首页 > 技术文章 > 关于分段估计思想的专题讨论

ly142857 2014-05-11 21:29 原文

$\bf命题:$设$f\left( x \right) \in C\left[ {0,1} \right]$,证明:$\lim \limits_{h \to \begin{array}{*{20}{c}}{{0^ + }} \end{array}} \int_0^1 {\frac{h}{{{x^2} + {h^2}}}} f\left( x \right)dx = \frac{\pi }{2}f\left( 0 \right)$

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$\bf命题:$设$\forall [\alpha ,\beta ] \subset \left[ {0, + \infty } \right),f\left( x \right) \in R[\alpha ,\beta ]$,且$\int_0^{ + \infty } {f\left( x \right)dx} $收敛,则对于常数$a>1$成立\[\mathop {\lim }\limits_{y \to \begin{array}{*{20}{c}}{{0^ + }} \end{array}} \int_0^{ + \infty } {{a^{ - xy}}f\left( x \right)dx} = \int_0^{ + \infty } {f\left( x \right)dx} \]

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$\bf命题:$设$\forall a \in \left( {0, + \infty } \right),f\left( x \right) \in R\left[ {0,a} \right]$,且$\mathop {\lim }\limits_{x \to  + \infty } f\left( x \right) = \alpha $,则

\[\mathop {\lim }\limits_{t \to \begin{array}{*{20}{c}}
{{0^ + }}
\end{array}} t\int_0^{ + \infty } {{e^{ - tx}}f\left( x \right)dx} = \alpha \]

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$\bf命题:$设$\forall \left[ {a,b} \right] \subset \left( { - \infty , + \infty } \right),f\left( x \right) \in R\left[ {a,b} \right]$,且$\int_{ - \infty }^{ + \infty } {\left| {f\left( x \right)} \right|dx} $,则$F\left( x \right) = \int_{ - \infty }^{ + \infty } {f\left( x \right)\cos \left( {2\pi tx} \right)dt} $在$\left( { - \infty , + \infty } \right)$上一致连续

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$\bf命题:$设$f\left( x \right) \in C\left[ {0, + \infty } \right)$,且$\int_0^{ + \infty } {\varphi \left( x \right)dx} $绝对收敛,则\[\mathop {\lim }\limits_{n \to \infty } \int_0^{\sqrt n } {f\left( {\frac{x}{n}} \right)\varphi \left( x \right)dx}  = f\left( 0 \right)\int_0^{ + \infty } {\varphi \left( x \right)dx} \]

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$\bf命题:$

附录

$\bf(Tauber定理)$设幂级数$\sum\limits_{n = 0}^\infty  {{a_n}{x^n}}  = S\left( x \right)$在$\left( { - 1,1} \right)$上成立,若$\lim \limits_{x \to \begin{array}{*{20}{c}}{{1^ - }}\end{array}} S\left( x \right) = S$,且$\lim \limits_{n \to \infty } n{a_n} = 0$,则$\sum\limits_{n = 0}^\infty  {{a_n}}  = S$

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$\bf(Riemman积分控制收敛定理)$设函数列$\left\{ {{f_n}\left( x \right)} \right\}$在$\left[ {a, + \infty } \right)$上内闭可积且一致收敛于$f(x)$,若存在函数$g(x)$,使得$\left| {{f_n}\left( x \right)} \right| \le g\left( x \right)$对每个$x$与$n$都成立,且$\int_a^{ + \infty } {g\left( x \right)dx} $收敛,则\[\mathop {\lim }\limits_{n \to \infty } \int_a^{ + \infty } {{f_n}\left( x \right)dx} {\rm{ = }}\int_a^{ + \infty } {f\left( x \right)dx} \]

 




 

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