How to prove this identity: $4\cdot(\frac12)!=\pi$

Question

Progress: I recently found this identity on an YouTube Video: $$4\cdot\left(\frac{1}{2}\right)!=\pi$$

But it didn't provide any kind of rigorous proof or simply a 'proof' in that regard. It only showed some Calculus identities to give some insight. Also importantly this is the only identity or equation I know that exactly equals to '$\pi$'. The part which is very interesting is $(\frac{1}{2})!$, the factorial of a fraction not an integer.

So if anyone can give a rigorous proof and even elementary proof or share more heuristics and insights, it would be useful. I'll really appreciate your efforts, thanks.

Answer 1

Note that gamma function is defined as$$\Gamma{(n)}=\int x^{n-1}e^{-x}dx$$ for $n>0$, $n\in\mathbb{R}$.

Gamma function is generalization of factorial function.

$$\Gamma (n)=(n-1)!.$$

Gamma function has property $$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$ and $$\Gamma(n+1)=n\Gamma(n).$$

So,

\begin{align} 4\left(\dfrac{1}{2}\right)!&=4\left(\dfrac{3}{2}-1\right)!\\ &=4\Gamma\left(\dfrac{3}{2}\right)\\ &=4\Gamma\left(1+\dfrac{1}{2}\right)\\ &=4\cdot \dfrac{1}{2} \Gamma\left(\dfrac{1}{2}\right)\\ &=2\sqrt{\pi}. \end{align}

I guess you want to prove $$4\left(\left(\dfrac{1}{2}\right)!\right)^2=\pi.$$

\begin{align} 4\left(\left(\dfrac{1}{2}\right)!\right)^2&=4\left(\left(\dfrac{3}{2}-1\right)!\right)^2\\ &=4\left(\Gamma\left(\dfrac{3}{2}\right)\right)^2\\ &=4\left(\Gamma\left(1+\dfrac{1}{2}\right)\right)^2\\ &=4\cdot \left(\dfrac{1}{2} \Gamma\left(\dfrac{1}{2}\right)\right)^2\\ &=4\cdot \left(\dfrac{1}{2}\sqrt{\pi}\right)^2\\ &=4\cdot \dfrac{1}{4}\pi\\ &=\pi. \end{align}

Answer 2

Not a answer, but important to know: As Qiaochu Yuan mentioned, the Gamma function gives your equality. But in the usual definition of the factorial it's undefined. That means the Gamma function is only one way to extend the factorial, even if it seems to be the most natural and useful way.