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

Question

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

Luke Alvarez · Accepted Answer

#=0#

see below

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]}

Isaiah Allen · Answer

undefined To determine if the improper integral $ \int_{-\infty}^{\infty} x^3 e^{-x^4} \, dx $ converges or diverges, we need to evaluate the integral separately for each half of the interval and then consider their combined behavior.

1. For the lower limit, $ -\infty $, we substitute $ t = -x^4 $, and as $ x $ approaches $ -\infty $, $ t $ approaches $ +\infty $. Thus, the integral becomes $ \int_{+\infty}^0 \frac{e^t}{-4t^{3/4}} \, dt $.

2. For the upper limit, $ +\infty $, we substitute $ u = x^4 $, and as $ x $ approaches $ +\infty $, $ u $ approaches $ +\infty $. Thus, the integral becomes $ \int_0^{+\infty} \frac{e^{-u}}{4u^{3/4}} \, du $.

Both integrals can be evaluated using techniques such as comparison tests or integration by parts. If both integrals converge, the original improper integral converges; if at least one of them diverges, the original integral diverges.