What is #int_-oo^oo (x^2)(e^(-x^3))dx#?

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Answer 1 · 2016-01-09T08:08:35

The integral diverges to infinity

First, we can use #u# substitution to find the indefinite integral:

Let #u = -x^3#
Then #du = -3x^2dx => x^2dx = -1/3du#

Then

#inte^(-x^3)x^2dx = inte^u(-1/3)du#

#= -1/3inte^udu#

#= -1/3e^u + C#

# = -1/3e^(-x^3) + C#

Now, evaluating the definite integral:

#int_(-oo)^oox^2e^(-x^3)dx = lim_(A->-oo)lim_(B->oo)int_A^Bx^2e^(-x^3)dx#

#= lim_(A->-oo)lim_(B->oo)[-1/3e^(-x^3)]_A^B#

#= lim_(A->-oo)lim_(B->oo)1/3(e^(-A^3)-e^(-B^3))#

#= lim_(A->-oo)1/3(e^(-A^3)-0)#

#= oo#

Thus the integral does not converge.