Integrating $\int\limits_{0}^{\infty} e^{-x^3}dx $ I have seen ways to evaluate this integral using the upper and lower incomplete gamma functions. I want to know if there are ways to calculate this integral using change of variables or tricks similar to the evaluation of 
$$ \int_{0}^{\infty} e^{-x^2}dx $$ using double integrals.
Thanks in advance
 A: Substituting $y=x^3$, we get
$$\int _0^\infty e^{-x^3}dx=\frac13\int_0^\infty e^{-y}y^{\frac13-1}dy=\frac13\Gamma(\frac13)=\Gamma(\frac43),$$
where $\Gamma(s)$ is the gamma function.
A: There isn't any better form for this integral than
$$\int_{0}^{\infty} e^{-x^3}\ dx = \Gamma\left(\frac{4}{3}\right)$$
You can find this rather simply by substitution, giving exactly the form of the gamma function. As far as it is known, I believe, there is no simpler way to write the values of the gamma function at third-integer arguments like there is for half-integers, giving the nice forms involving the square root of $\pi$ that you are thinking of.
More generally, we have the identity
$$\int_{0}^{\infty} e^{-x^\alpha}\ dx = \Gamma\left(\frac{\alpha+1}{\alpha}\right)$$
in the same way, for all real $\alpha > 0$.
A: If you apply the change of variables $x=\sqrt[3]{t}$ then the integral is equal to $$\frac{1}{3}\Gamma(\frac{1}{3})$$
A: As shown in the other answers, the integral can be expressed in terms of $\Gamma\left(\frac13\right)$. As no closed-form expression of this constant is known, this is a sure sign that you can't find an alternative resolution method.
