Let us first do some inital work. Here we first safely assume $r>1$. Then our integral turns into (We exclude the limits for now)
$ \int_{0}^{\infty} \frac{1}{x^r} \, \text{d}x = \left[ \frac{1}{1-r} x^{1-r} \right]_{0}^{\infty} $
Quite easily we can see that since $r>0$
$ \lim_{x \to \infty} \frac{1}{1-r}x^{1-r} = 0 $
This is quite good! Now we know that
$ \int_{0}^{\infty} \frac{1}{x^r} \, \text{d}x = \left[ \frac{1}{1-r} x^{1-r} \right]_{0}^{\infty} = 0 - \lim_{x \to 0} \frac{1}{1-r}x^{1-r}$
The problem is that this last limit does not exist for any values of r. It simply tends to $- \infty$ for all values of $r$.
Let us now plot the problem for a few values of r, to really strike home the point

As we can see the problem is how these functions behave at values very close to zero and very close to infinity. In a nuttshell, the problem with evaluating this integral is that
$ \frac{1}{x^r} $
Approaches $\infty$ at a faster rate when r> increases but, approaches 0 at a slower rate when r increases. Similarly
Approaches $\infty$ at a slower rate when r decreases but, approaches 0 at a faster rate when r increases. As can be seen from the graph.
Although a similar integral that does converge is
$ \int_{0}^{\infty} \frac{1}{k^r} \, \text{d}x = \frac{1}{r - 1} $
So to sum up
$ \int_{0}^{\infty} \frac{1}{x^r} \, \text{d}x = -\infty $
when $1