关于反常积分收敛的专题讨论

$f命题:$设$int_a^{ + infty } {fleft( x ight)dx} $收敛,若$lim limits_{x o egin{array}{*{20}{c}} {{ m{ + }}infty } end{array}} fleft( x ight)$存在,则$lim limits_{x o egin{array}{*{20}{c}} {{ m{ + }}infty } end{array}} fleft( x ight) = 0$

1

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

1

$f命题:$设$int_a^{ + infty } {fleft( x ight)dx} $收敛,且${fleft( x ight)}$在$left[ {a,{ m{ + }}infty } ight)$单调,则$lim limits_{x o egin{array}{*{20}{c}}{ + infty }end{array}} xfleft( x ight) = 0$,进而$lim limits_{x o egin{array}{*{20}{c}}{ + infty }end{array}} fleft( x ight) = 0$

1

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

1

$f命题:$设$int_a^{ + infty } {fleft( x ight)dx} $收敛,且$frac{{fleft( x ight)}}{x}$在${left[ {a, + infty } ight)}$上单调递减,则$lim limits_{x o egin{array}{*{20}{c}}{ + infty }end{array}} xfleft( x ight) = 0$

1

$f命题:$设$fleft( x ight)$单调且$lim limits_{x o egin{array}{*{20}{c}}
{{0^ + }}
end{array}} fleft( x ight) = + infty $,若$int_0^1 {fleft( x ight)dx} $收敛,则$lim limits_{x o egin{array}{*{20}{c}}
{{0^ + }}
end{array}} xfleft( x ight) = 0$

1

$f命题:$设$int_a^{ + infty } {fleft( x ight)dx} $收敛,且$xfleft( x ight)$在${left[ {a, + infty } ight)}$上单调递减,则$lim limits_{x oegin{array}{*{20}{c}} { + infty }end{array}} xfleft( x ight)ln x = 0$

1

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

1   2

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

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

1

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

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

1

$f命题:$设$fleft( x ight) in Cleft[ {a, + infty } ight)$,且$int_a^{ + infty } {fleft( x ight)dx} $收敛,则存在数列$left{ {{x_n}} ight} subset left[ {a, + infty } ight)$,使得$lim limits_{n oinfty } {x_n} = + infty ,lim limits_{n o infty } fleft( {{x_n}} ight) = 0$

1

$f命题:$设$int_a^{{ m{ + }}infty } {fleft( x ight)dx} $绝对收敛,且$lim limits_{x o egin{array}{*{20}{c}}{{ m{ + }}infty }end{array}} fleft( x ight) = 0$,则$int_a^{{ m{ + }}infty } {{f^2}left( x ight)dx} $收敛 

1

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

1

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

1

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

1

$f命题:$

附录

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

1

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

1

 

 




 

 



 关于反常积分收敛专题的练习题

原文地址:https://www.cnblogs.com/ly142857/p/3664523.html