关于反常积分收敛的专题讨论

$\bf命题：$设$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，若$\lim \limits_{x \to \begin{array}{*{20}{c}} {{\rm{ + }}\infty } \end{array}} f\left( x \right)$存在，则$\lim \limits_{x \to \begin{array}{*{20}{c}} {{\rm{ + }}\infty } \end{array}} f\left( x \right) = 0$

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$\bf命题：$设$f\left( x \right) \in {C^1}\left[ {a, + \infty } \right)$，若$\int_a^{ + \infty } {f\left( x \right)dx} ，\int_a^{ + \infty } {f'\left( x \right)dx}$均收敛，则$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} f\left( x \right) = 0$

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$\bf命题：$设$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，且${f\left( x \right)}$在$\left[ {a,{\rm{ + }}\infty } \right)$单调，则$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} xf\left( x \right) = 0$，进而$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} f\left( x \right) = 0$

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$\bf命题：$ 设$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，且可微函数${f\left( x \right)}$在$\left[ {a,{\rm{ + }}\infty } \right)$单调递减，则$\int_a^{ + \infty } {xf'\left( x \right)dx} $收敛

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$\bf命题：$设$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，且$\frac{{f\left( x \right)}}{x}$在${\left[ {a, + \infty } \right)}$上单调递减，则$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} xf\left( x \right) = 0$

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$\bf命题：$设$f\left( x \right)$单调且$\lim \limits_{x \to \begin{array}{*{20}{c}}
{{0^ + }}
\end{array}} f\left( x \right) = + \infty $，若$\int_0^1 {f\left( x \right)dx} $收敛，则$\lim \limits_{x \to \begin{array}{*{20}{c}}
{{0^ + }}
\end{array}} xf\left( x \right) = 0$

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$\bf命题：$设$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，且$xf\left( x \right)$在${\left[ {a, + \infty } \right)}$上单调递减，则$\lim \limits_{x \to\begin{array}{*{20}{c}} { + \infty }\end{array}} xf\left( x \right)\ln x = 0$

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$\bf命题：$设$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，且$f\left( x \right)$在${\left[ {a, + \infty } \right)}$上一致连续，则$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} f\left( x \right) = 0$

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$\bf命题：$设$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，且$f\left( x \right)$在${\left[ {a, + \infty } \right)}$上可导且导函数有界，则$\lim \limits_{x \to \begin{array}{*{20}{c}}
{ + \infty }
\end{array}} f\left( x \right) = 0$

$\bf命题：$设$\int_a^{ + \infty } {f\left( x \right)dx} $绝对收敛，且$f\left( x \right)$在${\left[ {a, + \infty } \right)}$上可导且导函数有界，则$\lim \limits_{x \to \begin{array}{*{20}{c}}
{ + \infty }
\end{array}} f\left( x \right) = 0$

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$\bf命题：$设$f\left( x \right)$在${\left[ {a, + \infty } \right)}$上可导且导函数有界，若$ \int_a^{ + \infty } {{f^2}\left( x \right)dx} < + \infty $，则$\lim \limits_{x \to \begin{array}{*{20}{c}}
{ + \infty }
\end{array}} f\left( x \right) = 0$

$\bf命题：$设$p \ge 1,f\left( x \right) \in {C^1}\left( { - \infty , + \infty } \right)$，且\[\int_{ - \infty }^{ + \infty } {{{\left| {f\left( x \right)} \right|}^p}dx} < + \infty ,\int_{ - \infty }^{ + \infty } {{{\left| {f'\left( x \right)} \right|}^p}dx} < + \infty \]
证明：$\lim \limits_{x \to \begin{array}{*{20}{c}}\infty \end{array}} f\left( x \right) = 0$，且$${\left| {f\left( x \right)} \right|^p} \le \frac{{p - 1}}{2}\int_{ - \infty }^{ + \infty } {{{\left| {f\left( t \right)} \right|}^p}dt} + \frac{1}{2}\int_{ - \infty }^{ + \infty } {{{\left| {f'\left( t \right)} \right|}^p}dt}$$

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$\bf命题：$设$f\left( x \right) \in C\left[ {a, + \infty } \right)$，且$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，则存在数列$\left\{ {{x_n}} \right\} \subset \left[ {a, + \infty } \right)$，使得$\lim \limits_{n \to\infty } {x_n} = + \infty ，\lim \limits_{n \to \infty } f\left( {{x_n}} \right) = 0$

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$\bf命题：$设$\int_a^{{\rm{ + }}\infty } {f\left( x \right)dx} $绝对收敛，且$\lim \limits_{x \to \begin{array}{*{20}{c}}{{\rm{ + }}\infty }\end{array}} f\left( x \right) = 0$，则$\int_a^{{\rm{ + }}\infty } {{f^2}\left( x \right)dx} $收敛

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$\bf命题：$设$f\left( x \right)$在$\left[ {0, + \infty } \right)$上可微，$f'\left( x \right)$在$\left[ {0, + \infty } \right)$上单调递增且无上界，则$\int_0^{ + \infty } {\frac{1}{{1 + {f^2}\left( x \right)}}dx} $收敛

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$\bf命题：$设$f\left( x \right) \in {C^1}\left[ {0, + \infty } \right),f\left( 0 \right) > 0,f'\left( x \right) \geqslant 0,\int_0^{ + \infty } {\frac{1}{{f\left( x \right) + f'\left( x \right)}}dx} < + \infty $，证明：$\int_0^{ + \infty } {\frac{1}{{f\left( x \right)}}dx} < + \infty $

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$\bf命题：$设正值函数$f\left( x \right)$在$\left[ {1, + \infty } \right)$上二阶连续可微，且$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} f''\left( x \right) = + \infty $，则$\int_1^{ + \infty } {\frac{1}{{f\left( x \right)}}dx} $收敛

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$\bf命题：$

[b]附录[/b]

$\bf(Dirichlet判别法)$设$\int_a^A {f\left( x \right)dx} $在$\left[ {a, + \infty } \right)$上有界，且$g(x)$在$\left[ {a, + \infty } \right)$上单调趋于$0$，则$\int_a^{ + \infty } {f\left( x \right)g\left( x \right)dx} $收敛

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$\bf(Abel判别法)$设$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，且$g(x)$在$\left[ {a, + \infty } \right)$上单调有界，则$\int_a^{ + \infty } {f\left( x \right)g\left( x \right)dx} $收敛

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[b] 关于反常积分收敛专题的练习题[/b]