Proving $\lim\limits_{x \to \infty} xf(x)=0$ if $\int_{0}^{\infty}f(x) dx$ converges. Let $f(x)$ be a monotone non-increasing function such that $\int_{0}^{\infty}f(x) dx$ converges. Prove: $\lim\limits_{x \to \infty} xf(x)=0$. My question is, why can't I simply contradict any other possibility by using the Integral Limit Comparison Test with $1\over x$? I am, after all, to show that $f(x)=o(x)$ as $x\to \infty$. I really don't understand why monotony is crucial here. I could use some help.
 A: Let
$F(x)
=\int_0^x f(t) dt
$.
We are given that
$\lim_{x \to \infty} F(x)$
exists.
Call this limit $L$.
Suppose it is not true that
$\lim_{x \to \infty} xf(x)
= 0
$.
Then there is a $c > 0$
such that
$x f(x) > c$
for arbitrarily large $x$.
Suppose
$x_0 f(x_0) > c$
for $x_0$ large enough that
$\int_{x_0}^{\infty} f(t) dt
< d
$.
Choose an $x_1 > x_0$ such that
$x_1 f(x_1) > c$.
Since $f$ is monotone decreasing,
$\int_{x_0}^{x_1} f(t) dt
> (x_1-x_0)f(x_1)
> (x_1-x_0)(c/x_1)
= c(1-x_0/x_1)
$.
If we choose
$x_1 > 2 x_0$,
then
$\int_{x_0}^{x_1} f(t) dt
> c/2
$.
So we have
$d
>\int_{x_0}^{\infty} f(t) dt
>\int_{x_0}^{x_1} f(t) dt
>c/2
$.
But since
$\lim_{x \to \infty} \int_{x}^{\infty} f(t) dt
=0
$,
we can choose $x_0$ large enough
so that $d$ is arbitrarily small.
In particular,
we can choose $d < c/2$,
which contradicts the
inequality above.
Therefore,
$\lim_{x \to \infty} xf(x)
= 0
$.
A: I assume that $f$ is non-negative (this is justified; see the end of the post for the somewhat tedious details). If $\int_0^{\infty}f(t)\,\mathrm dt$ converges, then, for each $\varepsilon>0$, there exists some $x_0>0$ such that $y>x_0$ implies that $$\int_y^{\infty}f(t)\,\mathrm dt<\frac{\varepsilon}{2}.\tag{1}$$ I will show that if $x>2x_0$, then $xf(x)<\varepsilon$, which will prove that claim that $\lim_{x\to\infty}xf(x)=0$.
Suppose that $x>2x_0$. Then, one has that
\begin{align*}
xf(x)=&\,2\frac{x}{2}f(x)=2\left(x-\frac{x}{2}\right)f(x)=2f(x)\int_{x/2}^x\,\mathrm dt=2\int_{x/2}^xf(x)\,\mathrm dt\\
\underset{\spadesuit}\leq&\,2\int_{x/2}^xf(t)\,\mathrm dt\leq2\int_{x/2}^{\infty}f(t)\,\mathrm dt\underset{\heartsuit}<2\frac{\varepsilon}{2}=\varepsilon.
\end{align*}
In this chain of inequalities,


*

*$\spadesuit$ follows from the fact that $f$ is non-increasing, so that $f(x)\leq f(t)$ for all $t\in[x/2,x]$; and

*$\heartsuit$ follows from (1), given that $x/2>x_0$.

The convergence of $\int_0^{\infty} f(t)\,\mathrm dt$ and the monotonicity of $f$ necessarily imply that $f(t)\geq0$ for all $t\geq0$. To see this, suppose, for the sake of contradiction, that $f(t_0)<0$ for some $t_0\geq 0$. Let
\begin{align*}
K\equiv&-f(t_0)>0.
\end{align*}
Let $M>0$ be an arbitrarily large positive number and choose some positive number $H$ sufficiently large so that $$H\geq\frac{t_0f(0)+M}{K}\tag{2}.$$
Finally, suppose that $x>t_0+H$. Then, one has that
\begin{align*}
\int_0^xf(t)\,\mathrm dt=&\,\int_0^{t_0}f(t)\,\mathrm dt+\int_{t_0}^xf(t)\,\mathrm  dt\underset{\clubsuit}{\leq}t_0f(0)+(x-t_0)f(t_0)=t_0 f(0)-(x-t_0)K\\
<&\,t_0 f(0)-HK\leq t_0 f(0)-[t_0 f(0)+M]=-M,
\end{align*}
where $\clubsuit$ follows again from the fact that $f$ is non-increasing. Since $M$ can be arbitrarily large, it follows that $$\lim_{x\to\infty}\int_0^xf(t)\,\mathrm dt=-\infty,$$ which contradicts the convergence of $\int_0^{\infty} f(t)\,\mathrm dt$.

As the counterexamples in the comments reveal, monotonicity is indispensable. Indeed, I used them at two crucial steps: at $\spadesuit$ and $\clubsuit$.
A: There are counterexamples otherwise: $f(x)$ is $n$ if $n<x<n+1/(n^3)$, $f(x)$ is 0 otherwise. The integral exists, but $f(x)$ diverges.
A: without loss of generality one can assume that $f(0)=a$. Then $\int_0^\infty f(x)dx=k=\int_0^af^{-1}(y)dy$ with first improper integral being type I and second type II. Since the second integral also converge, the assertion follows from $0<xf(x)<\int_0^{f(x)}f^{-1}(y)dy\to 0.$
(a picture might be helpful to see the simple geometric idea)
