Given the integral:
$\int_a^\infty \frac{1}{x^\alpha}\,\text{d}x$
and knowing that it converges when $\alpha >1$ and it diverges when $\alpha\le1$, I would like to know how I can transform the integral into
$\int_0^b \frac{1}{x^\beta}\,\text{d}x$
which converges when $\beta<1$ and diverges when $\beta\ge1$ by a couple of more or less simple steps. I can't really figure it out.