How to show that a strictly decreasing, continuous function which decays slower than 1/x is not integrable? I want to show that a function that decays slower than $1/x$ is not integrable and I tried it the following way:
Assume the positive, strictly decreasing and continuous function $g(x)$ decays slower than $1/x$, i.e. $$xg(x)\to\infty\,\, (x\to\infty).$$ 
Then there exists a large $c>0$ such that $g(x)\geq 1/x\,\, \forall x\geq c$. Hence,
$$\int\limits_0^\infty g(x) dx \geq \int\limits_c^\infty g(x) dx \geq \int\limits_c^\infty 1/x dx = \lim\limits_{x\to \infty} \ln x - \ln c = \infty.$$
However, I was given a heads up by a friend that $$xg(x)\to\infty\,\, (x\to\infty)$$ is not the negation of the limit being zero, which would correspond to a faster decay than $1/x$. Hence, what I need to show is:
If  $\limsup x g(x) > 0$ then $g(x)$ is not integrable. Can somebody give me a hint here?
 A: We show: if $g(x)$ is a positive, decreasing function satisfying $\limsup_{x\to\infty} xg(x) > 0$, then $\int_0^\infty g(x)\,dx$ diverges. (Continuity isn't required, other than as a way to ensure the partial integrals $\int_0^X g(x)\,dx$ are defined.)
By the lim sup hypothesis, there exists a constant $\delta>0$ and an increasing sequence $\{x_n\}$ tending to infinity such that $g(x_n) > \frac\delta{x_n}$. In particular, $g(x) > \frac\delta{x_n}$ for all $x\le x_n$, since $g$ is decreasing. Therefore (setting $x_0=0$ for convenience)
\begin{equation*}
\int_0^{\infty} g(x)\,dx = \sum_{j=1}^\infty \int_{x_{j-1}}^{x_j} g(x)\,dx > \sum_{j=1}^\infty \int_{x_{j-1}}^{x_j} \frac\delta{x_j} \,dx = \delta \sum_{j=1}^\infty \bigg( 1 - \frac{x_{j-1}}{x_j} \bigg).
\end{equation*}
If infinitely many of the $\frac{x_{j-1}}{x_j}$ are less than $\frac12$ then the right-hand side clearly diverges. Otherwise, $\frac{x_{j-1}}{x_j} \ge \frac12$ for $j\ge J$. Using the inequality $1-t \ge \frac{-\log t}{2\log2}$ for $t\in[\frac12,1]$, we see that
\begin{equation*}
\int_0^{\infty} g(x)\,dx > \delta \sum_{j=J}^\infty \bigg( 1 - \frac{x_{j-1}}{x_j} \bigg) \ge \frac\delta{2\log2} \sum_{j=J}^\infty \log\frac{x_j}{x_{j-1}} = +\infty,
\end{equation*}
since the sum is a telescoping series: $\sum_{j=J}^K \log\frac{x_j}{x_{j-1}} = \log\frac{x_K}{x_{J-1}}$.
