TITLE: Convergence of $\int_0^1 \frac{\sqrt{x-x^2}\ln(1-x)}{\sin{\pi x^2}} \mathrm{d}x.$
QUESTION [6 upvotes]: I would like to prove the convergence of the Newton integral
$$\int_0^1 f(x)\mathrm{d}x =\int_0^1 \frac{\sqrt{x-x^2}\ln(1-x)}{\sin{\pi x^2}} \mathrm{d}x.$$
I split this into two integrals $\displaystyle\int_0^\epsilon f(x)\mathrm{d}x$ and $\displaystyle\int_\epsilon^1 f(x)\mathrm{d}x$. It is easy to show that the integral is convergent on $(0, \epsilon]$ by limit comparison with $\displaystyle\int_0^\epsilon \frac{\sqrt{x}x}{\pi x^2}\mathrm{d}x$.
But I cannot find anything to compare with around $x = 1$ on $[\epsilon, 1)$.

REPLY [3 votes]: Regarding to Quotient test for integrals with non-negative integrands, we have $$\lim_{x\to 1^-}{(1-x)^{0.6}|f(x)|}=0<\infty$$ so the integral $$\int_{\epsilon>0}^1|f(x)|dx$$ converges