2014年浙江大学数学分析考研试题解答 - Eufisky - The lost book

# 2014年浙江大学数学分析考研试题解答

Eufisky posted @ 2015年9月19日 04:27 in 数学分析 with tags 考研 , 1467 阅读

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

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

\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*}

\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*}

7.$f(x)$在$[0,1]$单增,证明:

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

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

\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*}

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

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

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

\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*}

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