Convergence of $\int_0^1\frac{f(x)}{x}dx$. [duplicate]

The Problem:

Let $f:[0,1]\to[0,+\infty)$ be a continuous function such that $f(0)=0$. Is $\displaystyle \int_0^1 \frac{f(x)}{x}dx$ convergent?

The Context:

I've just come across this question (for a continuos function $f$, calculate $\lim_{n}\int_0^1\frac{nf(x)}{1+n^2x^2}dx$) in the special case where $f(0)=0$. My first try to was apply Lebesgue Dominated Convergence Theorem just by multiplying and dividing by $x$. So, in order to get a convergent bound, I need $f(x)/x$ to be convergent in $[0,1]$.

Actually, I'm not able neither to prove it, nor to disprove it. An, in order to be precise, I also have include some other extre hypothesis (such as $f\ge 0$).

Some remarks:

Of course, if $\lim_{x\to0+}f(x)/x=L>0$, the answer is clear (the integral converges).

If $\lim_{x\to0+}f(x)/x=0$, then (near $x=0$) $f(x)<x$ and we also get convergence.

So the only possibility to find a counterexample (such an $f$ for wich the integral diverges) is that $\nexists\lim_{x\to0^+}f(x)/x$. But, since $f(0)=0$, this is equivalent to "$f$ is not differentiable at $x=0$".

But, as I said befor, I am not able neither to find it, nor to prove the convergence.

Solution 1:

Hint: Draw a triangle with base $(\frac 1{n+1}, \frac 1 n)$ and height $\frac n {\ln n}$ for $n=2,3,...$ Combine these to get a graph of function $g$. Take $f(x)=xg(x)$. This function has the desired properties.