Closed form expression of $\Gamma(\frac{5}{3})$ So I made up this integral here: $$\int_0^\infty x^4e^{-x^3}dx$$And by substitution we get $$\int_0^\infty x^4e^{-x^3}dx=\frac{1}{3}\int_0^\infty u^{\frac{2}{3}}e^{-u}du=\frac{\Gamma(\frac{5}{3})}{3}$$Where $\Gamma(x)$ is the analytic continuation of $(x-1)!$. So is there any exact form for $\Gamma(\frac{5}{3})$? I tried using Euler's reflection formula and the Legendre multiplication formula but it led me no where.
 A: I doubt that an elementary closed form is possible (since $\Gamma(1/3)$ does not have an elementary expression), but here's I think the best you can do. Start by using Gauss' multiplication formula:
$$\prod_{i=0}^{k-1}\Gamma\left(z+\frac{i}{k}\right)=(2\pi)^{\frac{k-1}{2}}k^{\frac{1-2kz}{2}}\Gamma(kz)$$
with $z=1$ and $k=3$. One then gets that
$$\Gamma(5/3)=\frac{2\pi\cdot 3^{-5/2}\cdot 2!}{1!\cdot \Gamma(4/3)}=\frac{4\pi}{3\sqrt{3}\cdot \Gamma(1/3)}\approx 0.90274529295.$$
This may already be satisfactory, but if you want an expression which is more intuitive to understand, note that you can express $\Gamma(1/3)$ in terms of the arithmetic-geometric mean of $2$ and $\sqrt{2+\sqrt{3}}$ (see here):
$$\Gamma(1/3)=\frac{2^{7/9}\cdot \pi^{2/3}}{3^{1/12}\cdot \operatorname{AGM}\left(2,\sqrt{2+\sqrt{3}}\right)^{1/3}}.$$
As such, we obtain the result:
$$\Gamma(5/3)=\frac{2^{11/9}\pi^{1/3}}{3^{17/12}}\operatorname{AGM}\left(2,\sqrt{2+\sqrt{3}}\right)^{1/3}.$$
This expression is particularly nice because you could easily estimate the value of $\Gamma(5/3)$ if you don't have access to the $\Gamma$-function on a calculator. The AGM of two positive real numbers always satisfies $G\leq \operatorname{AGM}\leq A$, where $G$ is the geometric mean and $A$ is the arithmetic mean. In this case $G\approx 1.96563051084284$ and $A\approx 1.96592582628907$, which gives you a quite good estimate on the AGM.
