2014年浙江大学数学分析考研试题解答

6.设空间体积为$V$的任意$\Omega,X_0\in \Omega ,0<\alpha<3$.证明

\[\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \le C{V^{\alpha /3}}, \text{其中$C$只与$\alpha$有关}.\]

证:(Veer)由于$\alpha-3>-3$且$\displaystyle \int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX}$为三重积分,故积分广义可积.

在$X_0$处作一以$X_0$为圆心的球$D$,使其体积为$V_D=V$.设$D$的半径为$R$.记$D_1=D\cap \Omega,D_2=D/D_1,\Omega_2=\Omega/D_1$,则易知$V_{D_2}=V_{\Omega_2}$.显然

\begin{align*}\int_D {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= \int_{{D_1}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} + \int_{{D_2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \\\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= \int_{{D_1}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} + \int_{{\Omega _2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} .\end{align*}

由积分中值定理有

\begin{align*}\int_{{D_2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= {\left| {\xi - {X_0}} \right|^{\alpha - 3}}{V_{{D_2}}},\xi \in {D_2}\\\int_{{\Omega _2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= {\left| {\eta - {X_0}} \right|^{\alpha - 3}}{V_{{\Omega_2}}},\eta \in {\Omega _2}.\end{align*}

易知${\left| {\xi - {X_0}} \right|^{\alpha - 3}} \ge {\left| {\eta - {X_0}} \right|^{\alpha - 3}}$.又因为${V_{{D_2}}} = {V_{{\Omega _2}}}$,则\[\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \le \int_D {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} .\]由球坐标变换易得\[\int_D {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} = \int_0^\pi {d\varphi } \int_0^{2\pi } {d\theta } \int_0^R {{r^{\alpha - 1}}\sin \varphi dr} = 4\pi \frac{{{R^\alpha }}}{\alpha }.\]又因为$\displaystyle {V_D} = V = \frac{4}{3}\pi {R^3}$,则\[\int_D {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} = \frac{{{3^{\alpha /3}}{{\left( {4\pi } \right)}^{1 - \alpha /3}}}}{\alpha }{V^{\alpha /3}}.\]故\[\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \le \frac{{{3^{\alpha /3}}{{\left( {4\pi } \right)}^{1 - \alpha /3}}}}{\alpha }{V^{\alpha /3}},\]取$\displaystyle C = \frac{{{3^{\alpha /3}}{{\left( {4\pi } \right)}^{1 - \alpha /3}}}}{\alpha }$.

注:从上可以看出当$\Omega=D$时不等式可取等号,故$C$是最佳的,且此题可推广到$n$维上.

7.$f(x)$在$[0,1]$单增,证明:

\[\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {f\left( x \right)\frac{{\sin xy}}{x}dx} = \frac{\pi }{2}f\left( {{0_ + }} \right).\]

证:这是Dirichlet引理,菲赫金哥尔茨的《微积分教程》第三卷P358有详细的证明.另外,汪林的《数学分析问题研究与评注》P147上有他的推广及其证明.

对任意给出的$\varepsilon>0$, $\exists 0<\delta<1$,使得对于$0<t\leq \delta$,

\[0 \le g\left( t \right) - g\left( {{0_ + }} \right) < M_1\varepsilon ,\]

其中$M_1$是任意给定的常数.

考察积分

\begin{align*}\int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} &= \left( {\int_0^\delta {} + \int_\delta ^1 {} } \right)\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx\\&= {I_1} + {I_2}.\end{align*}

对于$I_1$,运用积分第二中值定理,我们有

\[{I_1} = \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\int_\eta ^\delta {\frac{{\sin xy}}{x}dx} = \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} ,\]

其中第二个因子对于一切值$y$一致有界.事实上,由反常积分$\displaystyle \int_0^\infty {\frac{{\sin z}}{z}dz}$的收敛性,可见当$z\to \infty$时, $z(z\geq 0)$的连续函数$\displaystyle \int_0^z {\frac{{\sin z}}{z}dz} $有有限的极限,并且对于一切值$z$有界

\[\left| {\int_0^z {\frac{{\sin z}}{z}dz} } \right| \le L\left( L \text{为常数}\right),\]从而

\[\left| {\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} } \right| = \left| {\int_0^{y\delta } {} + \int_0^{y\eta } {} } \right| \le 2L.\]

对于第一个因子,取$M_1=\frac{1 }{{4L}}$,则有$f\left( \delta \right) - f\left( {{0_ + }} \right) < \frac{\varepsilon }{{4L}}$.

因此\[\left| {{I_1}} \right| \le \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\left| {\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} } \right| < \frac{\varepsilon }{{4L}} \cdot 2L = \frac{\varepsilon }{2}.\]

至于$I_2$,由于$\displaystyle \int_\delta ^1 {\frac{{f\left( x \right) - f\left( {{0_ + }} \right)}}{x}dx} $存在,由Riemann-Lebesgue引理可知$\mathop {\lim }\limits_{y \to \infty } {I_2} = 0$,即对$\varepsilon >0,\exists M_2>0$,使得$y>M_2$时,有$\left| {{I_2}} \right| < \frac{\varepsilon }{2}$.

因此\[\left| {\int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} } \right| \le \left| {{I_1}} \right| + \left| {{I_2}} \right| < \varepsilon .\]

即\[\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} = 0.\]

从而

\begin{align*}&\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {f\left( x \right)\frac{{\sin xy}}{x}dx} = \frac{\pi }{2}f\left( {{0_ + }} \right)\\=& \mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} + f\left( {{0_ + }} \right)\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\frac{{\sin xy}}{x}dx} \\= &0 + f\left( {{0_ + }} \right)\int_0^{ + \infty } {\frac{{\sin z}}{z}dz} = \frac{\pi }{2}f\left( {{0_ + }} \right).\end{align*}