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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.

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  • $\begingroup$ Where is the problematic behaviour of the two integrands? $\endgroup$ – Daniel Fischer Feb 16 '16 at 17:03
  • $\begingroup$ in $x=0$, bot that doesn't tell me much $\endgroup$ – G.Fil Feb 16 '16 at 17:06
  • $\begingroup$ Knowing that allows you to analyse the problematic behaviour to see whether the improper integrals converge or not. The behaviour of the denominator is obvious. How does the numerator behave near $0$? $\endgroup$ – Daniel Fischer Feb 16 '16 at 17:08
  • $\begingroup$ $\log{\sin{x}}$ approaches $- \infty$ as $x$ approaches $0$ from the right. So the whole function approaches $- \infty $ . $\endgroup$ – G.Fil Feb 16 '16 at 17:12
  • $\begingroup$ Yes. The question is how fast? $\endgroup$ – Daniel Fischer Feb 16 '16 at 17:14
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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: $$ \int_{0}^{1}\frac{dx}{x}\cdot\log\left(\frac{\sin x}{x}\right)\,dx = -\frac{1}{2}\sum_{n\geq 0}\text{Li}_2\left(\frac{1}{n^2\pi^2}\right)=-\frac{1}{2}\left(\frac{\zeta(2)}{\pi^2}+\frac{\zeta(4)}{4\pi^4}+\frac{\zeta(6)}{9\pi^6}+\ldots\right).$$

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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.

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