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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$
$\bf证明$(1)不妨设${f\left( x \right)}$单调递减，则我们可以断言$f\left( x \right) \ge 0$，否则存在${x_0} \in \left[ {a, + \infty } \right)$，使得$f\left( {{x_0}} \right) < 0$，
于是当$x > {x_0}$时，由${f\left( x \right)}$的单调性知
\begin{align*}\int_a^x {f\left( t \right)dt} &= \int_a^{{x_0}} {f\left( t \right)dt} + \int_{{x_0}}^x {f\left( t \right)dt} \\&\le \int_a^{{x_0}} {f\left( t \right)dt} + f\left( {{x_0}} \right)\left( {x - {x_0}} \right) \to- \infty \left( {x \to+ \infty } \right)\end{align*}
这与$\int_a^{ + \infty } {f\left( x \right)dx} $收敛矛盾，故$f\left( x \right) \ge 0$

   (2)由于$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，则由$\bf{Cauchy收敛准则}$知，对任给$\varepsilon > 0$，存在正数$M>a$，使得当$x ,y> M$时，有
\[\left| {\int_x^y {f\left( t \right)dt} } \right| < \frac{\varepsilon }{2}\]
特别地，取$y=2x$，则由$\bf积分中值定理$知，存在$\xi \in \left[ {x,2x} \right]$，使得\[xf\left( \xi \right) = \int_x^{2x} {f\left( t \right)dt} < \frac{\varepsilon }{2}\]
从而由${f\left( x \right)}$单调递减及$f\left( x \right) \ge 0$知\[0 \le 2xf\left( {2x} \right) \le 2xf\left( \xi \right) = 2\int_x^{2x} {f\left( t \right)dt} < \varepsilon \]
所以我们有$\lim \limits_{x \to + \infty } xf(x) = 0$，进而由极限的定义即知$\lim \limits_{x \to + \infty }f\left( x \right) = 0$

$\bf命题：$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