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This question already has an answer here:

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?

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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
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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

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

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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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