# How do you test the improper integral int x^(-3/2) dx from [0, oo) and evaluate if possible?

Dec 4, 2017

The integral is divergent

#### Explanation:

The improper integral is calculated in $2$ parts :

${\lim}_{q \to {0}^{+}} {\int}_{q}^{1} {x}^{- \frac{3}{2}} \mathrm{dx} + {\lim}_{p \to \infty} {\int}_{1}^{p} {x}^{- \frac{3}{2}} \mathrm{dx}$

First part

${\lim}_{q \to {0}^{+}} {\int}_{q}^{1} {x}^{- \frac{3}{2}} \mathrm{dx} = {\lim}_{q \to {0}^{+}} {\left[{x}^{- \frac{3}{2} + 1} / \left(- \frac{3}{2} + 1\right)\right]}_{q}^{1}$

$= {\lim}_{q \to {0}^{+}} {\left[{x}^{- \frac{1}{2}} / \left(- \frac{1}{2}\right)\right]}_{q}^{1}$

$= {\lim}_{q \to {0}^{+}} {\left[2 {x}^{- \frac{1}{2}}\right]}_{1}^{q}$

$= {\lim}_{q \to {0}^{+}} \left(2 {q}^{- \frac{1}{2}} - 2\right)$

$= {\lim}_{q \to {0}^{+}} \left(\frac{2}{\sqrt{q}} - 2\right)$

$= + \infty$

This part is divergent, therefore the integral is divergent

Second part

${\lim}_{p \to \infty} {\int}_{1}^{p} {x}^{- \frac{3}{2}} \mathrm{dx} = {\lim}_{p \to \infty} {\left[{x}^{- \frac{3}{2} + 1} / \left(- \frac{3}{2} + 1\right)\right]}_{1}^{p}$

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

$= {\lim}_{p \to \infty} {\left[2 {x}^{- \frac{1}{2}}\right]}_{p}^{1}$

$= {\lim}_{p \to \infty} \left(2 - 2 {p}^{- \frac{1}{2}}\right)$

$= {\lim}_{p \to \infty} \left(2 - \frac{2}{\sqrt{p}}\right)$

$= 2$

This part is convergent.