Is there a close form for $\int_{0}^{\infty}e^{-x^3}dx$?

Question

Is there a close form for $\int_{0}^{\infty}e^{-x^3}dx$ ?

I tried numerical method and find that $$\int_{0}^{\infty}e^{-x^3}dx=0.892979$$ but, my question is that "does it have a closed form solution ?" thanks in advanced .

Answer 1

Let $u=x^3$ to get

$$I=\frac13\int_0^\infty u^{-2/3}e^{-u}~\mathrm du=\frac13\Gamma(1/3)=\Gamma(4/3)$$

where $\Gamma(x)$ is the Gamma function or an Euler integral of the first kind.

As given in this PDF we find that

$$\Gamma(4/3)=\frac{\pi^{1/3}2^{7/9}}{3^{13/12}}\mathrm K\left(\frac{\sqrt3-1}{2^{3/2}}\right)^{1/3}$$

Where $\mathrm K(x)$ is a complete elliptic integral of the first kind.