Does $\int_1^2\frac{\sin{x}}{\log{x}} \, dx$ converge?

Question

I tried to use the absolute convergence test. $$\int_1^2\left|\frac{\sin{x}}{\log{x}}\right| \, dx<\int_1^2\left|\frac{1}{\log{x}}\right| \, dx$$ but I'm not sure whether $\frac{1}{\log{x}}$ converges or not.

Thanks in advance!

Answer 1

No: for $x > 1$, $\log{x} < x-1$, so $\frac{1}{\log{x}} > \frac{1}{x-1}$. $\sin{1}$ is not zero, and $1<\pi/2$, so sine is increasing for $x$ near $1$. Therefore on the interval $(1,\epsilon)$, $$ \frac{\sin{x}}{\log{x}} > \frac{\sin{1}}{x-1}, $$ and the right-hand side does not have a convergent integral. Hence by comparison, neither does the left-hand side.