Prove the convergence of integral $$\int^{\pi /2}_{0} \frac{\ln(\sin x)}{\sqrt x}dx$$
Use the segment integral formula? The $\sqrt x$ is zero at $x=0$ and $\ln\sin x$ is $-\infty$ 
 A: Use a limit comparison test with the function $g(x) = x^{-2/3}$. 
We have $g$ integrable on $[0,\pi/2]$:
$$\int_{0}^{\pi/2}x^{-2/3}dx = 3(\pi/2)^{1/3} < \infty.$$
Now consider the limit
$$\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim_{x \rightarrow 0} \frac{x^{-1/2}\ln(\sin x)}{x^{-2/3}}=\lim_{x \rightarrow 0} \frac{\ln(\sin x)}{x^{-1/6}}.$$
Using L'Hospital's rule
$$\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)}=\lim_{x \rightarrow 0} \frac{-6x^{1/6} \cos x}{\frac{\sin x}x}=0$$
and the integral converges -- as there are $\delta$ and $\epsilon$ such that $|f(x)| < \epsilon x^{-2/3}$ for $0<x<\delta$.
A: 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.
A: Integrate it by parts
$$\int_{0}^{\pi/2}  \frac{\log(\sin (x))}{\sqrt x} dx  = \left [ \log( \sin x) (2\sqrt x)\right]_{0}^{\pi/2} - \int_{0}^{\pi/2} \frac{2 \sqrt x}{\sin(x)}\cos(x)dx$$taking limit on first you get you get zero. For the last integral, use Jordan's inequality for sine (lower bound) and $\cos(x) \le 1$. i.e.
$$\int_{0}^{\pi/2} \frac{2 \sqrt x}{\sin(x)}\cos(x)dx \le \int_0^{\pi/2}\frac{2 \pi\sqrt x}{2x}dx$$
you know that the left integral converges.
