Convergence of two improper integrals.

Question

I have two integrals:

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

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

Do they converge? I tried to settle by comparison, but nothing seems to work for me, hence I ask for your help.

Answer 1

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 $\alpha\in(0,3)$.

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By using the Weierstrass product for the sine function, we also have: $$ \int_{0}^{1}\frac{dx}{x}\cdot\log\left(\frac{\sin x}{x}\right)\,dx = -\frac{1}{2}\sum_{n\geq 0}\text{Li}_2\left(\frac{1}{n^2\pi^2}\right)=-\frac{1}{2}\left(\frac{\zeta(2)}{\pi^2}+\frac{\zeta(4)}{4\pi^4}+\frac{\zeta(6)}{9\pi^6}+\ldots\right).$$

Answer 2

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.