Integral $\int_0^1 \frac{\ln(\sin(x))}{\sqrt{x}}$

Question

For absolute convergence the integral is $\int_0^1 \frac{-\ln(\sin(x))}{\sqrt{x}}$.

Does this integral converges?

Answer 1

The problem is at $0$. Since $\sin(x) \sim x$ for small $x$ we expect that $\int_0^1 \frac{\ln(\sin(x))}{\sqrt{x}} dx$ converges if $\int_0^1 \frac{\ln(x)}{\sqrt{x}} dx$ converges.

Let $\epsilon>0$ be given. Since $\ln(\sin(x)) - \ln(x) = \ln(\sin(x)/x) \to \ln(1) = 0$ as $x \to 0$, there exists $\delta \in (0,1)$ such that $| \ln(\sin(x)) - \ln(x) | < \epsilon$ whenever $0 < x < \delta.$

Then,

1. $$\int_\delta^1 \frac{\ln(\sin(x))}{\sqrt{x}} dx < \infty,$$
2. $$ \int_\eta^\delta \frac{\ln(x)}{\sqrt{x}} dx = [ 2\sqrt{x} \cdot \ln(x) ]_\eta^\delta - \int_\eta^\delta 2\sqrt{x} \cdot \frac{1}{x} dx = [ 2\sqrt{x} \cdot \ln(x) ]_\eta^\delta - [ 4\sqrt{x} ]_\eta^\delta \\ \to 2\sqrt{\delta} \cdot \ln(\delta) - 4\sqrt{\delta} < \infty, $$ as $\eta\to 0$, so $$\int_0^\delta \frac{\ln(x)}{\sqrt{x}} dx < \infty,$$ 3. $$ \left| \int_0^\delta \frac{\ln(\sin(x))}{\sqrt{x}} dx - \int_0^\delta \frac{\ln(x)}{\sqrt{x}} dx \right| \leq \int_0^\delta \frac{|\ln(\sin(x)) - \ln(x)|}{\sqrt{x}} dx \leq \int_0^\delta \frac{\epsilon}{\sqrt{x}} dx \\ = \epsilon [2\sqrt{x}]_0^\delta = 2\sqrt{\delta}\epsilon < 2\epsilon < \infty . $$

Together these give $$ \int_0^1 \frac{\ln(\sin(x))}{\sqrt{x}} dx < \infty. $$