Let me try to explain the spirit of the question.
The functions $f(x)=1/x^{p}$ for $0<p<1$ and $f(x)=\ln x$ have the following properties: they are in some respect 'nice' on $\mathbb{R}^{+}$, but diverge at zero. Nevertheless, their improper Riemann integral $\int_{0}^{1}\,f(x)\,dx$ exists. The reason is, one may say, that they diverge 'sufficiently slowly' at the origin.
One might conjecture that the following: if $xf(x) \to 0$ as $x\to 0^{+}$, and $0$ is the only singularity of $f$ on $[0,\,1]$, then $\int_{0}^{1}\,f(x)\,dx$ exists as either a proper or an improper Riemann integral. We may assume that $\int_{\epsilon}^{1}\,f(x)\,dx$ exists as a proper integral for all $0<\epsilon<1$.
I very much doubt this conjecture is correct (because I expect I would have seen it somewhere by now), but have been unable to come up with a counterexample. For instance, no member of the following family is a counterexample: $f(x)=1/x^{p}(-\ln x/2)^{b}$, where $b$ is any real number and $p$ is as above (actually, to make it work for any $b$, the upper bound of the integral should be reduced from 1 to some number between $0$ and $1$---say to $\int_{0}^{1/2}$).
Thus my question: is this conjecture correct, or is there a counterexample?
I haven't carefully defined my class of functions because that delineation is, in fact, a part of the question. I imagine one would want to demand that some property (or a laundry list of properties) hold for $f$ on each closed interval $[a,\,1]$ for all $0<a<1$. And I would imagine that the only property that should really matter is that $f$ be (properly) Riemann integrable on all such intervals.
But maybe other conditions need to be applied in order to say something sensible. I would be OK, for example, with stipulating that $f$ can be analytically continued to the complex plane, and that, except for the singularity at the origin, it is then analytic on an open set that contains the closed disk of radius 1 centered at the origin. As far as the singularity at the origin, however, as indicated by my prototypical examples $1/x^{p}$ and $\ln x$, I would want to allow it to be as exotic as it wants to be---as long as $xf(x) \to 0$ as $x\to 0^{+}$. I would be surprised if, once we are in the complex plane, there is some trickery about this limit only existing along the positive real line but not in general, but I could live with it if that turns out to be the case.
If $f$ is positive and decreasing on $(0,1)$ and $\int_0^1 f < \infty,$ then yes, we must have $f(x)/x \to 0.$ But in general, no, there are examples where $f(x)/x \not \to 0.$ Here's a sketch of one: Over the points $1/n,$ put a triangular spike of height $n$ and base less than $1/n^3.$ The bases can be chosen to disjoint. Now define $f(x)$ to be one of the spikes whenever $x$ lies in one of the bases, with $f=0$ elsewhere. Because the sums of the areas of all these triangles is finite, we have $\int_0^1f<\infty.$ But $f(1/n)/(1/n) = 1$ for all $n.$