Best solutions to - "How to evaluate the integral ∫ex3dx"

abreviatsjw

Answered question

2022-01-07

How to evaluate the integral

\int e^{x^{3}} d x

Answer & Explanation

karton

Expert2022-01-11Added 613 answers

$\int e^{x^{3}} d x = \int \sum_{n = 0}^{\infty} \frac{x^{3 n}}{n!} d x \int \sum_{n = 0}^{\infty} \frac{x^{3 n}}{n!} d x = \sum_{n = 0}^{\infty} \frac{x^{3 n + 1}}{(3 n + 1) (n!)} + C \frac{1}{3 n + 1} = \frac{(\frac{1}{3})^{(n)}}{(\frac{4}{3})^{(n)}} \sum_{n = 0}^{\infty} \frac{x^{3 n + 1}}{(3 n + 1) (n!)} + c = x \sum_{n = 0}^{\infty} \frac{(\frac{1}{3})^{(n)} (x^{3})^{n}}{(\frac{4}{3})^{(n)} (n!)} + c x \sum_{n = 0}^{\infty} \frac{(\frac{1}{3})^{(n)} (x^{3})^{n}}{(\frac{4}{3})^{(n)} (n!)} + c = x 1 F 1 (\frac{1}{3}; \frac{4}{3}; x^{3}) + c s o \int e^{x^{3}} d x = x_{1} F_{1} (\frac{1}{3}; \frac{4}{3}; x^{3}) + C$

user_27qwe

Skilled2022-01-11Added 375 answers

another try you can solve it with Gamma function
$\int e^{x^{3}} d x = \frac{- 1}{3} \int e^{- t} t^{\frac{1}{3} - 1} d t \frac{- 1}{3} \int e^{- t} t^{\frac{1}{3} - 1} d t = \frac{- 1}{3} \int_{0}^{t} e^{- t} t^{\frac{1}{3}} d t + c \frac{- 1}{3} \int_{0}^{t} e^{- t} t^{\frac{1}{3} - 1} d t + c = \frac{1}{3} (\int_{0}^{\infty} e^{- t} t^{\frac{1}{3} - 1} d t - \int_{t}^{\infty} e^{- t} t^{\frac{1}{3} - 1} d t) + C \frac{- 1}{3} (\int_{0}^{\infty} e^{- t} t^{\frac{1}{3} - 1} d t - \int_{t}^{\infty} e^{- t} t^{\frac{1}{3} - 1} d t) + C = \frac{1}{3} Γ (\frac{1}{3}, t) + d \frac{1}{3} Γ (\frac{1}{3}, t) + d = \frac{1}{3} Γ (\frac{1}{3}, - x^{3}) + d s o \int e^{x^{3}} d x = \frac{1}{3} Γ (\frac{1}{3}, - x^{3}) + d$
where d and c are constant

nick1337

Expert2022-01-11Added 777 answers

The antiderivative of $e^{x^{3}}$ cannot be expressed in terms of elementary functions. We can, however, express it using power series. Since
$e^{x} = \sum_{n \geq 0} \frac{x^{n}}{n!}$
$e^{x^{3}} = \sum_{n \geq 0} \frac{(x^{3})^{n}}{n!} = \sum_{n \geq 0} \frac{x^{3 n}}{n!}$
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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