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

Apr 21, 2017

The improper integral:

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

is divergent.

#### Explanation:

Consider first that ${x}^{-} 2$ is an even function:

${\left(- x\right)}^{-} 2 = \frac{1}{- x} ^ 2 = \frac{1}{x} ^ 2 = {x}^{- 2}$

so that:

${\int}_{- 1}^{1} {x}^{- 2} \mathrm{dx} = 2 {\int}_{0}^{1} {x}^{-} 2 \mathrm{dx}$

Now pose:

$f \left(t\right) = {\int}_{t}^{1} {x}^{-} 2 \mathrm{dx} = {\left[{x}^{-} \frac{1}{- 1}\right]}_{t}^{1} = - 1 + \frac{1}{t} = \frac{1 - t}{t}$

So that:

${\int}_{0}^{1} {x}^{- 2} \mathrm{dx} = {\lim}_{t \to {0}^{+}} f \left(t\right) = {\lim}_{t \to {0}^{+}} \frac{1 - t}{t} = + \infty$

then the improper integral:

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

is divergent.