$\bf命题1：$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$
$\bf命题2：$设$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，且$\frac{{f\left( x
\right)}}{x}$在${\left[ {a, + \infty } \right)}$上单调递减，则$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}
} xf\left( x \right) = 0$
$\bf{证明}$ 我们类似$\bf命题1$容易证明$f\left( x \right) \ge 0$
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 = \frac{x}{2}$，则由积分中值定理知，存在$\xi \in \left[ {\frac{x}{2},x}
\right]$，使得f\left( \xi \right) \cdot
\frac{x}{2} = \int_{\frac{x}{2}}^x {f\left( t \right)dt} < \varepsilon
即$xf\left( \xi \right) < 2\varepsilon $，从而由$\frac{{f\left( x
\right)}}{x}$单调递减及$f\left( x \right) \ge 0$知0 \le
xf\left( x \right) = x \cdot \frac{{f\left( x \right)}}{x} \cdot x \le x \cdot
\frac{{f\left( \xi \right)}}{\xi } \cdot 2\xi = 2xf\left( \xi
\right) < 4\varepsilon
所以有$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}
} xf\left( x \right) = 0$，进而由极限的定义即知$\lim \limits_{x \to \begin{array}{*{20}{c}}{
+ \infty }\end{array}
} f\left( x \right) = 0$
$\bf命题3：$设$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，且$xf\left( x
\right)$在${\left[ {a, + \infty } \right)}$上单调递减，则$\lim \limits_{x \to\begin{array}{*{20}{c}} { + \infty
}\end{array}
} xf\left( x \right)\ln x = 0$
$\bf{证明}$ 我们类似$\bf命题1$容易证明$f\left( x \right) \ge 0$
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\ge1 $，使得当$x > \sqrt x >M$时，有\int_{\sqrt
x }^x {f\left( t \right)dt} < \frac{\varepsilon }{2}
从而由$xf(x)$单调递减及$f\left( x \right) \ge 0$知0 \le
xf\left( x \right)\ln x = 2xf\left( x \right)\int_{\sqrt x }^x {\frac{1}{t}dt}
\le 2\int_{\sqrt x }^x {\frac{{tf\left( t \right)}}{t}dt} =
2\int_{\sqrt x }^x {f\left( t \right)dt} < \varepsilon
所以有$\lim \limits_{x \to\begin{array}{*{20}{c}} { + \infty
}\end{array}
} xf\left( x \right)\ln x = 0$，进而由极限的定义即知$\lim \limits_{x \to \begin{array}{*{20}{c}}{
+ \infty }\end{array}
} f\left( x \right) = 0$