When does this integral converge?

Question

So I'd like to find out for which values of $a,b>0$ the following integral is well-defined and how that will change if the absolute value is removed? Thanks!

$I = \int_0^\infty \dfrac{|\sin(x^a)|}{x^b}dx $

Answer 1

I assume you really want to know for which values the improper integral converges, and you mean $f(a,b)$, not $f(x)$.

The change of variables $t = x^a$ gives you

$$f(a,b) = \dfrac{1}{a} \int_0^\infty |\sin(t)|\; t^p\; dt$$

where $p = (1-a-b)/a$. For convergence at $\infty$ you need $p < -1$, and for convergence at $0$ (where $\sin(t) \approx t$) you need $p > -2$. With $a > 0$ this is equivalent to $1 < b < a + 1$.

On the other hand, without the absolute value the integral converges as long as $-2 < p < 0$, using Dirichlet's test. In fact we can even get a closed form:

$$ \int_0^\infty \sin(t)\; t^p\; dt = \cos(\pi p/2)\; \Gamma(1+p)$$