Question

How do you determine if the improper integral converges or diverges `int 1 / [sqrt x]` from 0 to infinity?

Answer 1

The integral diverges.

Explanation:

We could use the comparison test for improper integrals, but in this case the integral is so simple to evaluate that we can just compute it and see if the value is bounded.

`int_0^oo1/sqrtx\ dx=int_0^oox^(-1/2)=[2sqrtx]_0^oo=lim_(x->oo)(2sqrtx)-2sqrt(0)=lim_(x->oo)(2sqrtx)=oo`

This means that the integral diverges.