One Step Forward from Gaussian Integral Now to solve the integral $ \int_0^\infty e^{-x^2} \, dx $ has become a simple task for us. But how can we solve this integral:
$$\int_0^\infty e^{-x^3} \, dx $$
 A: Note: Not an Answer but a Generalization
Consider the integral, $\displaystyle I = \int_0^\infty e^{-x^A} \text{ d}x $. Substitute $\displaystyle x = u^{\frac{1}{A}} \Rightarrow \text{ d}x = \frac{1}{A} u^{\frac{1-A}{A}} \text{ d}u $. 
$$ \therefore I = \int_0^\infty e^{-u} \frac{1}{A} u^{\frac{1-A}{A}} \text{ d}u$$
$\displaystyle I = \frac{1}{A} \int_0^\infty u^{\frac{1}{A} - 1} e^{-u} \text{ d}u $
$\displaystyle \Rightarrow I = \frac{1}{A} \Gamma \left( \frac{1}{A} \right) = \Gamma \left( \frac{1 + A}{A} \right) $
(Using the fact that $ t \Gamma \left(t \right) = \Gamma \left(t + 1 \right) $ ):
$$ \therefore \int_0^\infty e^{-x^A} \text{ d}x = \Gamma \left( \frac{1 + A}{A} \right) $$
A: $$\int_{0}^{+\infty}e^{-x^n}\,dx = \frac{1}{n}\int_{0}^{+\infty}z^{\frac{1}{n}-1}e^{-z}\,dz = \frac{1}{n}\,\Gamma\left(\frac{1}{n}\right) = \Gamma\left(1+\frac{1}{n}\right)$$
through a change of variable and the definition of the $\Gamma$ function.
A: Notice, the following Laplace transform $$\int_{0}^{\infty}e^{-st}t^ndt=\frac{\Gamma(n+1)}{s^{n+1}}$$
Now, we have $$\int_{0}^{\infty}e^{-x^n}dx$$ Let $x^n=t\implies nx^{n-1}dx=dt\iff dx=\frac{dt}{nt^{(n-1)/n}}$ 
$$\int_{0}^{\infty}e^{-t}\frac{dt}{nt^{(n-1)/n}}$$
$$=\frac{1}{n}\int_{0}^{\infty}e^{-t}t^{(1-n)/n}dt$$
$$=\frac{1}{n}L[t^{(1-n)/n}]_{s=1}$$
$$=\frac{1}{n}\left[\frac{\Gamma\left(\frac{1-n}{n}+1\right)}{s^{\frac{1-n}{n}+1}}\right]_{s=1}$$
$$=\frac{1}{n}\left[\frac{\Gamma\left(\frac{1}{n}\right)}{1^{\frac{1}{n}}}\right]$$ $$=\frac{1}{n}\Gamma\left(\frac{1}{n}\right)$$
Hence, as per given problem  $$\int_{0}^{\infty}e^{-x^3}dx$$
Here, $n=3$ hence we get 
$$\int_{0}^{\infty}e^{-x^3}dx=\frac{1}{3}\Gamma\left(\frac{1}{3}\right)$$
