Integral with $e$ , $ \int e^{x^3}\;dx$

Question

$ \int e^{x^3}\;dx$ so, i'm searching for answer this question, i think that this not is too easily, but i think this integral not exist solution undefined, this integral would be easy if had the derivative out but i'snt happen

Answer 1

There is no elementary function, say $\mathrm{f}$, with the property that $$\frac{\mathrm{df}}{\mathrm{d}x}=\mathrm{e}^{x^3}$$

Answer 2

Let $$I=\int\exp(x^3)\,\mathrm dx$$ Let $t=-x^3$ $$I=-\frac13\int t^{-2/3}\exp(-t)\,\mathrm dt$$

The lower incomplete gamma function is defined as $$\gamma(a,x)=\int_0^x t^{a-1}\exp(-t)\,\mathrm dt$$ Since when $a=1/3$ this integral is just the required integral with a fixed constant, we can assign. $$I=-\frac13\gamma\left(\frac13, -x^3\right)+\color{gray}{\text{constant}}$$ However the lower incomplete gamma function is not so widely used, but since the incomplete gamma function is defined as $$\Gamma(a,x)=\int_x^\infty t^{a-1}\exp(-t)\,\mathrm dt$$ Therefore it's true by definition that $$\gamma(a,x)+\Gamma(a,x)=\Gamma(a)$$ $$\gamma(a,x)=\Gamma(a)-\Gamma(a,x)$$ We can therefore rewrite $$I=-\frac13\left(\Gamma\left(\frac13\right)-\Gamma\left(\frac13, -x^3\right)\right)+\color{gray}{\text{constant}}$$ Which reduces to $$I=\frac13\Gamma\left(\frac13, -x^3\right)-\frac13\Gamma\left(\frac13\right)+\color{gray}{\text{constant}}$$ However since $-\frac13\Gamma(1/3)$ is a constant, this reduces to the following where the constant is another value $$I=\frac13\Gamma\left(\frac13, -x^3\right)+\color{gray}{\text{constant}}$$

See this page for more information on the functions $\gamma(a,x)$ and $\Gamma(a,x)$

Answer 3

the function of $\int e^{x^2} dx$ is undefined. most likely the function of $e^{x^3}$ is also undefined. there exists a error function ($erf$) which approximates the integral you can read more over that here http://en.wikipedia.org/wiki/Error_function