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For Gamma function's duplication formula $$\Gamma (z)\;\Gamma \left(z+{\frac {1}{2}}\right)=2^{{1-2z}}\;{\sqrt {\pi }}\;\Gamma (2z)$$ , if limit $z$ to real numbers, i.e. $z\in \mathbb R$, is it possible to prove it using basic analysis, i.e. not relying on complex analysis or beta, zeta etc functions?


marked as duplicate by Simply Beautiful Art, zhoraster, Shailesh, C. Falcon, S.C.B. Jan 16 '17 at 1:49

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  • $\begingroup$ Not within my lifetime. Especially if you meant the full duplication formula. $\endgroup$ – Simply Beautiful Art Jan 15 '17 at 15:10
  • $\begingroup$ not sure what is "full duplication formula", basically i'd like to find a proof of the formula above, within $\mathbb R$, not using complex analysis. @SimpleArt $\endgroup$ – athos Jan 15 '17 at 15:13
  • $\begingroup$ Use the limit definition along with $\Gamma(1/2)=\sqrt \pi$. $\endgroup$ – Mark Viola Jan 15 '17 at 15:13
  • $\begingroup$ @Dr.MV The one from Mohr-Bollerup theorem? Looks promising $\endgroup$ – Simply Beautiful Art Jan 15 '17 at 15:15
  • $\begingroup$ @Dr.MV would u pls elaborate? I only know $\Gamma(s) := \int_0^\infty x^{s-1} e^{-x} dx$ $\endgroup$ – athos Jan 15 '17 at 15:18

In THIS ANSWER, I showed using real analysis only that the Gamma function, $\Gamma(x)$, as defined by the integral, $\Gamma(x)=\int_0^\infty t^{x-1}\,e^{-t}\,dt$, $x>0$, can be represented by the limit

$$\bbox[5px,border:2px solid #C0A000]{\Gamma(x)=\lim_{n\to \infty}\frac{n^x\,n!}{x(x+1)(x+2)\cdots (x+n)}} \tag 1$$

Deonte by $G_n(x)$ the sequence of functions

$$G_n(x)=\frac{n^x\,n!}{x(x+1)(x+2)\cdots (x+n)} \tag2$$

Then, using $(2)$ we can write

$$\begin{align} G_n(x)G_n(x+1/2)&=\left(\frac{n^x\,n!}{x(x+1)\cdots (x+n)}\right)\left(\frac{2^{n+1}\,n^{x+1/2}\,n!}{(2x+1)(2x+3)\cdots (2x+(2n+1))}\right)\\\\ &=\frac{2^{2n+2}n^{2x+1/2}(n!)^2}{(2x)(2x+1)\cdots (2x+2n+1)}\\\\ &=\left(\frac{2^{2n+2}n^{1/2}\color{green}{(n!)^2}}{\color{red}{(2n)!}2^{2x}}\right)\left(\frac{1}{2x+2n+1}\right)\color{blue}{\overbrace{\left(\frac{(2n)^{2x}(2n)!}{(2x)(2x+1)\cdots(2x+2n)}\right)}^{G_{2n}(2x)}}\tag 3\\\\ &\sim \left(\frac{2^{2n+2}n^{1/2}\color{green}{(2\pi n)\left(\frac{n}{e}\right)^{2n}}}{\color{red}{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}}2^{2x}}\right)\left(\frac{1}{2x+2n+1}\right)\color{blue}{G_{2n}(2x)} \tag 4\\\\ &=\left(\frac{\sqrt \pi}{2^{2x-1}}\right)\left(\frac{2n}{2x+2n+1}\right)\,G_{2n}(2x) \end{align}$$

where Stirling's Approximation was used in going from $(3)$ to $(4)$. Finally, letting $n\to\infty$ and appealing to $(1)$ yields the coveted relationship



Let we consider $$ f(z) = \frac{\Gamma(2z)\,\Gamma(1/2)}{\Gamma(z)\,\Gamma(z+1/2)}.$$ Since $\Gamma(z)$ never vanishes, $f(z)$ is a continuous function on its domain. The singularity of the $\Gamma$ function are simple poles at the negative integers: in particular, the structure of the denominator and numerator of $f(z)$ implies that $f$ has no singularity and no zero on the real line. Since $\Gamma(z+1)=z\,\Gamma(z)$, we also have:

$$ \frac{f(z+1)}{f(z)} = \frac{(2z+1)(2z)}{z(z+1/2)} = 4 $$ hence it follows that $f(z)=C\cdot 4^z$. By computing $f(z)$ at $z=1$ we get the explicit value of $C$, hence Legendre's duplication formula through a real-analytic version of Herglotz' trick.

You may perform just the same trick to prove the full multiplication formula in the real case.

An efficient alternative is to consider $\frac{d}{dx}\log(\cdot )$ of both terms. Since $$ \frac{d}{dx}\log\Gamma(x) = \psi(x) = -\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+x-1}\right) $$ the duplication/multiplication formula for the $\Gamma$ function can be derived from the duplication/multiplication formula for the $\psi$ function, that is simple to prove through elementary series manipulations.

As a third alternative, Legendre duplication formula can be proved by computing $$ \int_{0}^{+\infty}\frac{d\theta}{(1+\cosh\theta)^n} $$ in two different ways, as done by me and Marco Cantarini here.


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