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

abreviatsjw 2022-01-07
How to evaluate the integral
$$\displaystyle\int{e}^{{{x}^{{3}}}}{\left.{d}{x}\right.}$$

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karton

$$\int e^{x^3}dx=\int\sum_{n=0}^\infty\frac{x^{3n}}{n!}dx \\\int\sum_{n=0}^\infty\frac{x^{3n}}{n!}dx=\sum_{n=0}^\infty\frac{x^{3n+1}}{(3n+1)(n!)}+C \\\frac{1}{3n+1}=\frac{(\frac{1}{3})^{(n)}}{(\frac{4}{3})^{(n)}} \\\sum_{n=0}^\infty\frac{x^{3n+1}}{(3n+1)(n!)}+c=x\sum_{n=0}^\infty\frac{(\frac13)^{(n)}(x^3)^n}{(\frac43)^{(n)}(n!)}+c \\x\sum_{n=0}^\infty\frac{(\frac13)^{(n)}(x^3)^n}{(\frac43)^{(n)}(n!)}+c=x1F1(\frac13;\frac43;x^3)+c \\so \\\int e^{x^3}dx=x_1F_1(\frac13;\frac43;x^3)+C$$

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user_27qwe

another try you can solve it with Gamma function
$$\int e^{x^3}dx=\frac{-1}{3}\int e^{-t}t^{\frac13-1}dt \\\frac{-1}{3}\int e^{-t}t^{\frac13-1}dt=\frac{-1}{3}\int_0^t e^{-t}t^{\frac13}dt+c \\\frac{-1}{3}\int_0^t e^{-t}t^{\frac13-1}dt+c=\frac13(\int_0^{\infty}e^{-t}t^{\frac13-1}dt-\int_t^\infty e^{-t}t^{\frac13-1}dt)+C \\\frac{-1}{3}(\int_0^\infty e^{-t}t^{\frac13-1}dt-\int_t^\infty e^{-t}t^{\frac13-1}dt)+C=\frac13\Gamma(\frac13,t)+d \\\frac13\Gamma(\frac13,t)+d=\frac13\Gamma(\frac13,-x^3)+d \\so \\\int e^{x^3}dx=\frac13\Gamma(\frac13,-x^3)+d$$
where d and c are constant

nick1337

The antiderivative of $$e^{x^3}$$ cannot be expressed in terms of elementary functions. We can, however, express it using power series. Since
$$e^x=\sum_{n\geq0}\frac{x^n}{n!}$$
$$e^{x^3}=\sum_{n\geq0}\frac{(x^3)^n}{n!}=\sum_{n\geq0}\frac{x^{3n}}{n!}$$
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).