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    $f引理:$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 + infty } xfleft( x ight) = 0$,进而$lim limits_{x o + infty }fleft( x ight) = 0$

     

    $f命题:$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

    $f证明$  对任意的$x in left[ {a, + infty } ight)$,由$f分部积分法$知
    [int_a^x {tf'left( t ight)dt} = tfleft( t ight)left| {egin{array}{*{20}{c}}x\a
    end{array}} ight. - int_a^x {fleft( t ight)dt} ]
    而由$int_a^{ + infty } {fleft( t ight)dt} $收敛知$lim limits_{x o + infty } int_a^x {fleft( t ight)dt} $存在,又由引理知$lim limits_{x o + infty }xfleft( x ight) = 0$,所以有$lim limits_{x o + infty }int_a^x {tf'left( t ight)dt}$存在,从而由反常积分收敛的定义即证

     

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  • 原文地址:https://www.cnblogs.com/ly758241/p/3787575.html
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