Convergence of $\int_0^1 \frac{\sqrt{x-x^2}\ln(1-x)}{\sin{\pi x^2}} \mathrm{d}x.$

Question

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 $[\epsilon, 1)$.

Answer 1

If you write $x = 1 - \delta$, you obtain

$$\int_0^{1-\varepsilon} \frac{\sqrt{(1-\delta)(1-(1-\delta))}\,\ln \delta}{\sin \bigl(\pi(1-\delta)^2\bigr)}\, d\delta = \int_0^{1-\varepsilon} \frac{\sqrt{\delta(1-\delta)}\,\ln \delta}{\sin \bigl(\pi(2\delta-\delta^2)\bigr)}\,d\delta,$$

and the integrand of that can be compared to the harmless

$$\frac{\sqrt{x}\,\ln x}{2\pi x}.$$

Answer 2

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