Convergence of two improper integrals. I have two integrals:


*

*$\displaystyle\int_0^1 \frac{\log(\sin x)}{x} \, \mathrm{d}x$

*$\displaystyle\int_0^1 \frac{\log(\sin x)}{\sqrt{x}} \, \mathrm{d}x$
Do they converge? I tried to settle by comparison, but nothing seems to work for me, hence I ask for your help.
 A: If you manage to prove that $\log\left(\frac{\sin x}{x}\right)$ is bounded between $-\frac{x^2}{5}$ and $-\frac{x^2}{6}$ over $(0,1)$, you just have to study the behaviour of:
$$ \int_{0}^{1}\frac{\log x}{x}\,dx,\qquad \int_{0}^{1}\frac{\log x}{\sqrt{x}}\,dx. $$
The first integral is obviously diverging, since the primitive of $\frac{\log x}{x}$ is $\frac{\log(x)^2}{2}$, while the second integral evaluates to $-4$. The integral 
$$ \int_{0}^{1}\frac{dx}{x^{\alpha}}\cdot\log\left(\frac{\sin x}{x}\right) $$
is between $-\frac{1}{5(3-\alpha)}$ and $-\frac{1}{6(3-\alpha)}$ for any $\alpha\in(0,3)$.

By using the Weierstrass product for the sine function, we also have:
$$ \int_{0}^{1}\frac{dx}{x}\cdot\log\left(\frac{\sin x}{x}\right)\,dx = -\frac{1}{2}\sum_{n\geq 0}\text{Li}_2\left(\frac{1}{n^2\pi^2}\right)=-\frac{1}{2}\left(\frac{\zeta(2)}{\pi^2}+\frac{\zeta(4)}{4\pi^4}+\frac{\zeta(6)}{9\pi^6}+\ldots\right).$$
A: Hint, first note that if $y\to 0^+$, we know by LHR that 
$$\log(y)=O((\frac1y)^\epsilon),\quad\forall \epsilon>0$$
And that
$$\log(1+y)=y+O(y)$$
Then plugging in 
$$\sin x=x+O(x^3)$$
as $x\to 0^+$, we get
$$\log(\sin x)=\log(x+O(x^3))=\log x+\log(1+O(x^2))=O((\frac1x)^\epsilon)+O(x^2)=O((\frac1x)^\epsilon)$$
Which means, near $x=0$, the two integrands belong under the classes 
$$O(x^{1-\epsilon})\quad\text{and}\quad O(x^{\frac12-\epsilon})$$
respectively. 
Now recall what you learned about $p-$ integral. 
