esiLimelawcuu

2023-02-22

What is the integral of ${e}^{{x}^{3}}$?

Hayley Rosario

This integral cannot be described in terms of simple functions.
You can select one integration method over another depending on what you need the integration for.
Integration via power series
Recall that ${e}^{x}$ is analytic on $R$, so $\forall x\in R$ the following equality holds
${e}^{x}=\sum _{n=0}^{+\infty }\frac{{x}^{n}}{n!}$
and this means that
${e}^{{x}^{3}}=\sum _{n=0}^{+\infty }\frac{{\left({x}^{3}\right)}^{n}}{n!}=\sum _{n=0}^{+\infty }\frac{{x}^{3n}}{n!}$
Now youcan integrate:
$\int {e}^{{x}^{3}}dx=\int \left(\sum _{n=0}^{+\infty }\frac{{x}^{3n}}{n!}\right)dx=c+\sum _{n=0}^{+\infty }\frac{{x}^{3n+1}}{\left(3n+1\right)n!}$
Integration via the Incomplete Gamma Function
First, substitute $t=-{x}^{3}$:
$\int {e}^{{x}^{3}}dx=-\frac{1}{3}\int {e}^{-t}{t}^{-\frac{2}{3}}dt$
The function ${e}^{{x}^{3}}$ is continuous. This means that its primitive functions are $F:R\to R$ such that
$F\left(y\right)=c+{\int }_{0}^{y}{e}^{{x}^{3}}dx=c-\frac{1}{3}{\int }_{0}^{-{y}^{3}}{e}^{-t}{t}^{-\frac{2}{3}}dt$
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 ${\int }_{0}^{s}f\left(t\right)dt$ is finite (I call $s=-{y}^{3}$).
So you have that
$\int {e}^{{x}^{3}}dx=c-\frac{1}{3}{\int }_{0}^{s}f\left(t\right)dt$
Remark that ${t}^{-\frac{2}{3}}<1⇔t>1$. 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 $|{\int }_{1}^{+\infty }f\left(t\right)dt|<|{\int }_{1}^{+\infty }{e}^{-t}dt|=e$. So following improper integral of $f\left(t\right)$ is finite:
$c\prime ={\int }_{0}^{+\infty }f\left(t\right)dt={\int }_{0}^{+\infty }{e}^{-t}{t}^{\frac{1}{3}-1}dt=\Gamma \left(\frac{1}{3}\right)$.
We can write:
$\int {e}^{{x}^{3}}dx=c-\frac{1}{3}\left({\int }_{0}^{+\infty }f\left(t\right)dt-{\int }_{s}^{+\infty }f\left(t\right)dt\right)$
that is
$\int {e}^{{x}^{3}}dx=c-\frac{1}{3}c\prime +\frac{1}{3}{\int }_{s}^{+\infty }{e}^{-t}{t}^{\frac{1}{3}-1}dt$.
In the end we get
$\int {e}^{{x}^{3}}dx=C+\frac{1}{3}\Gamma \left(\frac{1}{3},t\right)=C+\frac{1}{3}\Gamma \left(\frac{1}{3},-{x}^{3}\right)$

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