Intuition behind $(-\frac{1}{2})! = \sqrt{\pi}$ It can be shown that using the definition of the Gamma function as:
$$\Gamma(t) = \int_0^\infty x^{t-1} e^{-x} dx $$ that $$\Gamma(\tfrac{1}{2}) = \sqrt{\pi}$$ or slightly abusing notation, that $(-\frac{1}{2})! = \sqrt{\pi}$. Is there an intuitive explanation to this?
I want to make clear that I am not per se interested in a proof of this fact (most often these are clever technical manipulations) but in insight into this phenomenon.  
 A: 
Is there an intuitive explanation to this ?

Yes. There is an umbilical connection between $\bigg(\dfrac1n\bigg){\large!}$ or $\Gamma\bigg(\dfrac1n\bigg)$, and geometric shapes  of the form $X^n+Y^n=R^n\iff x^n+y^n=1\iff y=\sqrt[n]{1-x^n}$, whose area is  $\displaystyle\int_0^1\sqrt[n]{1-x^n}~dx$, which is nothing else than the beta function in disguise. Ultimately, it's  all related to Newton's binomial theorem. The latter expands the power of a sum into a  sum of powers, with the help of binomial coefficients, or beta functions, which are then  expressed in terms of factorials, or $\Gamma$ functions.
A: Consider the area of the surface of the $n-$Ball with radius $1$ . 
It is given by:

$$
A_{n}=2\frac{\pi^{n/2}}{\Gamma(n/2)}
$$

Our intuition tells us that for $n=1$ the surface "area" (or to be mathematically more precise the, Hausdorff measure as @Michael Galuza pointed out correctly)should be 2, because it consist of two points. To make this consistent with the above formula we have to demand that

$$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$

A: $\color{red}{\text{beta-gamma function relation}}$
Notice, we know from beta function that $$\int_{0}^{\pi/2}\sin^m(\theta) \cos^n(\theta)d\theta$$ $$=\frac{\Gamma\left(\frac{m+1}{2}\right)\Gamma\left(\frac{n+1}{2}\right)}{2\Gamma\left(\frac{m+n+2}{2}\right)}$$
Substituting $m=n=0$, we get 
 $$\int_{0}^{\pi/2}(\sin(\theta))^{0} (\cos^n(\theta))^{0}d\theta=\frac{\Gamma\left(\frac{0+1}{2}\right)\Gamma\left(\frac{0+1}{2}\right)}{2\Gamma\left(\frac{0+0+2}{2}\right)}$$
$$\int_{0}^{\pi/2}d\theta=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(1\right)}$$
$$[\theta]_{0}^{\pi/2}=\frac{\left(\Gamma\left(\frac{1}{2}\right)\right)^2}{2\times 1}$$
$$\frac{\pi}{2}=\frac{\left(\Gamma\left(\frac{1}{2}\right)\right)^2}{2}$$ $$\left(\Gamma\left(\frac{1}{2}\right)\right)^2=\pi$$ $$\Gamma\left(\frac{1}{2}\right)=\sqrt \pi$$
