Integration of x^3 * e^(-x^2) from zero to infinity?

Question

This should equal .5, but I can't find out how to do it. I keep getting infinity. I think you have to use the fact that the integration (e^(-x^2)) from o to infinity is sqrt(pi)/2 (from the error function), but I'm not entirely sure

Alam Ko Iyan · Accepted Answer

let u = -x^2
du = -2x dx

∫ x^3 e^(-x^2) dx
= -1/2 ∫ x^2 e^(-x^2) (-2x dx)
= 1/2 ∫ u e^u du ..... integration by parts is needed here .
= 1/2 (u e^u - ∫ e^u du)
= 1/2 (u - 1) e^u 
= 1/2 (-x^2 - 1) e^(-x^2)

evaluated at infinity:
this becomes negative infinity * zero which is indeterminate
= lim 1/2 (-x^2 - 1)/e^(x^2) ... then we can use L'Hopitals rule
= lim 1/2 (-2x) / (2x e^(x^2))
= 0


evaluate at zero:
= (1/2) (0 - 1) e^0 = -1/2

thus the evaluation is:
= 0 - (-1/2)
= 1/2
the error function is not needed here .
instead, improper integrals are applied here ...

Integration of x^3 * e^(-x^2) from zero to infinity?

1 Answer