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

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 $m a t h \boldsymbol{R}$, so $\forall x \in \setminus m a t h \boldsymbol{R}$ 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 $t = - {x}^{3}$:
$\int {e}^{{x}^{3}} \mathrm{dx} = - \frac{1}{3} \int {e}^{- t} {t}^{- \frac{2}{3}} \mathrm{dt}$
The function ${e}^{{x}^{3}}$ is continuous. This means that its primitive functions are $F : \setminus m a t h \boldsymbol{R} \to \setminus m a t h \boldsymbol{R}$ such that
$F \left(y\right) = c + {\int}_{0}^{y} {e}^{{x}^{3}} \mathrm{dx} = c - \frac{1}{3} {\int}_{0}^{- {y}^{3}} {e}^{- t} {t}^{- \frac{2}{3}} \mathrm{dt}$
and this is well defined because the function $f \left(t\right) = {e}^{- t} {t}^{- \frac{2}{3}}$ is such that for $t \to 0$ it holds $f \left(t\right) \approx {t}^{- \frac{2}{3}}$, so that the improper integral ${\int}_{0}^{s} f \left(t\right) \mathrm{dt}$ is finite (I call $s = - {y}^{3}$).
So you have that
$\int {e}^{{x}^{3}} \mathrm{dx} = c - \frac{1}{3} {\int}_{0}^{s} f \left(t\right) \mathrm{dt}$

Remark that ${t}^{- \frac{2}{3}} < 1 \Leftrightarrow t > 1$. This means that for $t \to + \infty$ we get that $f \left(t\right) = {e}^{- t} \cdot {t}^{- \frac{2}{3}} < {e}^{- t} \cdot 1 = {e}^{- t}$, so that $| {\int}_{1}^{+ \infty} f \left(t\right) \mathrm{dt} | < | {\int}_{1}^{+ \infty} {e}^{- t} \mathrm{dt} | = e$. So following improper integral of $f \left(t\right)$ is finite:
$c ' = {\int}_{0}^{+ \infty} f \left(t\right) \mathrm{dt} = {\int}_{0}^{+ \infty} {e}^{- t} {t}^{\frac{1}{3} - 1} \mathrm{dt} = \Gamma \left(\frac{1}{3}\right)$.

We can write:
$\int {e}^{{x}^{3}} \mathrm{dx} = c - \frac{1}{3} \left({\int}_{0}^{+ \infty} f \left(t\right) \mathrm{dt} - {\int}_{s}^{+ \infty} f \left(t\right) \mathrm{dt}\right)$
that is
$\int {e}^{{x}^{3}} \mathrm{dx} = c - \frac{1}{3} c ' + \frac{1}{3} {\int}_{s}^{+ \infty} {e}^{- t} {t}^{\frac{1}{3} - 1} \mathrm{dt}$.
In the end we get
$\int {e}^{{x}^{3}} \mathrm{dx} = C + \frac{1}{3} \Gamma \left(\frac{1}{3} , t\right) = C + \frac{1}{3} \Gamma \left(\frac{1}{3} , - {x}^{3}\right)$