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.
Best Answer
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.