$$ \mbox{Find out if the following improper integral exists:}\quad \int_{0}^{1}{\left\vert\,\log\left(x\right)\,\right\vert\over \sin\left(\sqrt{\,x\,}\,\right)}\,{\rm d}x $$
We have that ${\left\vert\,\log\left(x\right)\,\right\vert\over \sin\left(\sqrt{\,x\,}\,\right)} \sim_0 \frac{\left\vert\,\log\left(x\right)\,\right\vert}{ \sqrt{x}}$ and $\left\vert\,\log\left(x\right)\,\right\vert =_0 o(\sqrt{x})$, so if I had to guess I'd say the integral exists.
Is this true? How do I rigorously show it?
Over the interval $[0,\pi/2]$ the sine function is concave, hence $\sin x\geq\frac{2}{\pi}x$. This gives: $$0\leq I=\int_{0}^{1}\frac{-\log x}{\sin\sqrt{x}}\,dx\leq\frac{\pi}{2}\int_{0}^{1}\frac{-\log x}{\sqrt{x}}\,dx=2\pi\int_{0}^{1}(-\log t)\,dt$$ by substituting $x=t^2$. Since the logarithm function belongs to $L^1((0,1))$, because: $$\int_{0}^{1}(-\log t)\,dt=\int_{0}^{+\infty}ue^{-u}\,du=\Gamma(2)=1,$$ (by substituing $t=e^{-u}$) this gives that your integral exists and it is bounded by $2\pi$.
By convexity, it is possible to improve this bound up to: $$ I\in\left(4,\frac{4}{\sin 1}\right).$$