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$\bf引理：$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 + \infty } xf\left( x \right) = 0$，进而$\lim \limits_{x \to + \infty }f\left( x \right) = 0$

 

$\bf命题：$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

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