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Prove convergence of $$\int^{\pi /2}\limits_{0} \frac{\ln(\sin x)}{\sqrt x}\mathrm dx = I$$

Looking for other answers I got that $$\int\limits_0^\frac{\pi}{2}\frac{\ln x\ \mathrm dx}{\sqrt{x}} < I <-2\sqrt{2\pi}$$

How conclude from here that $I$ is convergent? I was mainly looking at this answer.

Edit.

In this answer can you explain why integrand is monotonous and why from that follows convergence?

unit 1991
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  • Do these answer your question? (1) https://math.stackexchange.com/questions/815065/prove-the-convergence-of-integral?noredirect=1 or (2) https://math.stackexchange.com/questions/313437/convergence-of-definite-integral?noredirect=1 – Sumanta Feb 18 '22 at 16:41
  • @User My link is the same as your first linked question.I am asking specific question how continue from there. – unit 1991 Feb 18 '22 at 17:02
  • If the integral diverged, then $I$ would be $-\infty$, which is against the inequality. – xpaul Feb 18 '22 at 17:27
  • Why $I$ would be $-\infty$ if integral diverged? @xpaul – unit 1991 Feb 18 '22 at 17:29
  • Use the definition of divergence of improper integrals. – xpaul Feb 18 '22 at 17:31
  • @xpaul For definition $lim_{c\to b} \int_a ^c f(x)dx$ must be finite. But for that you need to compute $I$ or I am doing something wrong? – unit 1991 Feb 18 '22 at 17:34
  • Note $$\int_\epsilon^{\pi/2}\frac{\ln(\sin x)}{\sqrt x}dx$$ is monotone w.r.t $\epsilon\in(0,\pi/2)$. – xpaul Feb 18 '22 at 17:46
  • @xpaul Can you give hints how show that? – unit 1991 Feb 18 '22 at 17:48

1 Answers1

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Define $$ I(\epsilon)=\int_\epsilon^{\pi/2}\frac{\ln(\sin x)}{\sqrt x}dx. $$ Clearly $I(\epsilon)<0$. For $0<\epsilon<\epsilon'<\pi/2$, one has $$ I(\epsilon)-I(\epsilon')=\int_\epsilon^{\epsilon'}\frac{\ln(\sin x)}{\sqrt x}dx \le 0 $$ which implies that $I(\epsilon)$ is increasing in $\epsilon\in(0,\pi/2)$. Then when $\epsilon\to 0^+$, the limit of $I(\epsilon)$ is either finite or $-\infty$.

xpaul
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