$f引理：$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$
$f证明$(1)不妨设${fleft( x
ight)}$单调递减，则我们可以断言$fleft( x
ight) ge 0$，否则存在${x_0} in left[ {a, + infty }
ight)$，使得$fleft( {{x_0}}
ight) < 0$，
于是当$x > {x_0}$时，由${fleft( x
ight)}$的单调性知
egin{align*}int_a^x {fleft( t
ight)dt} &= int_a^{{x_0}} {fleft( t
ight)dt} + int_{{x_0}}^x {fleft( t
ight)dt} \&le int_a^{{x_0}} {fleft( t
ight)dt} + fleft( {{x_0}}
ight)left( {x - {x_0}}
ight) o- infty left( {x o+ infty }
ight)end{align*}
这与$int_a^{ + infty } {fleft( x
ight)dx} $收敛矛盾，故$fleft( x
ight) ge 0$
(2)由于$int_a^{ + infty } {fleft( x
ight)dx} $收敛，则由$f{Cauchy收敛准则}$知，对任给$varepsilon > 0$，存在正数$M>a$，使得当$x ,y> M$时，有
[left| {int_x^y {fleft( t
ight)dt} }
ight| < frac{varepsilon }{2}]
特别地，取$y=2x$，则由$f积分中值定理$知，存在$xi in left[ {x,2x}
ight]$，使得[xfleft( xi
ight) = int_x^{2x} {fleft( t
ight)dt} < frac{varepsilon }{2}]
从而由${fleft( x
ight)}$单调递减及$fleft( x
ight) ge 0$知[0 le 2xfleft( {2x}
ight) le 2xfleft( xi
ight) = 2int_x^{2x} {fleft( t
ight)dt} < varepsilon ]
所以我们有$lim limits_{x o + infty } xf(x) = 0$，进而由极限的定义即知$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