Gaussian integral in 3rd dimensions I have been wondering about computing $(1/3)!$ and using the Gamma function. After substituting for $x=t^{\frac{1}{3}}$, I got $\int_{0}^{\infty}e^{-x^{3}}dx$. May I know if there is any way to proceed, as using spherical coordinates just makes it much messier and $x^{2}+y^{2}+z^{2}=r^{2}$ (3D pythogorean theorem) does not help unlike the original gaussian integral.
Or is there no actual method to define (1/3)!, or is (1/3)! not a number that can be represented by normal mathematical functions?
Would complex analysis help?
 A: 
Is there no actual method to define $\frac13!$ ?

The Gamma function is the definition of non-integer factorials. Explicitly
$$
\frac13! : = \int_{0}^{\infty} x^{\frac13}e^{-x}\, \mathrm{d}x \tag{1}
$$
is the definition of $\frac13!$.
Your use of the word "actual" leads me to believe that you're not comfortable with numbers being defined by integrals, but this is very common. In fact, $\ln(2)$ which is one of the most famous constants in math, has an integral definition given by $\ln(2) := \int_1^2 \frac{1}{x}\, \mathrm{d}x$. So equation $(1)$ is a perfectly valid definition of $\frac13!$.

Is $\frac13!$ not a number that can be represented by normal mathematical functions?

Since $\frac{1}{3}! = \frac{1}{3}\Gamma\left(\frac{1}{3}\right)$, we'll just focus on examining possible representations for $\Gamma\left(\frac{1}{3}\right)$.
I. J. Zucker showed here (Table 1) that
$$
K(\lambda^*(3)) = K\left( \frac{\sqrt{3}-1}{2\sqrt{2}}\right) = \frac{3^{\frac14}}{\pi 2^\frac73} \Gamma\left(\frac13\right)^3 \tag{2}
$$
where $K(k)$ is the elliptic integral of the first kind and $\lambda^*(\cdot)$ is the modular lambda-star function. This means that $\Gamma\left(\frac{1}{3}\right)$, and in turn $\frac13!$, can be expressed in terms of the function $K(k)$ using equation $(2)$ as follows:
$$
\frac13! = \frac{2^{\frac{7}{9}}}{3^{\frac{13}{12}}}\sqrt[3]{\pi K\left( \frac{\sqrt{3}-1}{2\sqrt{2}}\right)} \tag{3}
$$
Also of note is that in this paper Yu. V. Nesterenko shows (in Corollary 5) that $\Gamma\left(\frac13\right)$ is algebraically independent from $\pi$ and $e^{\pi \sqrt{3}}$. This means that it's impossible to get a nice representation of $\frac13!$ in terms of $\pi$ like you can do with the half-integer values of the Gamma function $\Gamma\left( \frac{n}{2}\right)=\sqrt{\pi}\frac{(n-2)!!}{2^{\frac{n-1}{2}}}$.
In short, the simplest way to express $\frac13!$ appears to just be writing $\frac13!$.
