Examine convergence of $\int_0^{\infty} \frac{1}{x^a \cdot |\sin(x)| ^b}dx$ Examine convergence of $\int_0^{\infty} \frac{1}{x^a \cdot |\sin(x)| ^b}dx$ for $a, b > 0$. There are 2 problems. $|\sin(x)|^b = 0$ for $x = k \pi$ and $x^a = 0$ for $x = 0$. We can write $\int_0^{\infty} \frac{1}{x^a \cdot |\sin(x)| ^b}dx = \int_0^{1} \frac{1}{x^a \cdot |\sin(x)| ^b}dx + \int_1^{\infty} \frac{1}{x^a \cdot |\sin(x)| ^b}dx$ but what to do next?

Note that $\lim_{x \to 0} x^{-1} \sin x = 1$ and on $[0,1]$ we have $\sin 1 \leqslant x^{-1} \sin x \leqslant 1.$
Hence, with $a,b > 0$,
$$(\sin 1)^b x^{a+b} \leqslant x^a |\sin x|^b = x^{a+b}|x^{-1} \sin x |^b \leqslant x^{a+b},$$
and
$$\frac{1}{x^{a+b}} \leqslant \frac{1}{x^a |\sin x|^b} \leqslant \frac{(\sin 1)^{-b}}{x^{a+b} }.$$
By the comparison test, the integral over $[0,1]$ converges if $a+b < 1$ and diverges if $a+b \geqslant 1.$
You can show the integral over $[1, \infty)$ diverges if $a \leqslant 1$ by comparing with a divergent $p$-series, using
$$\int_1^\infty \frac{1}{x^a |\sin x|^b} \, dx \geqslant \sum_{k=1}^\infty \int_{2k \pi + \pi/4}^{2k \pi + \pi/2} \frac{1}{x^a |\sin x|^b} \, dx. $$
Since $0 < 1/\sqrt{2}\leqslant \sin x  \leqslant 1$ on $[2k\pi + \pi/4, 2k \pi + \pi/2],$ we have 
$$\int_1^\infty \frac{1}{x^a |\sin x|^b} \, dx \geqslant \frac{\pi}{4}\sum_{k=1}^\infty \frac{1}{(2\pi)^a(k + 1/4)^a } \geqslant \frac{\pi}{4}\sum_{k=1}^\infty \frac{1}{(4\pi)^ak^a } . $$
