Is $\prod_{m=1}^{\infty}\frac{\Gamma(2mx+x+1)}{(2mx)^{2x+1}\Gamma(2mx-x)}=\frac{2x^{x+\frac 12}}{\pi^x\ \Gamma(x+1)}$ also for $x\notin\mathbb Z^+$?

Question

By calling upon the roots of unity, it can be proved that for $n\in\mathbb Z^+$, $$\prod_{k=1}^{n-1} \sin\frac{k\pi}{2n}= \frac{\sqrt n}{2^{n-1}} $$ (see, for example, this.)

Using the infinite product representation for $\sin x$, the LHS becomes $$\prod_{k=1}^n \frac{k\pi}{2n}\prod_{m=1}^{\infty}\left( 1-\frac{k^2}{4m^2n^2}\right)\\ = n!\left(\frac{\pi}{2n} \right)^n \prod_{m=1}^{\infty}\prod_{k=1}^n \frac{(2mn-k)(2mn+k)}{4m^2n^2}\\ = n!\left(\frac{\pi}{2n}\right)^n \prod_{m=1}^{\infty}\frac{\Gamma(2mn+n+1)}{(2mn)^{2n+1}\Gamma(2mn-n)} $$ and therefore

$$ \prod_{m=1}^{\infty}\frac{\Gamma(2mx+x+1)}{(2mx)^{2x+1}\Gamma(2mx-x)}= \frac{2x^{x+\frac 12}}{\pi^x \ \Gamma(x+1)} $$ for $x\in\mathbb Z^+$. Naturally, I wondered whether this result also holds for other $x\notin\mathbb Z^+$, and indeed on checking I saw that this is true for $x=\frac 12, \frac 32, \frac 13$ or even $x=\pi$, so I suspect that it holds atleast for all $x\gt 0$. How can this be proven/disproven?

Answer 1

Suppose that $|\arg x|<\pi$. Then, by http://dlmf.nist.gov/5.11.E13, $$ \frac{{\Gamma (2mx + x + 1)}}{{(2mx)^{2x + 1} \Gamma (2mx - x)}} = 1 - \frac{{(2x + 1)(x + 1)}}{{24x}}\frac{1}{{m^2 }}+\mathcal{O}_x\!\left(\frac{1}{m^3}\right) $$ as $m\to +\infty$. Thus, the product on the LHS converges uniformly on compact subsets of $|\arg x|<\pi$ and defines an analytic function on $|\arg x|<\pi$.

Now assume that $|\arg x|\leq\frac{\pi}{2}$. By Stirling's formula \begin{align*} & \frac{{\Gamma (2mx + x + 1)}}{{(2mx)^{2x + 1} \Gamma (2mx - x)}} \sim \frac{{(2m + 1)^{2mx + x + 1/2} }}{{(2m)^{2x + 1} (2m - 1)^{2mx - x - 1/2} e^{2x} }} \\ &= \left( {1 - \frac{1}{{2m}}} \right)^{2x + 1} \left( {1 + \frac{2}{{2m - 1}}} \right)^{x + 1/2} \left( {\left( {1 + \frac{2}{{2m - 1}}} \right)^m \frac{1}{e}} \right)^{2x} \end{align*} as $x\to \infty$, uniformly in $m\geq 1$. The relative error in this approximation is $$ \frac{{1 + \frac{1}{{12x(2m + 1)}} + \mathcal{O}\!\left( {\frac{1}{{\left| x \right|^2 m^2 }}} \right)}}{{1 + \frac{1}{{12x(2m - 1)}} + \mathcal{O}\!\left( {\frac{1}{{\left| x \right|^2 m^2 }}} \right)}} = 1 + \mathcal{O}\!\left( {\frac{1}{{\left| x \right|m^2 }}} \right). $$ Thus \begin{align*} & \left| {\prod\limits_{m = 1}^\infty {\frac{{\Gamma (2mx + x + 1)}}{{(2mx)^{2x + 1} \Gamma (2mx - x)}}} } \right| \\ & \ll \prod\limits_{m = 1}^\infty {\left( {1 - \frac{1}{{2m}}} \right)^{2\Re x + 1} \left( {1 + \frac{2}{{2m - 1}}} \right)^{\Re x + 1/2} \left( {\left( {1 + \frac{2}{{2m - 1}}} \right)^m \frac{1}{e}} \right)^{2\Re x} } \\ & \le \prod\limits_{m = 1}^\infty {\exp \left( { - \frac{{2\Re x + 1}}{{2m}}} \right)\exp \left( {\frac{{2\Re x + 1}}{{2m - 1}}} \right)\exp \left( {\frac{{2\Re x}}{{10m^2 }}} \right)} \\ &= \prod\limits_{m = 1}^\infty {\exp \left( {\frac{{2\Re x + 1}}{{2m(2m - 1)}}} \right)\exp \left( {\frac{{2\Re x}}{{10m^2 }}} \right)} \\ & = 2\exp \left( {2\Re x\left( {\log 2 + \frac{{\pi ^2 }}{{60}}} \right)} \right) = \mathcal{O}(e^{2\left| x \right|} ). \end{align*} The RHS of your expression, which is analytic for $|\arg x|<\pi$, is $$ \frac{{2x^{x + \frac{1}{2}} }}{{\pi ^x \Gamma (x + 1)}} \sim \sqrt {\frac{2}{{\pi x}}} \left( {\frac{e}{\pi }} \right)^x = \mathcal{O}(e^{\left| x \right|} ) $$ for $|\arg x|\leq\frac{\pi}{2}$, by Stirling's formula. Thus, the LHS minus the RHS is $\mathcal{O}(e^{2\left| x \right|} )$ for $|\arg x|\leq\frac{\pi}{2}$ and is identically zero on the positive integers. By a theorem of Carlson (see below), the difference is identically zero on $|\arg x|\leq\frac{\pi}{2}$ too. By analytic continuation, the difference is zero on the whole of $|\arg x|<\pi$.

In summary, $$ \prod_{m=1}^{\infty}\frac{\Gamma(2mx+x+1)}{(2mx)^{2x+1}\Gamma(2mx-x)}= \frac{2x^{x+\frac 12}}{\pi^x \ \Gamma(x+1)} $$ for $|\arg x|<\pi$.

Carlson's theorem: If $f(z)$ is holomorphic in the sector $|\arg z|\leq \alpha$ with $\alpha\geq \frac{\pi}{2}$, $|f(z)|=\mathcal{O}(e^{c|z|})$ with some $c < \pi$ in this sector, and if $f(n) = 0$ when $n = 1, 2, 3,\ldots$, then $f(z) \equiv 0$.

For more on this theorem, see G. H. Hardy, On two theorems of F. Carlson and S. Wigert, Acta Math. 42 (1920), pp. 327–339.

Answer 2

This is not a proof of anything.

The computation of the lhs being extremely expensive, I limited the computation to the first $30$ terms.

Making $x=a+ib$, some results

$$\left( \begin{array}{cccc} a & b & \text{rhs} & \text{lhs} \\ 0 & 1 & 0.795813 -0.045963 \,i\text{ } & 0.800772 -0.044617 \,i \\ 0 & 2 & 0.773587 -0.195399 \,i\text{ } & 0.779696 -0.191053 \,i \\ 0 & 3 & 0.732927 -0.315339 \,i\text{ } & 0.741014 -0.308754 \,i \\ 0 & 4 & 0.676840 -0.422501 \,i\text{ } & 0.687590 -0.414307 \,i \\ 1 & 1 & 0.658251 -0.068584 \,i\text{ } & 0.665858 -0.067326 \,i \\ 1 & 2 & 0.656654 -0.171940 \,i\text{ } & 0.664883 -0.168907 \,i\\ 1 & 3 & 0.628090 -0.272345 \,i\text{ } & 0.637779 -0.267813 \,i \\ 1 & 4 & 0.582309 -0.364496 \,i\text{ } & 0.594064 -0.358934 \,i \\ 2 & 1 & 0.573048 -0.073927 \,i\text{ } & 0.582002 -0.072918 \,i \\ 2 & 2 & 0.563992 -0.155310 \,i\text{ } & 0.573548 -0.153249 \,i \\ 2 & 3 & 0.539657 -0.237748 \,i\text{ } & 0.550351 -0.234794 \,i \\ 2 & 4 & 0.501184 -0.315789 \,i\text{ } & 0.513483 -0.312278 \,i \\ 3 & 1 & 0.499438 -0.068583 \,i\text{ } & 0.509233 -0.067926 \,i \\ 3 & 2 & 0.487726 -0.138505 \,i\text{ } & 0.498050 -0.137239 \,i \\ 3 & 3 & 0.465351 -0.207995 \,i\text{ } & 0.476580 -0.206271 \,i \\ 3 & 4 & 0.432020 -0.274233 \,i\text{ } & 0.444496 -0.272316 \,i \\ 4 & 1 & 0.434265 -0.061133 \,i\text{ } & 0.444538 -0.060791 \,i \\ 4 & 2 & 0.422558 -0.122038 \,i\text{ } & 0.433262 -0.121417 \,i \\ 4 & 3 & 0.402176 -0.181636 \,i\text{ } & 0.413593 -0.180873 \,i \\ 4 & 4 & 0.372961 -0.238275 \,i\text{ } & 0.385341 -0.237588 \,i \end{array} \right)$$