# Convergence of Integral near 0

I am trying to determine the convergence of the integral $$\int_0^1 \frac{f(x)}{x}\, dx$$ given that $f(x)$ is bounded and continuous on $[0,1]$, and that $f(x)=0$. The boundedness is just so that the question of convergence is only at the point $x=0$. I specifically want $f$ to be only continuous on $[0,1]$ and not differentiable in a neighborhood of the origin as I could just use a Taylor expansion of $f$ to solve the problem then.

I believe that the integral should converge but I can't figure out exactly how to write it down. Since $1/x$ is the critical exponent of convergence near $0$ it seems that multiplying $1/x$ by any function which vanishes at the origin should be enough to make the integral converge. More concretely, if $f(x)=x^{1/n} log(x)^m$ then $\lim_{x \to 0} f(x)=0$ for all positive values of $n$ and $m$, and $\int_0^1 f(x)/x \, dx < \infty$. The derivative of these $f$ become infinite as $x\to 0$, and at faster rates for larger $m$ and $n$, so they are good candidates for $\int f(x)/x$ to not converge, yet the integral still converges.

Any suggestions for a proof, or a counterexample to show the integral does not always converge would be much appreciated.

• One nice approach is to use the Weierstrass approximation theorem to show that there is a sequence of polynomials $f_n(x)$ such that $f_n \to f$ uniformly and $f_n(0) = 0$. From there, it follows that $f_n(x)/x \to f(x)/x$ uniformly, so that the limit of the integrals is the integral of the limit. – Omnomnomnom Aug 31 '14 at 18:11
• Try $f(x)=\frac{1}{1+|\log x|}$ for $x>0$ and $f(0)=0$. – Kelenner Aug 31 '14 at 18:11
• That counter example works Kelenner, thanks! – Ryan Hunter Aug 31 '14 at 18:18

A counterexample was given in comments by Kelenner: $f(x)=\frac{1}{1+|\log x|}$ for $x>0$ and $f(0)=0$.
You may be interested in the concept of Dini continuity. If a function is Dini-continuous, and $f(0)=0$, then $\int_0^1 \frac{f(x)}{x} \,dx$ converges.
Dini continuity can be awkward to work with, but it's the weakest assumption that makes the logarithmic divergence in $1/x$ go away. As a result, it comes up in the context of singular integral operators.