Question

How do you test the improper integral `int x^(-1/2) dx` from `[1,oo)` and evaluate if possible?

Answer 1

`int_1^oo x^(-1/2)dx = oo`

Explanation:

Based on the integral test the convergence of the integral:

`int_1^oo x^(-1/2)dx`

is equivalent to the convergence of the series:

`sum_(n=1)^oo 1/n^(1/2)`

which we can immediately see is divergent based on the `p`-series criteria.

In fact, the integrand function is defined and continuous in `[1,oo)`, so:

`int_1^oo x^(-1/2)dx = lim_(t->oo) int_1^t x^(-1/2)dx`

`int_1^oo x^(-1/2)dx = lim_(t->oo) [x^(1/2)/(1/2)]_1^t`

`int_1^oo x^(-1/2)dx = lim_(t->oo) 2sqrt(t)-2 = oo`