<s>[INST] How to prove this identity: $4\cdot(\frac12)!=\pi$ 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.
 [/INST] 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}

</s>