# What is int_-oo^oo (x^2)(e^(-x^3))dx?

Jan 9, 2016

The integral diverges to infinity

#### Explanation:

First, we can use $u$ substitution to find the indefinite integral:

Let $u = - {x}^{3}$
Then $\mathrm{du} = - 3 {x}^{2} \mathrm{dx} \implies {x}^{2} \mathrm{dx} = - \frac{1}{3} \mathrm{du}$

Then

$\int {e}^{- {x}^{3}} {x}^{2} \mathrm{dx} = \int {e}^{u} \left(- \frac{1}{3}\right) \mathrm{du}$

$= - \frac{1}{3} \int {e}^{u} \mathrm{du}$

$= - \frac{1}{3} {e}^{u} + C$

$= - \frac{1}{3} {e}^{- {x}^{3}} + C$

Now, evaluating the definite integral:

${\int}_{- \infty}^{\infty} {x}^{2} {e}^{- {x}^{3}} \mathrm{dx} = {\lim}_{A \to - \infty} {\lim}_{B \to \infty} {\int}_{A}^{B} {x}^{2} {e}^{- {x}^{3}} \mathrm{dx}$

$= {\lim}_{A \to - \infty} {\lim}_{B \to \infty} {\left[- \frac{1}{3} {e}^{- {x}^{3}}\right]}_{A}^{B}$

$= {\lim}_{A \to - \infty} {\lim}_{B \to \infty} \frac{1}{3} \left({e}^{- {A}^{3}} - {e}^{- {B}^{3}}\right)$

$= {\lim}_{A \to - \infty} \frac{1}{3} \left({e}^{- {A}^{3}} - 0\right)$

$= \infty$

Thus the integral does not converge.