Show that the improper integral $\int\limits_0^{\frac{\pi}{2}}\ln(\sin(x)) dx$ exists.
My attempt:
We know that for all $\alpha$ with $0<\alpha<\frac{1}{2}$ the expression $\int\limits_{\alpha}^{\frac{\pi}{2}}\ln(\sin(x)) dx$ is a well defined Riemann-integral (continuity of composition of functions). We split the integral into $$ \int\limits_{\alpha}^{\frac{\pi}{2}}\ln(\sin(x)) dx=\underset{\text{improper integral if }\alpha\to 0}{\underbrace{\int\limits_{\alpha}^{\frac{1}{2}}\ln(\sin(x)) dx}}+\underset{\text{Riemann-integral, existence o.k.}}{\underbrace{\int\limits_{\frac{1}{2}}^{\frac{\pi}{2}}\ln(\sin(x)) dx}}~~~~~~~~~~~(1) $$
Since $h:]0,\frac{1}{2}]\to\mathbb{R}$, with $h(x):=\sin(x)- \frac{x}{2}$, is monotonously increasing (see first derivative) it follows $\sin(x)\geq \frac{x}{2}$ for all $x\in~]0,\frac{1}{2}]$. We know that $-\ln(x)$ is a monotonously decreasing function, so
$$ 0\leq-\ln(\sin(x))=\left|\ln(\sin(x))\right|\leq -\ln\left(\frac{x}{2}\right), \text{ for all } x\in~]0,\frac{1}{2}]. $$ Further, $$ \int\limits_{\alpha}^{\frac{1}{2}}-\ln\left(\frac{x}{2}\right) dx =-\left(x\ln\left(\frac{x}{2}\right)-x\right)\Big|_{\alpha}^{\frac{1}{2}}=-\frac{1}{2}\ln\left(\frac{1}{4}\right)-\frac{1}{2}+\alpha \ln\left(\frac{\alpha}{2}\right)+\alpha. $$ As we know that $\lim\limits_{\alpha\to 0}\alpha \ln\left(\frac{\alpha}{2}\right)=0$ (rule of L'Hospital) it follows $$ 0\leq \lim\limits_{\alpha\to 0}\int\limits_{\alpha}^{\frac{1}{2}}-\ln\left(\frac{x}{2}\right) dx=\cdots=\lim\limits_{\alpha\to 0}\left(-\frac{1}{2}\ln\left(\frac{1}{4}\right)-\frac{1}{2}+\alpha \ln\left(\frac{\alpha}{2}\right)+\alpha\right)= -\frac{1}{2}\ln\left(\frac{1}{4}\right)-\frac{1}{2}. $$ By comparison test, the improper integral $\int\limits_{0}^{\frac{1}{2}}\ln(\sin(x)) dx= \lim\limits_{\alpha\to 0}\int\limits_{\alpha}^{\frac{1}{2}}\ln(\sin(x)) dx$ exists. The addition of two limits yields a well defined value so equation $(1)$ shows the existence of $\int\limits_{0}^{\frac{\pi}{2}}\ln(\sin(x)) dx$.
Is this correct? Maybe someone has another, quicker approach to show the existence?
PS: There are already a few related questions (see, Integration of $\ln\sin x$ from 0 to$ \frac{\pi}{2}$by DUIS ) where the value of the improper integral is calculated. But those questions doesn't say anything about its existence.
