How to evaluate the integral $\int e^{x^3}dx $

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).

$$\int e^{x^3}dx=\int \sum_{n=0}^{\infty }\frac{x^{3n}}{n!}dx$$

$$\int \sum_{n=0}^{\infty }\frac{x^{3n}}{n!}dx=\sum_{n=0}^{\infty }\frac{x^{3n+1}}{(3n+1)(n!)}+c$$

$$\frac{1}{3n+1}=\frac{(\frac{1}{3})^{(n)}}{(\frac{4}{3})^{(n)}}$$

$$\sum_{n=0}^{\infty }\frac{x^{3n+1}}{(3n+1)(n!)}+c=x\sum_{n=0}^{\infty }\frac{(\frac{1}{3})^{(n)}(x^3)^n}{(\frac{4}{3})^{(n)}(n!)}+c$$

$$x\sum_{n=0}^{\infty }\frac{(\frac{1}{3})^{(n)}(x^3)^n}{(\frac{4}{3})^{(n)}(n!)}+c=\ x\ 1F1(\frac{1}{3};\frac{4}{3};x^3)+c$$

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

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