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

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]}