 abreviatsjw

2022-01-07

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

$\int {e}^{{x}^{3}}dx=\int \sum _{n=0}^{\mathrm{\infty }}\frac{{x}^{3n}}{n!}dx\phantom{\rule{0ex}{0ex}}\int \sum _{n=0}^{\mathrm{\infty }}\frac{{x}^{3n}}{n!}dx=\sum _{n=0}^{\mathrm{\infty }}\frac{{x}^{3n+1}}{\left(3n+1\right)\left(n!\right)}+C\phantom{\rule{0ex}{0ex}}\frac{1}{3n+1}=\frac{\left(\frac{1}{3}{\right)}^{\left(n\right)}}{\left(\frac{4}{3}{\right)}^{\left(n\right)}}\phantom{\rule{0ex}{0ex}}\sum _{n=0}^{\mathrm{\infty }}\frac{{x}^{3n+1}}{\left(3n+1\right)\left(n!\right)}+c=x\sum _{n=0}^{\mathrm{\infty }}\frac{\left(\frac{1}{3}{\right)}^{\left(n\right)}\left({x}^{3}{\right)}^{n}}{\left(\frac{4}{3}{\right)}^{\left(n\right)}\left(n!\right)}+c\phantom{\rule{0ex}{0ex}}x\sum _{n=0}^{\mathrm{\infty }}\frac{\left(\frac{1}{3}{\right)}^{\left(n\right)}\left({x}^{3}{\right)}^{n}}{\left(\frac{4}{3}{\right)}^{\left(n\right)}\left(n!\right)}+c=x1F1\left(\frac{1}{3};\frac{4}{3};{x}^{3}\right)+c\phantom{\rule{0ex}{0ex}}so\phantom{\rule{0ex}{0ex}}\int {e}^{{x}^{3}}dx={x}_{1}{F}_{1}\left(\frac{1}{3};\frac{4}{3};{x}^{3}\right)+C$ user_27qwe

another try you can solve it with Gamma function
$\int {e}^{{x}^{3}}dx=\frac{-1}{3}\int {e}^{-t}{t}^{\frac{1}{3}-1}dt\phantom{\rule{0ex}{0ex}}\frac{-1}{3}\int {e}^{-t}{t}^{\frac{1}{3}-1}dt=\frac{-1}{3}{\int }_{0}^{t}{e}^{-t}{t}^{\frac{1}{3}}dt+c\phantom{\rule{0ex}{0ex}}\frac{-1}{3}{\int }_{0}^{t}{e}^{-t}{t}^{\frac{1}{3}-1}dt+c=\frac{1}{3}\left({\int }_{0}^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{1}{3}-1}dt-{\int }_{t}^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{1}{3}-1}dt\right)+C\phantom{\rule{0ex}{0ex}}\frac{-1}{3}\left({\int }_{0}^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{1}{3}-1}dt-{\int }_{t}^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{1}{3}-1}dt\right)+C=\frac{1}{3}\mathrm{\Gamma }\left(\frac{1}{3},t\right)+d\phantom{\rule{0ex}{0ex}}\frac{1}{3}\mathrm{\Gamma }\left(\frac{1}{3},t\right)+d=\frac{1}{3}\mathrm{\Gamma }\left(\frac{1}{3},-{x}^{3}\right)+d\phantom{\rule{0ex}{0ex}}so\phantom{\rule{0ex}{0ex}}\int {e}^{{x}^{3}}dx=\frac{1}{3}\mathrm{\Gamma }\left(\frac{1}{3},-{x}^{3}\right)+d$
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
${e}^{x}=\sum _{n\ge 0}\frac{{x}^{n}}{n!}$
${e}^{{x}^{3}}=\sum _{n\ge 0}\frac{\left({x}^{3}{\right)}^{n}}{n!}=\sum _{n\ge 0}\frac{{x}^{3n}}{n!}$
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).

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