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.
Best Answer
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.