Calculus Proving Question How do you prove that
$$\int_0^{\infty} \exp\left(-x^3\right)\, dx = \Gamma\left(\frac43\right)$$
I tried to integrate it and I got: 
$$\int_0^\infty \frac{\exp\left(-x^3\right)}{-3x^2}\, dx$$
I don't see how this turns out to be $0.89$ or $\dfrac13!$
Need advice
 A: If you change variable $x=\sqrt[3]{t}$ and your integrand becomes $$\frac{e^{-t}}{3 t^{2/3}}$$ where you probably recognize the integrand in the incomplete $\Gamma$ function. So, the antiderivative is $$-\frac{\Gamma \left(\frac{1}{3},t\right)}{3}$$ and the value of the integral is $\Gamma \left(\frac{4}{3}\right)$, the approximate value of which being $0.892979511569249$.
A: Substituting $t=x^{1/3}$ gives
$$
\begin{align}
\int_0^\infty e^{-t^3}\,\mathrm{d}t
&=\int_0^\infty e^{-x}\tfrac13x^{-2/3}\,\mathrm{d}x\\
&=\tfrac13\Gamma\left(\tfrac13\right)\\
&=\Gamma\left(\tfrac43\right)
\end{align}
$$
using the recursive relation $\Gamma(x+1)=x\Gamma(x)$.
To be precise, the factorial is defined for non-negative integers. In that case, $n!=\Gamma(n+1)$. However, if we extend this to non-integral values, we would get $\Gamma\left(\frac43\right)=\frac13!$
A: Generally speaking, $$n!=\int_0^\infty e^{-\sqrt[n]x}dx\qquad\iff\qquad\int_0^\infty e^{-x^n}dx=\left(\tfrac1n\right){\large!}$$ This can be shown through a simple change in variable, by letting $x=t^n$ in the former expression, thus arriving at the familiar integral $n\displaystyle\int_0^\infty t^{n-1}e^{-t}dt=n\Gamma(n)=\Gamma(n+1)=n!$
A: put $$ z=x^3 ,dx = \frac{dz}{3x^2}=\frac{dz}{3z^{2/3}}$$
then  $$\int_{0}^{\infty}e^{-x^3}dx= \int_{0}^{\infty}\frac{e^{-z}}{3z^{2/3}}dz =\frac{1}{3}\int_{0}^{\infty}z^{-2/3}e^{-z}= \frac{1}{3}Γ(1-\frac{2}{3})=\frac{1}{3}Γ(\frac{1}{3})=Γ(1+\frac{1}{3})$$$$=Γ(\frac{4}{3})$$ 
