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?
 A: To even have a chance at convergence at $\infty$, you need $a > 1$. However, then near $x = 0$, we have $x^a \lvert \sin(x) \rvert^b \approx x^{a+b}$ and since $a+b > 1$, we will have divergence near $x=0$. Thus the integral diverges for all $a,b > 0$.
A: $f(x) = \frac{1}{xsin(x)}$.  $f$ is strictly positive.  Let's start by looking at some interval on which $f(x)$ is discontinuous.  How about $[3,\pi]$.  To show that $f$ diverges we need to find a lower bound and show that it diverges.  Because $x$ is bounded we can pick $g(x) = \frac{1}{4sin(x)}$.  If you integrate that function you will find that it goes off to infinity.  Therefore the integral diverges.
We now have to show that if there is any open interval on which the integral diverges the entire integral diverges.  Luckily this is easy.  As we have already established $f$ is strictly positive therefore the integral of $f$ is strictly positive.
If you want to show that this is true for all $a, b$ just recognize you can choose an interval in which $x$ is bounded and that there is some interval that contains a discontinuity and $csc(x)^n$ diverges for all $n$.
I hope this helps.  I really like this problem.  I'm going to get a class to do this tomorrow for extra credit or something tomorrow.
EDIT: I forgot to mention that if $b=0$ then it's different because $sin(x)^b=1$ so you just have a basic integral that is integrable everywhere.
A: 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 
$$\int_1^\infty \frac{1}{x^a |\sin x|^b} \, dx \geqslant \frac{\pi}{4}\sum_{k=1}^\infty \frac{1}{(2\pi)^a(k + 1/4)^a } \geqslant \frac{\pi}{4}\sum_{k=1}^\infty \frac{1}{(4\pi)^ak^a } . $$
