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

Jun 19, 2017

${\int}_{1}^{\infty} {x}^{- \frac{1}{2}} \mathrm{dx} = \infty$

#### Explanation:

Based on the integral test the convergence of the integral:

${\int}_{1}^{\infty} {x}^{- \frac{1}{2}} \mathrm{dx}$

is equivalent to the convergence of the series:

${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ \left(\frac{1}{2}\right)$

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

In fact, the integrand function is defined and continuous in $\left[1 , \infty\right)$, so:

${\int}_{1}^{\infty} {x}^{- \frac{1}{2}} \mathrm{dx} = {\lim}_{t \to \infty} {\int}_{1}^{t} {x}^{- \frac{1}{2}} \mathrm{dx}$

${\int}_{1}^{\infty} {x}^{- \frac{1}{2}} \mathrm{dx} = {\lim}_{t \to \infty} {\left[{x}^{\frac{1}{2}} / \left(\frac{1}{2}\right)\right]}_{1}^{t}$

${\int}_{1}^{\infty} {x}^{- \frac{1}{2}} \mathrm{dx} = {\lim}_{t \to \infty} 2 \sqrt{t} - 2 = \infty$