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

Question

1863 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-09T08:08:35

The integral diverges to infinity

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

Let $u = -x^3$
Then $du = -3x^2dx => x^2dx = -1/3du$

Then

$inte^(-x^3)x^2dx = inte^u(-1/3)du$

$= -1/3inte^udu$

$= -1/3e^u + C$

$= -1/3e^(-x^3) + C$

Now, evaluating the definite integral:

$int_(-oo)^oox^2e^(-x^3)dx = lim_(A->-oo)lim_(B->oo)int_A^Bx^2e^(-x^3)dx$

$= lim_(A->-oo)lim_(B->oo)[-1/3e^(-x^3)]_A^B$

$= lim_(A->-oo)lim_(B->oo)1/3(e^(-A^3)-e^(-B^3))$

$= lim_(A->-oo)1/3(e^(-A^3)-0)$

$= oo$

Thus the integral does not converge.

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