Studying the convergence of the integral $$\int_0^\pi \frac{\ln(\sin(x))}{x}dx$$
I'm studying for an exam in calculus and I saw this problem and I'm having trouble showing that this integral diverges (That's what says in the solution )
I tried separating the integral like this $$\int_0^\pi \frac{\ln(\sin(x))}{x}dx=\int_0^{\frac{\pi}{2}}\frac{\ln(\sin(x))}{x}dx+\int_\frac{\pi}{2}^\pi\frac{\ln(\sin(x))}{x}dx$$
Then I substituted $x=\pi-t$ for the second integral and I get
$$\int_0^\pi \frac{\ln(\sin(x))}{x}dx=\pi\int_0^{\frac{\pi}{2}}\frac{\ln(\sin(x))}{x(\pi-x)}dx$$
But I don't know how to procede.
I also tried a different method with the limit test but I don't know with what function to compare. Can someone help me with this problem?
Best Answer
I assume it's an improper integral (from both sides). So let's split it into two integrals as you did and let's show that the first integral on RHS $$ \int_0^{\frac{\pi}{2}}\frac{\ln(\sin(x))}{x}dx $$ doesn't converge.
The integrand is negative and we know $\sin x < x$ for $x>0$. Using the fact that $\ln x$ is increasing, we have $$ \int_0^{\frac{\pi}{2}}\frac{\ln(\sin(x))}{x}dx \leq \int_0^{\frac{\pi}{2}}\frac{\ln(x)}{x}dx = \int_{-\infty}^{\ln \pi/2} t \, dt. $$ The last integral obviously diverges to $-\infty$.