Does $\int_{0}^{\pi/2} \ln(\sin(x))dx$ converge or diverge?

Question

Problem:

Does $\int_{0}^{\pi/2} \ln(\sin(x))dx$ converge or diverge?

Solution:

If $\int_{0}^{\pi/2} \ln(\sin(x))dx$, is absolute convergent, then it's convergent. Hence:

$0\leq\int_{0}^{\pi/2} |\ln(\sin(x))|dx \leq \int_{0}^{\pi/2} |\ln(\sin(\pi/2))|dx$ since $\sin(x)$ is increasing on that interval, as well as $\ln(x)$.

What we get is that:

$0\leq\int_{0}^{\pi/2} |\ln(\sin(x))|dx \leq \int_{0}^{\pi/2} 0 dx$

Hence:

$0\leq\int_{0}^{\pi/2} |\ln(\sin(x))|dx \leq 0$

Conclusion: Because it's absolute convergent, it must converge.

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So, how's my solution? Is it correct? If not, what's the proper way of solving it?

Thanks.

Answer 1

It is wrong, since you actually have $\bigl|\ln\bigl(\sin(x)\bigr)\bigr|\geqslant \ln\bigl(\sin(x)\bigr)$, and therefore$$\int_0^{\pi/2}\bigl|\ln\bigl(\sin(x)\bigr)\bigr|\,\mathrm dx\geqslant\int_0^{\pi/2}\ln\bigl(\sin(x)\bigr)\,\mathrm dx.$$Actually, in this case, convergence and absolute convergence mean the same thing, because$$\left(\forall x\in\left(0,\frac\pi2\right)\right):\left|\ln\bigl(\sin(x)\bigr)\right|=-\ln\bigl(\sin(x)\bigr),$$since$$\left(\forall x\in\left(0,\frac\pi2\right)\right):0<\sin(x)<1.$$

Answer 2

Substitution might help. Let $u=\sin{x}$.

Then $\int_0^{\pi/2}\ln(\sin{x})dx=\int_0^1\frac{\ln{u} \ du}{\sqrt{1-u^2}}$

The integrand is bounded at the right end point of the interval by l'hopital's rule.

For any $0<\Delta<1$, $\ 0\le u \le\Delta \implies \frac{|\ln{u}|}{\sqrt{1-u^2}}\le \frac{|\ln u|}{\sqrt{1-\Delta^2}}$ so convergence at the left endpoint can be established by comparison.