Theorem about improper integral convergence by comparison I have this theorem that says
Suppose $|f(x)| \leq g(x)$ for  $a \leq x \leq +\infty$. Then from the convergence of $\int_a^{+\infty} g(x)\,dx$ follows the convergence of $\int_a^{+\infty}f(x)\,dx$.
Now, it follows from this theorem that
a) if $\exists \lim_{x \rightarrow +\infty}{|f(x)|x^\lambda} = c$ and $\lambda > 1$ then $\int_a^{+\infty}f(x)\,dx$ converges.
b) if $\exists \lim_{x \rightarrow +\infty}{f(x)x^\lambda} = c > 0$ and $\lambda \leq 1$ then $\int_a^{+\infty}f(x)\,dx$ diverges.
Now my question. What is the analogue of this theorem for $\int_0^a f(x)\,dx$ improper integral of first type? In particular I was trying to solve $\int_0^1 \dfrac {dx}{x-\sin x}$ and I encountered this problem.
Edit: This example better illustrates my question.
When using this theorem to show the convergence of $\int_0^1 {\dfrac{dx}{\sqrt{x^3-3x^2+3x}}}$, I can multiply the integrand by $\sqrt{x}$, where $\lambda \leq 1$.
$\lim_{x \rightarrow 0} {\dfrac{\sqrt{x}} {{\sqrt{x^3-3x^2+3x}}}} = \lim_{x \rightarrow 0} \dfrac {1} {\sqrt{x^2 - 3x +3}} = \dfrac{1}{\sqrt{3}}$. This limit converges, but the improper integral converges as well. Whereas according to the theorem the integral had to diverge.
 A: It is simpler to use equivalence: the improper integrals of two equivalent functions with constant signs both converge or both diverge.
Here, Taylor-Young's formula asserts that $x-\sin x=\frac{x^3}6+o(x^3)$, so near $0$, $x-\sin x\sim_0\frac{x^3}6$, and therefore
$$\frac 1{x-\sin x}\sim_0\frac 6{x^3},$$
and the improper integral of the latter function on $[0,1]$ diverges.
A: Same thing: If $|f(x)|\le g(x)$ for every $x$ in some set, and the integral of $g(x)\,dx$ over that set is finite, then so is the integral of $f(x)\,dx,$ and any unboundedness, such as a vertical asymptote, does not disturb that conclusion.
A: Okay, now I found the answer to the question.
Let's consider the integrals $\int_a^{+\infty} {\dfrac{p}{x^\lambda}}dx$ and $\int_0^1 {\dfrac{p}{x^\lambda}}dx$, $p = const$, $\lambda = const$.
The first integral converges when $\lambda > 1$ and diverges when $0 < \lambda \leq 1$.
But in the second case, it's the contrary, thus, the second integral converges if $0 < \lambda \leq 1$ and diverges if otherwise. That's the reason why from the convergence of $\lim_{x \rightarrow 0} {\dfrac{\sqrt{x}} {{\sqrt{x^3-3x^2+3x}}}}$ follows the convergence of $\int_0^1 {\dfrac{dx}{\sqrt{x^3-3x^2+3x}}}dx$, because $\int_0^1{\dfrac{1}{\sqrt{x}}}dx$ converges.
This is just a direct consequence of the comparison test theorem, that's why I will mark Michael Hardy's answer as the solution, because in order to solve such problems only the knowledge of that theorem is sufficient.
