For monotonic $f$: if the improper integral $\int_0^\infty f(x)dx$ converges, then $\lim_{x\to \infty}xf(x)=0$ I need to prove (as I wrote in the title): 
for monotonic $f$: if the improper integral $\int_0^\infty f(x)dx$ converges, then $\lim_{x\to\infty}xf(x)=0$
hints please? tried to think of Cauchy's.....
 A: For large $x$ $f$ will either be nonnegative or nonpositive. So assume WLOG $f(x)\ge 0$. Then it clearly must be monotone decreasing. Note that since  $\int_0^\infty f(x)\,dx<\infty$ and $f\ge 0$ we have $\lim\limits_{t\to\infty}\int_{\frac{t}{2}}^{t} f(x)\,dx=0$. Then since $f$ is monotone decreasing we have $0\le \displaystyle \frac{tf(t)}{2}\le \int_{\frac{t}{2}}^{t} f(x)\,dx\to 0$ at $t\to\infty$, giving the result.
A: Let $F(x) = \int_{0}^xf(x)dx$, Note that $F'(x) = f(x)$. Because $f(x)$ is monotonic at $[0,+\infty)$, it implies $f(x)$ is ultimately either $\lt 0$ or $\gt 0$ after a specific point $x_0$ as $x\rightarrow +\infty$. Therefore, its antiderivative $F(x)$ is ultimately either decreasing or increasing for all $x \gt x_0$. Moreover since $\int_{0}^{\infty}f(x)dx$ is convergent, then $F(x)$ must be bound. To sum up, $F(x)$ is ultimately either monotonic increasing with a upper bound or monotonic decreeing with a lower bound at $[x_0,+\infty）$, where $x_0 \gt 0$.   
Without losing generality, we assume F(x) is monotone increasing at $[x_0, +\infty)$, it implies $f(x) \gt 0$ at $[x_0, +\infty)$. Thus, if $lim_{x\rightarrow\infty}xf(x) = d$, Note $d \gt 0$ since $f(x) \gt 0$ and monotonic at $[0,+\infty)$. Hence by the definition of limit:
$$\text{There exist X, when } x\gt x_0 \gt X \text{ such that, }|xf(x) - d| \lt \frac{d}{2}\tag{1}$$
Note (1) implies 
$$xf(x) \gt \frac{d}{2} \Longrightarrow f(x) \gt \frac{N}{x}\space\space(\text{where  }N=\frac{d}{2})\tag{2}$$
Also since $f(x) \gt 0$, (2) implies:
$$\int_{x}^{\infty}f(x)dx \gt \int_{x}^{\infty}\frac{N}{x}dx~~(x \gt x_0 \gt X)\tag{3}$$
Note, $F(x) = \int_{0}^{x}f(x)dx + \int_{x}^{\infty}f(x)dx$. Since $\int_{x}^{\infty}\frac{N}{x}dx$ is divergent, By (3), so does $\int_{x}^{\infty}f(x)dx$, which contradict the fact F(x) is convergent. Hence, $lim_{x\rightarrow\infty}xf(x) = 0$.
