Question

How do you show whether the improper integral `int (x^2)(e^(-x^3)) dx` converges or diverges from negative infinity to infinity?

Answer 1

See the explanation section, below.

Explanation:

We need (among other things):

`lim_(urarr-oo)e^u = 0` and `lim_(urarroo)e^u = oo`.

Let's note that `int x^2 e^(-x^3)dx= -1/3e^(-x^3) +C`.
(By substitution with `u=x^3`.)

We also note that to (attempt to) evaluate an integral that is improper at both limits of integration, we need to breack the interval into two pieces using `int_a^b f(x) dx = int_a^c f(x) dx + int_c^b f(x) dx`.

Let's use `c=0`, (Because that makes the exponential easy to evaluate.)

`int_-oo^oo (x^2)(e^(-x^3)) dx = int_-oo^0 (x^2)(e^(-x^3)) dx + int_0^oo (x^2)(e^(-x^3)) dx` `" "` (If both integrals exist.)

`int_-oo^0 (x^2)(e^(-x^3)) dx = lim_(ararr-oo)int_a^0 (x^2)(e^(-x^3)) dx`

`= lim_(ararr-oo) ((-1/3e^(-x^3))]_a^0)`

`= lim_(ararr-oo) (-1/3+1/3e^(-a^3))`

As `ararr-oo`, the exponent `-a^3 rarr oo` so the integral diverges.

We conclude that the original integral, `int_-oo^oo (x^2)(e^(-x^3)) dx` diverges.

Note
By the way, the other integral, `int_0^oo (x^2)(e^(-x^3)) dx = 1/3`

Also, we might have noted before integrating, that
as `xrarr-oo`,
the integrand `x^2e^(-x^3) rarroo`.
So. there was no chance of the integral left of zero being finite.

Here is the graph of the function:

graph{x^2 e^(-x^3) [-12.87, 32.78, -4.08, 18.77]}