TITLE: Prove the convergence of integral
QUESTION [6 upvotes]: $$\int^{\pi /2}_{0} \frac{\ln(\sin x)}{\sqrt x}dx$$
Use the segment integral formula? The $\sqrt x$ is zero at $x=0$ and $\ln\sin x$ is $-\infty$

REPLY [3 votes]: We know that $\displaystyle\underbrace{\int_0^\tfrac\pi2\frac{\ln x}{\sqrt x}dx}_A=\sqrt{2\pi}\bigg(\ln\frac\pi2-2\bigg)$. This can easily be proven using integration by 
parts. Now, let's show that $\displaystyle\int_0^\tfrac\pi2\frac{\ln(\sin x)}{\sqrt x}dx$ is bounded : $\displaystyle\int_0^\tfrac\pi2\frac{\ln(\sin x)}{\sqrt x}dx-\int_0^\tfrac\pi2\frac{\ln x}{\sqrt x}dx=$
$=\displaystyle\int_0^\tfrac\pi2\frac{\ln(\sin x)-\ln x}{\sqrt x}dx=\int_0^\tfrac\pi2\frac{\ln\dfrac{\sin x}x}{\sqrt x}dx$. But $\dfrac{\sin x}x$ is strictly decreasing on this interval, 
being bounded between $1$ and $\dfrac2\pi$ . So our last integral is also bounded in between $\displaystyle\int_0^\tfrac\pi2\frac{\ln1}{\sqrt x}dx=$
$=0$, and $\displaystyle\int_0^\tfrac\pi2\frac{\ln\dfrac2\pi}{\sqrt x}dx=\bigg(\ln\frac2\pi\bigg)\bigg[2\sqrt x\bigg]_0^\frac\pi2=\underbrace{\sqrt{2\pi}\cdot\ln\frac2\pi}_B$ , since the logarithm is monotonous 
on the interval $\bigg[\dfrac2\pi,1\bigg]$. Therefore, our initial integral is also bounded in between $A+0=A$, and 
$A+B=-2\sqrt{2\pi}$ . Then, since the integrand is monotonous on the interval $\bigg[0,\dfrac\pi2\bigg]$, convergence 
immediately follows.