Prove that $2^{2z-1}\Gamma(z)\,\Gamma(z+\frac{1}{2})=\sqrt{\pi}\,\Gamma(2z)$ using Gauss's identity.

Question

I'm trying to derive the functional equation $2^{2z-1}\Gamma(z)\,\Gamma(z+\frac{1}{2})=\sqrt{\pi}\,\Gamma(2z)$ using Gauss's formula: $$\Gamma(z)=\lim_{n\to\infty}\frac{n!\,\,n^z}{z(z+1)\cdots(z+n)}\,,$$ but it's tricky business.

I've started by just looking at the product $$\frac{n!\,\,n^z}{z(z+1)\cdots(z+n)}\cdot\frac{n!\,\,n^{z+1/2}}{(z+1/2)(z+3/2)\cdots(z+(2n+1)/2)}$$ and the numerator very nicely becomes $(n!\, n^{2z})(n!\,n^{1/2})$, so we have the beginning stages of $\Gamma(2z)$ and $\Gamma(1/2)=\sqrt{\pi}$. But the denominator is killing me! Any advice?

Also, where in the world does this $2^{2z-1}$ come from?

Thanks!

Answer 1

With Pochhammer symbol notation, we write

$$\Gamma(z)\Gamma\left(z+\frac{1}{2}\right)=\lim_{n\to\infty}\frac{n!\,n^z}{(z)_{n+1}}\frac{n!\,n^{z+1/2}}{(z+1/2)_{n+1}}. \tag{$\circ$}$$

Notice that

$$(z)_{n+1}(z+1/2)_{n+1}=z(z+1)\cdots(z+n)~\times~(z+1/2)(z+3/2)\cdots\left(z+\frac{2n+1}{2}\right) $$

$$=\frac{2z+0}{2}\frac{2z+2}{2}\cdots\frac{2z+2n}{2}~\times~\frac{2z+1}{2}\frac{2z+3}{2}\cdots\frac{2z+(2n+1)}{2}=\frac{(2z)_{2(n+1)}}{2^{2(n+1)}}$$

Therefore for $(\circ)$ we rewrite

$$2^{2(n+1)}\frac{n!\,n^{2z}}{(2z)_{2(n+1)}}\frac{n!\,n^{1/2}}{1}. \tag{$\bullet$}$$

We have an extra power of $2$, the Pochhammer's $\color{Red}2(n+1)$ is not compatible with the numerator's simple $(\color{Red}1\cdot n)!$ and $(\color{Red}1\cdot n)^{2z}$, and we are missing a $(1/2)_{n+1}$ down below. Now we notice that

$$\left(\frac{1}{2}\right)_{n+1}=\frac{1+0}{2}\cdots\frac{1+2n}{2}=\frac{1}{2^{n+1}}\frac{(2n+1)!}{(2\cdot1)\cdots(2\cdot n)}=\frac{(2n+1)!}{2^{2n+1}n!}.$$

We choose to further rewrite $(\bullet)$ as

$$2^{2(n+1)}\frac{n!\,n^{2z}}{(2z)_{2(n+1)}}\frac{n!\,n^{1/2}}{\color{Purple}{(1/2)_{n+1}}}\color{DarkBlue}{\frac{(2n+1)!}{2^{2n+1}n!}}=\color{LimeGreen}2\frac{(2n+1)!\,n^{2z}}{(2z)_{2(n+1)}}\frac{n!\,n^{1/2}}{(1/2)_{n+1}}. \tag{$\triangle$}$$

There is but one thing left to do to make $(\triangle)$ into the form of $\Gamma(2z)\Gamma(1/2)$ (granted, the first $\Gamma$ will have a dummy variable $2n+1$ while the second simply has $n$). The $n^{2z}$ in the left numerator needs to be $(2n+1)^{2z}$, or something asymptotically $\sim$ to it, like - I don't know - $(\color{Orange}2n)^{2z}$?

Now can you tell why the $\color{Orange}2^{\color{Orange}{2z}\color{LimeGreen}{-1}}$ needs to be there for it to work out? ;-)

Answer 2

Recall that

$$B(x,y)= \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}=2\int_0^{\frac {\pi}2}\sin ^{2y-1}\theta\cos^{2x-1}\theta d\theta$$

We now choose $x=y=z$

\begin{align} B(z,z)& = \frac{{\Gamma {{(z)}^2}}}{{\Gamma (2z)}} = 2\int_0^{\frac{\pi }{2}} {{{(\sin \theta \cos \theta )}^{2z - 1}}} d\theta \\ &= 2\int_0^{\frac{\pi }{2}} {{{\left( {\frac{{\sin 2\theta }}{2}} \right)}^{2z - 1}}} d\theta = \frac{2}{{{2^{2z - 1}}}}\int_0^{\frac{\pi }{2}} {{{\left( {\sin 2\theta } \right)}^{2z - 1}}} d\theta \\ &= \frac{1}{{{2^{2z - 1}}}}\int_0^{\frac{\pi }{2}} {{{\left( {\sin 2\theta } \right)}^{2z - 1}}} d\left( {2\theta } \right)= \frac{1}{{{2^{2z - 1}}}}\int_0^\pi {{{\sin }^{2z - 1}}\phi } d\phi \\ & = \frac{2}{{{2^{2z - 1}}}}\int_0^{\pi /2} {{{\sin }^{2z - 1}}\phi } d\phi = \frac{1}{{{2^{2z - 1}}}}B\left( {z,\frac{1}{2}} \right) \\ \\ &= \frac{1}{{{2^{2z - 1}}}}\frac{{\Gamma (z)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma \left( {z + \frac{1}{2}} \right)}} \end{align} where $B(z,\frac{1}{2})=\frac{{\Gamma (z)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma \left( {z + \frac{1}{2}} \right)}}$ is used.

$$\frac{1}{{{2^{2z - 1}}}}\frac{{\Gamma (z)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma \left( {z + \frac{1}{2}} \right)}} = \frac{{\Gamma {{(z)}^2}}}{{\Gamma (2z)}}$$

$$\sqrt \pi \Gamma (2z) = {2^{2z - 1}}\Gamma (z)\Gamma \left( {z + \frac{1}{2}} \right)$$

As desired. This works (given the integral definition works) for $\Re(z) >0$.

I can't think of a solution in terms of Gauss' identity, but it seems quite laborious to use it. Also, you'll have to consider a different variable for each of the limits to make it rigorous, so maybe it is not a good idea using it after all.

Answer 3

It can be proved using Wallis's product formula : $$ \sqrt{\frac{\pi}{2}} = \lim_{n\rightarrow\infty}\frac{2^{2n}(n!)^{2}}{(2n+1)!}(2n+1)^{1/2} $$ By Gauss's formula, $$ \Gamma(s)\Gamma(s+1/2)=\lim_{n\rightarrow\infty}\frac{n^{2s+1/2}(n!)^2}{s(s+1)\dots(s+n)(s+1/2)(s+3/2)\dots(s+n+1/2)}=\lim_{n\rightarrow\infty}\frac{2^{2n+2}n^{2s+1/2}(n!)^2}{2s(2s+1)(2s+2)\dots(2s+2n)(2s+2n+1)}=\lim_{n\rightarrow\infty}\frac{(2n)^{2s}(2n)!}{2s(2s+1)(2s+2)\dots(2s+2n)}\frac{2^{2n+2}n^{2s+1/2}(n!)^2}{(2s+2n+1)(2n)^{2s}(2n)!} $$ Note that the first fraction converges to $\Gamma(2s)$. For the second one, $$ \frac{2^{2n+2}n^{2s+1/2}(n!)^2}{(2s+2n+1)(2n)^{2s}(2n)!}=\frac{2^{2n+2-2s}n^{1/2}(n!)^2}{(2s+2n+1)(2n)!}=\frac{2^{2n}(n!)^2}{(2n+1)!}(2n+1)^{1/2}2^{2-2s}\frac{2n+1}{2s+2n+1}\frac{n^{1/2}}{(2n+1)^{1/2}} $$ And by Wallis's product formula, this converges to $$ \sqrt{\frac{\pi}{2}}2^{2-2s}2^{-1/2}=\sqrt{\pi}2^{1-2s} $$ Hence we proved that $$ \Gamma(s)\Gamma(s+1/2)=\sqrt{\pi}2^{1-2s}\Gamma(2s) $$