Proving the convergence of the improper integral $\int_0^1 \operatorname{ln}(\operatorname{sin}x)dx$

Question

I'm trying to prove that

\begin{equation*} \int_0^1 \operatorname{ln}(\operatorname{sin}x)dx \end{equation*}

converges.

I tried to show this by decomposing

\begin{equation*} \operatorname{ln}(\operatorname{sin}x)=\operatorname{ln}(\frac{\operatorname{sin}x}{x}x)=\operatorname{ln}(\frac{\operatorname{sin}x}{x})+\operatorname{ln}x. \end{equation*}

However, I can't tell whether

\begin{equation*} \int_0^1 \operatorname{ln}(\frac{\operatorname{sin}x}{x})dx \end{equation*}

converges.

I would appreciate any help on this problem.

Answer 1

HINT: You only have a singularity at $x=0$. Try to compare the integrated function with another one that is asymptotically equivalent and that has an explicit primitive.

P.S. Alternatively, keep following your route and observe that $\log\frac{\sin x}{x}$ has no singularities on $[0, 1]$.

Answer 2

For $x \in [0,1]$, we have $$\dfrac{x}2 \leq \sin(x) \leq x$$ Hence, we have $$\log\left(\dfrac{x}2\right) \leq \log\left(\sin(x)\right) \leq \log\left(x \right)$$ This gives us that $$\int_0^1\log\left(\dfrac{x}2\right)dx \leq \int_0^1\log\left(\sin(x)\right)dx \leq \int_0^1\log\left(x \right)dx$$ Hence, we have $$-1-\log2 \leq \int_0^1\log\left(\sin(x)\right)dx \leq -1$$

EDIT: In case this was not clear, the integrability of $\log\left(\sin(x)\right)$ on $(0,1]$ is trivial, since the function is continuous on $(0,1]$.