How to evaluate the integral $\int e^{x^3}dx $ How to evaluate the integral
$$\int e^{x^3}dx \quad ?$$
I've tried to set $t=x^3$, but it seems to be a blind alley; I don't know what to do with $\int\frac{e^t}{3\sqrt[3]{t^2}}dt$.
 A: The integral cannot be evaluated. We have to use power series of exponent and then integral term by term. $$e^{x}=\sum_{n \geq 0}{\frac{x^n}{n!}}$$
A: The antiderivative of $e^{x^3}$ cannot be expressed in terms of elementary functions. We can, however, express it using power series. Since 
$$ e^x = \sum_{n \geq 0} \frac{x^n}{n!}, $$
$$ e^{x^3} = \sum_{n \geq 0} \frac{(x^3)^n}{n!} = \sum_{n \geq 0} \frac{x^{3n}}{n!}.$$
You can integrate term by term to find a series representation of the antiderivative (which converges on the entire complex plane, since $e^{x^3}$ is an entire function). 
A: $$\int e^{x^3}dx=\int \sum_{n=0}^{\infty }\frac{x^{3n}}{n!}dx$$
$$\int \sum_{n=0}^{\infty }\frac{x^{3n}}{n!}dx=\sum_{n=0}^{\infty }\frac{x^{3n+1}}{(3n+1)(n!)}+c$$
$$\frac{1}{3n+1}=\frac{(\frac{1}{3})^{(n)}}{(\frac{4}{3})^{(n)}}$$
$$\sum_{n=0}^{\infty }\frac{x^{3n+1}}{(3n+1)(n!)}+c=x\sum_{n=0}^{\infty }\frac{(\frac{1}{3})^{(n)}(x^3)^n}{(\frac{4}{3})^{(n)}(n!)}+c$$
$$x\sum_{n=0}^{\infty }\frac{(\frac{1}{3})^{(n)}(x^3)^n}{(\frac{4}{3})^{(n)}(n!)}+c=\ x\ 1F1(\frac{1}{3};\frac{4}{3};x^3)+c$$
so $$\int e^{x^3}dx=\ x\ {}_1F_1(\frac{1}{3};\frac{4}{3};x^3)+c$$
where ${}_1F_1$ is Hypergeometric Function of the First Kind
A: Interestingly, the definite integral
$$
\int_0^\infty e^{-x^n}dx
$$
can be evaluated for any $n>0$, and is equal to $\Gamma((n+1)/n)$.
A: another try you can solve it with Gamma function
$$\int e^{x^3}dx=\frac{-1}{3}\int e^{-t}t^{\frac{1}{3}-1}dt$$
$$\frac{-1}{3}\int e^{-t}t^{\frac{1}{3}-1}dt=\frac{-1}{3}\int_{0}^{t}e^{-t}t^{\frac{1}{3}-1}dt+c$$
$$\frac{-1}{3}\int_{0}^{t}e^{-t}t^{\frac{1}{3}-1}dt+c=\frac{1}{3}(\int_{0}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt-\int_{t}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt)+c$$
$$\frac{-1}{3}(\int_{0}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt-\int_{t}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt)+c=\frac{1}{3}\Gamma (\frac{1}{3},t)+d$$
$$\frac{1}{3}\Gamma (\frac{1}{3},t)+d=\frac{1}{3}\Gamma (\frac{1}{3},-x^3)+d$$
so
$$\int e^{x^3}dx=\frac{1}{3}\Gamma (\frac{1}{3},-x^3)+d$$
where d and c are constant
A: I'd like to give step by step solution to @Eckhard 's  answer .
Substitude $y=x^n$ :
$\int_0^\infty e^{-x^n}dx = \frac1n\int_0^\infty e^{-y}y^{1/n-1}dy = \frac1n \Gamma(\frac 1n) = \Gamma(\frac 1n+1) = {\frac 1n}! $
$x^n=y \Rightarrow nx^{n-1}dx=dy \Rightarrow dx=\frac1nx^{1-n}dy \Rightarrow dx=\frac 1n y^{\frac 1n -1}dy$
The last integral is taken due to the main definition of gamma function . 
Where :
$\Gamma(z) = \int_o^\infty e^{-x}x^{z-1}dx$
-Hope it was helpful.
