Question

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

Answer 1

`=0`

see below

Explanation:

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]}