# How do you test the improper integral int absx(x^2+1)^-3 dx from (-oo, oo) and evaluate if possible?

Jun 5, 2018

$= \frac{1}{2}$

#### Explanation:

${\int}_{- \infty}^{\infty} \setminus \left\mid x \right\mid {\left({x}^{2} + 1\right)}^{-} 3 \setminus \mathrm{dx}$

The integrand is even:

• $f \left(- x\right) = f \left(x\right)$

... and for an even function:

• ${\int}_{- a}^{a} \setminus f \left(x\right) \setminus \mathrm{dx} = 2 {\int}_{0}^{a} \setminus f \left(x\right) \setminus \mathrm{dx}$

$\implies 2 {\int}_{0}^{\infty} \setminus \left\mid x \right\mid {\left({x}^{2} + 1\right)}^{-} 3 \setminus \mathrm{dx}$

$= 2 {\int}_{0}^{\infty} \setminus \textcolor{red}{x} {\left({x}^{2} + 1\right)}^{-} 3 \setminus \mathrm{dx}$

$= 2 {\int}_{0}^{\infty} \setminus d \left(- \frac{1}{4} {\left({x}^{2} + 1\right)}^{-} 2\right)$

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

Jun 5, 2018

${\int}_{- \infty}^{\infty} \left\mid x \right\mid {\left({x}^{2} + 1\right)}^{-} 3 \text{d} x = \frac{1}{2}$

#### Explanation:

We note that $| - x | {\left({\left(- x\right)}^{2} + 1\right)}^{-} 3 = | x | {\left({x}^{2} + 1\right)}^{-} 3$ for all $x$ so

${\int}_{- \infty}^{\infty} \left\mid x \right\mid {\left({x}^{2} + 1\right)}^{-} 3 \text{d"x=int_0^oo2x(x^2+1)^-3"d} x$

Now let $u = {x}^{2} + 1$ and $\text{d"u=2"d} x$; $u \left(0\right) = 1$ and $u \left(\infty\right) = \infty$

Then

${\int}_{0}^{\infty} 2 x {\left({x}^{2} + 1\right)}^{-} 3 \text{d"x=int_1^oou^-3"d} u = {\lim}_{a \to \infty} {\left[- \frac{1}{2 {u}^{2}}\right]}_{1}^{a} = {\lim}_{a \to \infty} - \frac{1}{2 {a}^{2}} + \frac{1}{2} = \frac{1}{2}$