I fail to see how this can be achieved:
Define the improper integral (of a non-blocked function) as a limit, and calculate or prove that the integral diverges;
$\large \int_0^1 \frac{dx}{ \sqrt{x}}, $
its a definite integral so what's the relation here to improper ones?
The integrated function isn't defined and, in fact, not even bounded in some (right) neighborhood of zero, thus
$$\int_0^1\frac{dx}{\sqrt x}=\left.\lim_{\epsilon\to 0} 2\sqrt x\right|_\epsilon^1=2$$