Prove convergence of $\int_0^{\pi/2} \log\sin x\,\mathrm dx$ I've seen many proof of
$$\int_0^{\pi/2} \log\sin x\,\mathrm dx= -\dfrac{\pi}{2} \log 2$$
where result was obtained through algebraic manipulations...
But to do this, we must to prove that the integral converges.
My question is : how to prove that $\int_0^{\pi/2} \log\sin x\,\mathrm dx$ converges? 
Thanks for answers.
 A: The integrand is negative so you can use the comparison test (if you prefer to work with positive integrands, rewrite it as $-\int_0^{\frac{\pi}{2}} (-\log \sin x) \, dx$). Thus, we can compare your integral with the integral $\int_0^1 \log(x) \, dx$ which can be evaluated explicitly:
$$ \int_0^1 \log(x) \, dx = \lim_{\delta \to 0} \int_{\delta}^1 \log(x) \, dx = \lim_{\delta \to 0} \left[ x \log(x) - x\right]^{x = 1}_{x = \delta} = -1 -\delta \log \delta + \delta \xrightarrow[\delta \to 0]{} -1$$
Since
$$ \lim_{x \to 0} \frac{\log \sin x}{\log x} = \lim_{x \to 0} \frac{\frac{\cos x}{\sin x}}{\frac{1}{x}} = \left( \lim_{x \to 0} \cos x \right) \left( \lim_{x \to 0} \frac{x}{\sin x} \right) = 1$$
the integrals must converge/diverge together, and in this case, they converge together.
A: Thanks to Sangchul Lee comment : on $[0, \pi/2]$
$$ \begin{split} x\geq \sin x \geq \frac{2}{\pi}  x  \Rightarrow \int_0^{\pi/2} \log x\,\mathrm dx = \frac{\pi}{2}\log(\pi/2)-\frac{\pi}{2} \geq  \int_0^{\pi/2} \log \sin x\,\mathrm dx &\geq \frac{\pi}{2}\log(2/\pi)+ \int_0^{\pi/2} \log x\,\mathrm dx\\&= \frac{\pi}{2}\log(2/\pi) + \frac{\pi}{2}\log(\pi/2)-\frac{\pi}{2}
\end{split}$$
The integral $\int_0^{\pi/2} \log \sin x\,\mathrm dx$ is bounded, thus converges.
A: For $0<x<1$,
$\log(\sin(x))=$
$\log(x)(1+\frac{1}{\log(x)}\log(\frac{\sin(x)}{x}))=$
$\log(x)(1+\epsilon(x))$ 
with $\lim_{x\to 0^+}\epsilon(x)=0$.
So, by equivalence criteria, near $0^+$, our integral is convergent since $\int_0log(x)dx $ is concergent.
To compute its value, let
$$S=\int_0^\frac{\pi}{2}\log(\sin(x))dx$$
And $$C=\int_0^\frac{\pi}{2}\log(\cos(x))dx$$
With the change $y=\frac{\pi}{2}-x$, we get $C=S$.
On the other hand
$$S+C=\int_0^\frac{\pi}{2}(\log(\sin(2x)-\log(2))dx$$
or, with $t=2x$
$$=\frac{1}{2}\int_0^\pi \log(\sin(t))dt-\frac{\pi}{2}\log(2)$$
$$\frac{1}{2}(S+\int_\frac{\pi}{2}^\pi \log(\sin(t))dt)-\frac{\pi}{2}\log(2)$$
and finally, with $u=\pi-t$, we get
$$\color{green}{S=C=-\frac{\pi}{2}\log(2)}$$
