}Rightarrow ln f(t)=frac{ln (1+t)}{t}Rightarrow frac{f'(t)}{f(t)}=frac{t-(1+t)ln (1+t)}{{{t}^{2}}(1+t)}Rightarrow f'(t)=frac{t-(1+t)ln (1+t)}{{{t}^{2}}(1+t)}cdot {{(1+t)}^{frac{1}{t}}}]于是

[underset{tto {{0}^{+}}}{mathop{lim }},f'(t)=underset{tto {{0}^{+}}}{mathop{lim }},frac{t-(1+t)ln (1+t)}{{{t}^{2}}(1+t)}cdot {{(1+t)}^{frac{1}{t}}}=ecdot underset{tto {{0}^{+}}}{mathop{lim }},frac{t-(1+t)ln (1+t)}{{{t}^{2}}}=ecdot underset{tto {{0}^{+}}}{mathop{lim }},frac{1-ln (1+t)-1}{2t}=-frac{e}{2}]于是令$x=n$，于是[underset{nto +infty }{mathop{lim }},n[{{(1+frac{1}{n})}^{n}}-e]=-frac{e}{2}]

（4）求不定积分[int{frac{(1+sin x){{e}^{-x}}}{(1-cos x)}}dx]

解：由于[int{frac{(1+sin x){{e}^{-x}}}{(1-cos x)}}dx=int{frac{1}{1-cos x}}dx+int{frac{sin xcdot {{e}^{-x}}}{1-cos x}}dx]

而[int{frac{sin xcdot {{e}^{-x}}}{1-cos x}}dx=-{{e}^{-x}}frac{sin x}{1-cos x}-int{frac{1}{1-cos x}}dx]

于是[int{frac{(1+sin x){{e}^{-x}}}{(1-cos x)}}dx=-{{e}^{-x}}frac{sin x}{1-cos x}+C]（其中$C$为常数）

（5）计算曲面积分：$int_{C}{xyds}$，其中$C$为球面${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=9$和平面$x+y+z=0$的交线

解：由于平面$x+y+z=0$过原点$(0,0,0)$，从而交线的半径为$3$，且交线关于$x,y,z$轮换对称

从而$int_{C}{xyds}=int_{C}{yzds}=int_{C}{xzds}=frac{1}{6}int_{C}{2(xy+yz+xz)ds}$

而$2(xy+yz+xz)={{(x+y+z)}^{2}}-({{x}^{2}}+{{y}^{2}}+{{z}^{2}})$

于是[int_{C}{xyds}=frac{1}{6}int_{C}{[{{(x+y+z)}^{2}}-({{x}^{2}}+{{y}^{2}}+{{z}^{2}})]ds=}-frac{3}{2}int_{C}{ds}=-frac{3}{2}cdot 2pi cdot 3=-9pi ]

二、已知$f(x)$在$[-1,1]$内二阶可微，且$f(0)=0$，求证：存在$xi in [-1,1]$，使得$f''(xi )=3int_{-1}^{1}{f(x)dx}$

证明：由于$f(x)$在$[-1,1]$内二阶可微，于是

     $f(x)=f(0)+f'(0)x+frac{f''(xi )}{2!}{{x}^{2}}=f'(0)x+frac{1}{2}f''(xi ){{x}^{2}},0<xi <x$

于是  [int_{-1}^{1}{f(x)dx=f'(0)int_{-1}^{1}{xdx+frac{1}{2}}}f''(xi )int_{-1}^{1}{{{x}^{2}}}dx]

即存在$xi in (0,x)subset [-1,1]$，使得$f''(xi )=3int_{-1}^{1}{f(x)dx}$

三、已知${{x}_{n}}=f(frac{1}{{{n}^{2}}})+f(frac{2}{{{n}^{2}}})+cdots +f(frac{n}{{{n}^{2}}}),n=1,2,cdots $，其中$f(x)$在$x=0$的附近可导，且$f(0)=0,f'(0)=1$，求证：$underset{nto infty }{mathop{lim }},{{x}_{n}}$存在，并求其值。

证明：由于$f(x)$在$x=0$附近可导，于是

    $f(x)=f(0)+f'(0)x+alpha (x)x=x+alpha (x)x$

其中$alpha (x)$是关于$x$的函数，$alpha (0)=0$且$alpha (x)to 0$，当$xto 0$

于是$f(frac{i}{{{n}^{2}}})=frac{i}{{{n}^{2}}}+alpha (frac{i}{{{n}^{2}}})cdot frac{i}{{{n}^{2}}}Rightarrow sumlimits_{i=1}^{n}{f(frac{i}{{{n}^{2}}})=sumlimits_{i=1}^{n}{frac{i}{{{n}^{2}}}+sumlimits_{i=1}^{n}{alpha (frac{i}{{{n}^{2}}})}}}cdot frac{i}{{{n}^{2}}},i=1,2,cdots ,n$

由[underset{nto +infty }{mathop{lim }},alpha (frac{i}{{{n}^{2}}})=0Rightarrow ]对任意的$varepsilon >0$，存在$N>0$，当$n>N$时，有$left| alpha (frac{i}{{{n}^{2}}}) right|<varepsilon $

于是当$n>N$时，$left| sumlimits_{i=1}^{n}{alpha (frac{i}{{{n}^{2}}})}cdot frac{i}{{{n}^{2}}} right|<frac{varepsilon }{2}(1+frac{1}{n})$

于是$underset{nto +infty }{mathop{lim }},sup sumlimits_{i=1}^{n}{alpha (frac{i}{{{n}^{2}}})}cdot frac{i}{{{n}^{2}}}le frac{varepsilon }{2}$及$underset{nto +infty }{mathop{lim }},inf sumlimits_{i=1}^{n}{alpha (frac{i}{{{n}^{2}}})}cdot frac{i}{{{n}^{2}}}ge -frac{varepsilon }{2}$

由的任意性知：$underset{nto infty }{mathop{lim }},{{x}_{n}}=frac{1}{2}$，从而$underset{nto infty }{mathop{lim }},{{x}_{n}}$存在

四、已知$Omega $.是由方程$left{begin{array}{ll}
3{{x}^{2}}+{{y}^{2}}=1 \
z=0
end{array}
right.$绕y轴旋转所生成的椭球面，$Sigma $为$Omega $的上半表面，$zge 0$，
$(lambda ,u,v)$为其方向余弦，求曲面积分$iint_{Sigma }{z(lambda x+uy+vz)dS}$

解：由题可知：椭球面$Omega $的方程为：$3{{x}^{2}}+{{y}^{2}}+3{{z}^{2}}=1$

由于$iint_{sum }{z(lambda x+uy+vz)dS}=iint_{sum }{xzdydz+yzdzdx+{{z}^{2}}dxdy}$

取平面${{sum }_{1}}:3{{x}^{2}}+{{y}^{2}}+3{{z}^{2}}=1,z=0$

于是$iint_{{{sum }_{1}}}{xzdydz+yzdzdx+{{z}^{2}}dxdy}=0$

于是$iint_{sum }{z(lambda x+uy+vz)dS}=iint_{sum +{{sum }_{1}}}{xzdydz+yzdzdx+{{z}^{2}}dxdy}$

令[V:3{{x}^{2}}+{{y}^{2}}+3{{z}^{2}}le 1,zge 0]

由高斯公式可知：

$iint_{sum }{z(lambda x+uy+vz)dS}=iint_{sum +{{sum }_{1}}}{xzdydz+yzdzdx+{{z}^{2}}dxdy}=4iiint_{V}{zdxdydz}$

于是令$x=frac{1}{sqrt{3}}Rsin varphi cos theta ,y=Rsin varphi sin theta ,z=frac{1}{sqrt{3}}Rcos varphi $

其中$0le Rle 1,0le theta le 2pi ,0le varphi le frac{pi }{2},left| J right|=frac{1}{3}{{R}^{2}}sin varphi $

于是

[iint_{sum }{z(lambda x+uy+vz)dS}=4iiint_{V}{zdxdydz}=frac{4}{3}int_{0}^{2pi }{dtheta }int_{0}^{frac{pi }{2}}{dvarphi int_{0}^{1}{frac{1}{sqrt{3}}}}Rcos varphi cdot {{R}^{2}}sin varphi dR=frac{sqrt{3}}{9}pi ]

五、已知$z=f(x,y)$在区域$D$存在二阶连续偏导数，已知$=left{begin{array}{ll}
u=x+2y \
v=x+ay
end{array}
right.$变换，求$a$使得方程$2frac{{{partial }^{2}}z}{partial {{x}^{2}}}+frac{{{partial }^{2}}z}{partial xpartial y}-frac{{{partial }^{2}}z}{partial {{y}^{2}}}=0$变为$frac{{{partial }^{2}}z}{partial upartial v}=0$

解：由于$left{begin{array}{ll}
u=x+2y \
v=x+ay
end{array}
right.$

$Rightarrow {{u}_{x}}=1,{{u}_{y}}=2,{{v}_{x}}=1,{{v}_{z}}=a$

于是$frac{partial z}{partial x}=frac{partial z}{partial u}cdot frac{partial u}{partial x}+frac{partial z}{partial v}cdot frac{partial v}{partial x}={{z}_{u}}+{{z}_{v}}$

$frac{partial z}{partial y}=frac{partial z}{partial u}cdot frac{partial u}{partial y}+frac{partial z}{partial v}cdot frac{partial v}{partial y}=2{{z}_{u}}+a{{z}_{v}}$

于是$frac{{{partial }^{2}}z}{partial {{x}^{2}}}=[frac{partial ({{z}_{u}})}{partial u}cdot frac{partial u}{partial x}+frac{partial ({{z}_{u}})}{partial v}cdot frac{partial v}{partial x}]+[frac{partial ({{z}_{v}})}{partial u}cdot frac{partial u}{partial x}+frac{partial ({{z}_{v}})}{partial v}cdot frac{partial v}{partial x}]={{z}_{uu}}+2{{z}_{uv}}+{{z}_{vv}}$

$frac{{{partial }^{2}}z}{partial {{y}^{2}}}=2[frac{partial ({{z}_{u}})}{partial u}cdot frac{partial u}{partial y}+frac{partial ({{z}_{v}})}{partial v}cdot frac{partial v}{partial x}]+a[frac{partial ({{z}_{v}})}{partial u}cdot frac{partial u}{partial y}+frac{partial ({{z}_{v}})}{partial v}cdot frac{partial v}{partial y}]=4{{z}_{uu}}+4a{{z}_{uv}}+{{a}^{2}}{{z}_{vv}}$

$frac{{{partial }^{2}}z}{partial xpartial y}=[frac{partial ({{z}_{u}})}{partial u}cdot frac{partial u}{partial y}+frac{partial ({{z}_{u}})}{partial v}cdot frac{partial v}{partial y}]+[frac{partial ({{z}_{v}})}{partial u}cdot frac{partial u}{partial y}+frac{partial ({{z}_{v}})}{partial v}cdot frac{partial v}{partial y}]=2{{z}_{uu}}+(a+2){{z}_{uv}}+a{{z}_{vv}}$

由$2frac{{{partial }^{2}}z}{partial {{x}^{2}}}+frac{{{partial }^{2}}z}{partial xpartial y}-frac{{{partial }^{2}}z}{partial {{y}^{2}}}=0Rightarrow (6-3a){{z}_{uv}}+(2+a-{{a}^{2}}){{z}_{vv}}=0$

于是令

$left{begin{array}{ll}
6-3a=0 \
2+a-{{a}^{2}}=0
end{array}
right.$

$Rightarrow a=-1$

即当$a=-1$时，$frac{{{partial }^{2}}z}{partial upartial v}=0$

六、已知$alpha >-1$，求证$I=int_{0}^{+infty }{frac{1-{{e}^{-alpha x}}}{x{{e}^{x}}}}dx$收敛，并求$I$的值。

证明：由于$I=int_{0}^{+infty }{frac{1-{{e}^{-alpha x}}}{x{{e}^{x}}}}dx=int_{0}^{1}{frac{1-{{e}^{-alpha x}}}{x{{e}^{x}}}}dx+int_{1}^{+infty }{frac{1-{{e}^{-alpha x}}}{x{{e}^{x}}}}dx={{I}_{1}}+{{I}_{2}}$

其中${{I}_{1}}=int_{0}^{1}{frac{1-{{e}^{-alpha x}}}{x{{e}^{x}}}}dx,{{I}_{2}}=int_{1}^{+infty }{frac{1-{{e}^{-alpha x}}}{x{{e}^{x}}}}dx$

对于${{I}_{1}}$：考虑到$underset{xto {{0}^{+}}}{mathop{lim }},{{x}^{p}}cdot frac{1-{{e}^{-alpha x}}}{x{{e}^{x}}}dx=0,0<p<1$

于是${{I}_{1}}$在$alpha >-1$时收敛

对于${{I}_{2}}$：考虑到$alpha >-1$时，$underset{xto +infty }{mathop{lim }},{{x}^{2}}cdot frac{1-{{e}^{-alpha x}}}{x{{e}^{x}}}=0$

于是${{I}_{2}}$在$alpha >-1$时收敛

从而$I$在$alpha >-1$时收敛

（2）不妨设$I(alpha )=int_{0}^{+infty }{frac{1-{{e}^{-alpha x}}}{x{{e}^{x}}}}dx,alpha >-1$

于是$I'(alpha )=int_{0}^{+infty }{frac{x{{e}^{-alpha x}}}{x{{e}^{x}}}}dx=int_{0}^{+infty }{{{e}^{-(alpha +1)x}}}dx=-frac{1}{alpha +1}{{e}^{-(alpha +1)x}}|_{0}^{+infty }=frac{1}{alpha +1},alpha >-1$

且$I(0)=0$

于是$I(alpha )=int_{0}^{alpha }{frac{1}{t+1}dt=ln (1+alpha ),alpha >-1}$

七、已知级数$sumlimits_{n=1}^{+infty }{frac{{{n}^{n+2}}}{{{(nx+1)}^{n}}}}$，求证

（1）      该级数在$(1,+infty )$上收敛

（2）       该级数在$(1,+infty )$上非一致收敛，但是$f(x)=sumlimits_{n=1}^{+infty }{frac{{{n}^{n+2}}}{{{(nx+1)}^{n}}}}$在$(1,+infty )$上连续

证明：（1）用比较判别法考虑$underset{nto +infty }{mathop{lim }},frac{{{n}^{n+4}}}{{{(nx+1)}^{n}}}$

由于$nx+1>nxRightarrow frac{{{n}^{n+4}}}{{{(nx+1)}^{n}}}<frac{{{n}^{4}}}{{{x}^{n}}}$

考虑到$xin (1,+infty )$时，$underset{nto +infty }{mathop{lim }},frac{{{n}^{4}}}{{{x}^{n}}}=0Rightarrow underset{nto +infty }{mathop{lim }},frac{{{n}^{n+4}}}{{{(nx+1)}^{n}}}=0$

从而由比较判别法知：该级数在$(1,+infty )$上收敛；

（2）由于取${{x}_{n}}=frac{n+1}{n}$，此时[underset{nto +infty }{mathop{lim }},frac{{{n}^{n+3}}}{{{(n+2)}^{n}}}=+infty ]且$sumlimits_{n=1}^{+infty }{frac{1}{n}}$发散

    于是该级数在$(1,+infty )$上非一致收敛

另一方面 ，由于对任意的$[a,b]subset (1,+infty )$

此时$frac{{{n}^{n+4}}}{{{(nx+1)}^{n}}}<frac{{{n}^{4}}}{{{a}^{n}}}$，$a>1$

而$underset{nto infty }{mathop{lim }},frac{{{n}^{4}}}{{{a}^{n}}}=0$，则当$n$充分大时，有$frac{{{n}^{2}}}{{{a}^{n}}}<frac{1}{{{n}^{2}}}$

而[sumlimits_{n=1}^{+infty }{frac{1}{{{n}^{2}}}}]收敛

由$M$判别法可知，该级数在$[a,b]$上一致收敛

由$[a,b]$的任意性及开区间覆盖定理知：该级数在$(1,+infty )$上内闭一致收敛

从而$f(x)=sumlimits_{n=1}^{+infty }{frac{{{n}^{n+2}}}{{{(nx+1)}^{n}}}}$在$(1,+infty )$上连续