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
Answered 2022-01-11 Author has 8658 answers

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

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user_27qwe
Answered 2022-01-11 Author has 9557 answers

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

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nick1337
Answered 2022-01-11 Author has 9671 answers

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).

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