How do you determine if the improper integral converges or diverges #int [(x^3)( e^(-x^4) )] dx# from negative infinity to infinity?

We have: #int_(-oo)^(oo) x^3 e^(-x^4) \ dx#

For starters, it's actually a fairly tame function that you are looking to integrate

it is the product of 2 continuous functions, and that is a very important factor

for #f(x) = x^3 e^(-x^4)#, we have #f(0) = 0# and, so there should be no singularities.

it is an odd function as #f(x) = - f(-x)# so it is also symmetric in terms of a #180^o# rotation about the origin.

it is also non-periodic should it should also have some "value" as #-oo to x to +oo#..

we could explore the derivative to check for weird behaviour, but we may as well look at the [indefinite] integral as it is surprisingly easy:

#int x^3 e^(-x^4) \ dx#

there is a pattern:

as #d/dx( e^(-x^4) ) = - 4x^3 e^(-x^4) #

so #d/dx(- 1/4 e^(-x^4) ) = x^3 e^(-x^4) #

so as an indefinite integral #int x^3 e^(-x^4) dx = int d/dx(- 1/4 e^(-x^4) ) \ dx = - 1/4 e^(-x^4) + C#

Now applying the limits of integration:

#int_(-oo)^(oo) x^3 e^(-x^4) \ dx = - 1/4( e^(-x^4))_(x to -oo)^(x to oo) = 0#

not surprising once you see this

graph{x^3 e^(-x^4) [-2.002, 2.003, -1.001, 1.002]}