I am trying to determine the convergence of the integral \begin{equation} \int_0^1 \frac{f(x)}{x}\, dx \end{equation} 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.

  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Aug 31 '14 at 18:11
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    $\begingroup$ Try $f(x)=\frac{1}{1+|\log x|}$ for $x>0$ and $f(0)=0$. $\endgroup$ – Kelenner Aug 31 '14 at 18:11
  • $\begingroup$ That counter example works Kelenner, thanks! $\endgroup$ – 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.

The better known assumption of Hölder continuity implies Dini continuity, and thus is sufficient for your purpose. Both of these are much weaker than differentiability.

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    $\begingroup$ Thanks for your reply. I think Holder continuity is precisely the distinction I was looking for. $\endgroup$ – Ryan Hunter Sep 1 '14 at 14:16

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