What is the integral of e^(x^3)ex3?

1 Answer

You can't express this integral in terms of elementary functions.

Depending on what you need the integration for, you may choose a way of integration or another.

Integration via power series
Recall that e^xex is analytic on mathbb{R}, so forall x in \mathbb{R} the following equality holds
e^x=sum_{n=0}^{+infty}x^n/{n!}
and this means that
e^{x^3}=sum_{n=0}^{+infty}(x^3)^n/{n!}=sum_{n=0}^{+infty}{x^{3n}}/{n!}
Now youcan integrate:
int e^{x^3} dx=int (sum_{n=0}^{+infty}{x^{3n}}/{n!}) dx=c+sum_{n=0}^{+infty}{x^{3n+1}}/{(3n+1)n!}

Integration via the Incomplete Gamma Function
First, substitute t=-x^3:
int e^{x^3} dx = - 1/3 int e^{-t} t^{-2/3} dt
The function e^{x^3} is continuous. This means that its primitive functions are F:\mathbb{R} to \mathbb{R} such that
F(y) = c + int_0^y e^{x^3}dx=c- 1/3 int_0^{-y^3} e^{-t} t^{-2/3} dt
and this is well defined because the function f(t)=e^{-t}t^{-2/3} is such that for t to 0 it holds f(t) ~~ t^{-2/3}, so that the improper integral int_0^s f(t) dt is finite (I call s=-y^3).
So you have that
int e^{x^3} dx=c- 1/3 int_0^s f(t)dt

Remark that t^{-2/3}< 1 hArr t>1. This means that for t to +infty we get that f(t)=e^{-t} * t^{-2/3} < e^{-t} * 1 = e^{-t}, so that |int_1^{+ infty} f(t)dt|<|int_1^{+infty} e^{-t}dt|=e. So following improper integral of f(t) is finite:
c'=int_0^{+infty}f(t)dt=int_0^{+infty} e^{-t}t^{1/3 -1}dt=Gamma(1/3).

We can write:
int e^{x^3} dx=c-1/3 (int_0^{+infty} f(t)dt -int_s^{+infty} f(t)dt)
that is
int e^{x^3} dx=c-1/3 c' +1/3 int_s^{+infty} e^{-t}t^{1/3 -1}dt.
In the end we get
int e^{x^3} dx=C+1/3 Gamma(1/3,t) =C+1/3 Gamma(1/3,-x^3)