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?

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 $a,b > 0$.
