Duplication formula for gamma function

Question

Using the Weierstrass definition for $\Gamma(x)$ and $\Gamma\Big(x + \frac12\Big)$, how can I prove the duplication formula? This is problem $10.7.3$ in the book Irresistible Integrals, by Boros and Moll.

Any help is highly appreciated.

Answer 1

The duplication formula can be written as

$$\frac{\Gamma(x)\Gamma(x+\frac1{2})}{\Gamma(2x)}= \frac{\Gamma(\frac1{2})}{2^{2x-1}}= \frac{\sqrt{\pi}}{2^{2x-1}}.$$

We want to derive this formula using the Weierstrass definition for the gamma function,

$$\frac1{\Gamma(x)}=xe^{\gamma x}\prod_{k=1}^{\infty}\left(1+\frac{x}{k}\right)e^{-x/k}.$$

We have

$$\frac{\Gamma(x)\Gamma(x+\frac1{2})}{\Gamma(2x)}=\frac{2xe^{2\gamma x}}{xe^{\gamma x}(x+\frac1{2})e^{\gamma x}e^{\gamma/2}}\frac{\prod_{k=1}^{\infty}\left(1+\frac{2x}{k}\right)e^{-2x/k}}{\prod_{k=1}^{\infty}\left(1+\frac{x}{k}\right)e^{-x/k}\prod_{k=1}^{\infty}\left(1+\frac{x}{k}+\frac{1}{2k}\right)e^{-x/k}e^{-1/2k}}\\ =\frac{1}{e^{\gamma/2}}\lim_{n \rightarrow \infty}\frac{2x\prod_{k=1}^{2n}\left(1+\frac{2x}{k}\right)}{x(x+\frac1{2})\prod_{k=1}^{n}\left(1+\frac{x}{k}\right)\prod_{k=1}^{n}\left(1+\frac{x}{k}+\frac{1}{2k}\right)}\frac{\prod_{k=1}^{2n}e^{-2x/k}}{(\prod_{k=1}^{n}e^{-x/k})^2\prod_{k=1}^{n}e^{-1/2k}}\\ =\frac{1}{e^{\gamma/2}}\lim_{n \rightarrow \infty}P_n(x)Q_n(x).$$

First simplify $P_n(x)$ as follows:

$$P_n(x)=\frac{2x\prod_{k=1}^{2n}\left(1+\frac{2x}{k}\right)}{x(x+\frac1{2})\prod_{k=1}^{n}\left(1+\frac{x}{k}\right)\prod_{k=1}^{n}\left(1+\frac{x}{k}+\frac{1}{2k}\right)}\\=\frac{(n!)^2}{(2n)!\left(x+n+\frac1{2}\right)}\frac{\prod_{k=0}^{n}\left(2x+2k\right)\prod_{k=0}^{n-1}\left(2x+2k+1\right)}{\prod_{k=0}^{n}\left(x+k\right)\prod_{k=0}^{n-1}\left(x+k+\frac1{2}\right)}\\=\frac{(n!)^22^{2n+1}}{(2n)!\left(x+n+\frac1{2}\right)}$$

Next consider $Q_n(x)$:

$$Q_n(x)=\frac{\prod_{k=1}^{2n}e^{-2x/k}}{(\prod_{k=1}^{n}e^{-x/k})^2\prod_{k=1}^{n}e^{-1/2k}}\\=\frac{n^{1/2}}{2^{2x}}\frac{(2n)^{2x}\prod_{k=1}^{2n}e^{-2x/k}}{(n^x\prod_{k=1}^{n}e^{-x/k})^2n^{1/2}\prod_{k=1}^{n}e^{-1/2k}}$$

Reassembling we get

$$\frac{\Gamma(x)\Gamma(x+\frac1{2})}{\Gamma(2x)}=\frac{1}{e^{\gamma/2}}\lim_{n \rightarrow \infty}\frac{(n!)^22^{2n+1}}{(2n)!\left(x+n+\frac1{2}\right)}\frac{n^{1/2}}{2^{2x}}\frac{(2n)^{2x}\prod_{k=1}^{2n}e^{-2x/k}}{(n^x\prod_{k=1}^{n}e^{-x/k})^2n^{1/2}\prod_{k=1}^{n}e^{-1/2k}}\\=\frac{1}{2^{2x-1}}\lim_{n \rightarrow \infty}\frac{n}{\left(x+n+\frac1{2}\right)}\frac{(n!)^22^{2n}}{(2n)!n^{1/2}}\frac{(2n)^{2x}\prod_{k=1}^{2n}e^{-2x/k}}{e^{\gamma/2}(n^x\prod_{k=1}^{n}e^{-x/k})^2n^{1/2}\prod_{k=1}^{n}e^{-1/2k}}.$$

We can evaluate the limit in three parts.

First,

$$\lim_{n \rightarrow \infty}\frac{n}{\left(x+n+\frac1{2}\right)}=1.$$

Second, using a well-known identity for the Euler-Mascheroni constant,

$$\lim_{n \rightarrow \infty}\frac{(2n)^{2x}\prod_{k=1}^{2n}e^{-2x/k}}{e^{\gamma/2}(n^x\prod_{k=1}^{n}e^{-x/k})^2n^{1/2}\prod_{k=1}^{n}e^{-1/2k}}=\frac{e^{-2\gamma x}}{(e^{-\gamma x})^2e^{-\gamma /2}e^{\gamma /2}}=1.$$

Third using Stirlings's asymptotic formula $n! \sim \sqrt{2\pi}n^{n+1/2}e^{-n},$

$$\lim_{n \rightarrow \infty}\frac{(n!)^22^{2n}}{(2n)!n^{1/2}}=\sqrt{\pi},$$

and finally we get

$$\frac{\Gamma(x)\Gamma(x+\frac1{2})}{\Gamma(2x)}=\frac{\sqrt{\pi}}{2^{2x-1}}.$$

Answer 2

For people looking for an easier way to prove this without using the Weierstrass definition, an alternative is to use Bohr-Mollerup's theorem to the function $F:[0,\infty)\rightarrow [0,\infty)$ $$F(x)=\frac{1}{\sqrt{\pi}}2^{x-1}\Gamma\left(\frac{x}{2}\right)\Gamma\left(\frac{x+1}{2}\right)$$ Noting that this function is log-convex and satisfies $F(1)=1$ and $F(x+1)=xF(x)$ shows that $F(x)=\Gamma(x)$, therefore as analytic functions they must coincide on the complex plane outside of the poles of $\Gamma$, in particular: $\Gamma(2z)=F(2z)$ as desired.