improper improper integral?

∫ x^3*e^(-x^2) dx , from x = 0 to x = ∞
Sol :
Let u = - x^2 , then du = - 2x*dx

∫ x^3*e^(-x^2) dx
= ∫ x^3*e^u * du / (-2x)
= (1/2) * ∫ - x^2 e^u du
= (1/2) * ∫ u*e^u du
= (1/2) * ∫ u * d e^u
= (1/2) * [ u*e^u - ∫ e^u du ] , 分部積分
= (1/2)*[ u*e^u - e^u ] + c
= (1/2)*e^u*( u - 1 ) + c
= (1/2)*e^(-x^2)*( - x^2 - 1 ) + c
= (-1/2)*e^(-x^2)*( x^2 + 1 ) + c

∫ x^3*e^(-x^2) dx , from x = 0 to x = ∞
= lim b→∞ [ ∫ x^3*e^(-x^2) dx , from x = 0 to x = b ]
= lim b→∞ [ (-1/2)*e^(-x^2)*( x^2 + 1 ) ] , from x = 0 to x = b
= (-1/2) * lim b→∞ [ e^(-x^2)*( x^2 + 1 ) ] , from x = 0 to x = b
= (-1/2) * lim b→∞ [ e^(-b^2)*( b^2 + 1 ) - 1 ]
= (-1/2) * ( 0 - 1 ) , 參考註解
= 1/2 ..... Ans

註解.
lim b→∞ e^(-b^2)*( b^2 + 1 ) = 0
pf :
當 b → ∞
e^(-b^2)*( b^2 + 1 )
= ( b^2 + 1 ) / e^(b^2)
→ 2b / [ 2b * e^(b^2) ] , by the L'Hopital's Rule
→ 1 / e^(b^2)
→ 0
Q.E.D.