Convergence of definite integral I have to find out the convergence of the next integral:
$$\int^{\pi/2}_0{\frac{\ln(\sin(x))}{\sqrt{x}}}dx$$
Any help? Thanks
 A: The only problem is at $0$.
We have, as $x$ approaches $0$,
$$
\sin x\sim x
$$
so
$$
\frac{\log(\sin x)}{\sqrt{x}}\sim\frac{\log x}{\sqrt{x}}
$$
and the integral of the latter converges at $0$.
If you did not know that, compare for instance with $\frac{1}{x^{2/3}}$.
Since 
$$
x^{2/3}\frac{\log x}{\sqrt{x}}\longrightarrow 0
$$
we have
$$
\lvert \frac{\log x}{\sqrt{x}}\rvert\leq \frac{C}{x^{2/3}}.
$$
Now it is easily seen that the integral of $1/x^{2/3}$ converges at $0$.
So your integral converges.
A: If you integrate by parts, then you find
$$\int_0^{\pi/2}dx\,\frac{\ln\sin x}{\sqrt{x}}=2\sqrt{x}\ln\sin x\Bigg|_0^{\pi/2}+2\int_0^{\pi/2}dx\,\sqrt{x}\frac{\cos x}{\sin x}.$$
It is fairly clear at this point that the integral converges because $\sqrt{x}/\sin x\sim x^{-1/2}$ at $x=0$.
A: The tricky part of the integral is near $x=0$.  There, note that $\sin{x} \sim x$, and consider 
$$\int dx \frac{\log{x}}{\sqrt{x}}$$
Substitute $x=u^2$, $dx=2 u du$ and this integral is equal to
$$2 \int du u \frac{1}{u} \log{u^2} = 4\int du \log{u} = 4 (u \log{u}-u) = 2 (\sqrt{x} \log{x} - 2 \sqrt{x}) $$
Thus, near $x=0$, the singularity is integrable, and the integral converges.
