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

Question

William Abdalla · Accepted 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^x# 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)#

Luke Alvarez · Answer

undefined The integral of $ e^{x^3} $ with respect to $ x $ does not have a closed-form expression in terms of elementary functions. Therefore, it cannot be expressed using standard functions like polynomials, exponentials, logarithms, trigonometric functions, or their inverses. However, it is possible to express the integral using special functions such as the exponential integral function or the error function.