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How does one prove the following identity?

$$\sqrt{(-1)^n\frac{\Gamma(n+1/2)}{\Gamma(1/2-n)}} = \frac{(2n-1)!!}{2^n}$$

I attempted to prove this using the definition of the double factorial, however I couldn't continue and feel like there is a better method. I would appreciate it if somebody could show me a proof. Thank you

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  • $\begingroup$ I think you may use the functional equation $\Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin\pi s}$. $\endgroup$
    – Riemann
    Nov 26, 2022 at 13:27
  • $\begingroup$ @Riemann I ended up getting $$\Gamma(n+1/2)^2 = (-1)^n\frac{\Gamma(n+1/2)}{\Gamma(1/2-n)}\pi$$ What next? $\endgroup$
    – user1107557
    Nov 26, 2022 at 13:34
  • $\begingroup$ Then you may start from the value of $\Gamma(1/2)$ with $\Gamma(s+1)=s\Gamma(s)$ to derive the formula $$\frac{\Gamma(n+1/2)}{\sqrt{\pi}} = \frac{(2n-1)!!}{2^n}$$ or I think simply induction might work as well. $\endgroup$
    – Riemann
    Nov 26, 2022 at 13:46
  • $\begingroup$ @Riemann That is an interesting identity I have never seen before. Could you please link a proof, thanks! $\endgroup$
    – user1107557
    Nov 26, 2022 at 13:51
  • $\begingroup$ It is actually quite elementary, and standard proof of $\Gamma(n+1)=n!$ rely on this equation. The proof is also one line: you will see this if you to integration by parts with the definition $$\Gamma(s) = \int_0^\infty t^{s-1}e^{-t}dt.$$ $\endgroup$
    – Riemann
    Nov 26, 2022 at 13:57

1 Answer 1

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We use the identity \begin{align*} \Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin (\pi z)}\qquad\qquad z\notin \mathbb{Z} \end{align*} evaluated at $z=n-\frac{1}{2}$ and obtain \begin{align*} \Gamma\left(1-\left(n+1/2\right)\right)\Gamma\left(n+1/2\right)&=\frac{\pi}{\sin\left(\pi\,\frac{2n+1}{2}\right)}\\ \color{blue}{\Gamma(1/2-n)\Gamma(n+1/2)}&\color{blue}{=(-1)^n\,\pi}\tag{1} \end{align*}

We obtain with (1) and application of the identities \begin{align*} \Gamma(z+1)&=z\Gamma(z)\qquad\qquad\qquad z\in\mathbb{C}\setminus\{0,-1,-2,\ldots\}\tag{2}\\ \Gamma(1/2)&=\sqrt{\pi}\tag{3} \end{align*}

\begin{align*} \color{blue}{\sqrt{(-1)\frac{\Gamma(n+1/2)}{\Gamma(1/2-n)}}} &=\frac{1}{\sqrt{\pi}}\Gamma(n+1/2)\tag{$\to$ (1)}\\ &=\frac{1}{\sqrt{\pi}}\left(n-\frac{1}{2}\right)\Gamma(n-1/2)\tag{$\to$ (2)}\\ &=\frac{1}{\sqrt{\pi}}\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\Gamma(n-3/2)\tag{$\to$ (2)}\\ &=\cdots\\ &=\frac{1}{\sqrt{\pi}}\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\cdots\left(\frac{1}{2}\right)\Gamma(1/2)\tag{$\to$ (2)}\\ &=\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\cdots\left(\frac{1}{2}\right)\tag{$\to$ (3)}\\ &=\frac{1}{2^n}(2n-1)(2n-3)\cdots 1\\ &\,\,\color{blue}{=\frac{1}{2^n}(2n-1)!!} \end{align*} and the claim follows.

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  • $\begingroup$ This answer is incredible! Thank you $\endgroup$
    – user1107557
    Nov 27, 2022 at 9:28
  • $\begingroup$ @MakoJ: You're welcome. :-) $\endgroup$ Nov 27, 2022 at 9:51

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