A Partial product involving the Gamma function

I previously asked this question about the shape of the following infinite product involving the Gamma function.

$$\prod_{n=1}^\infty\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\Gamma\left(n-x+1\right)}=\prod_{n=1}^\infty\prod_{k=1}^\infty\left(1-\frac{x^2}{\left(n+k\right)^2}\right)$$

It appeared to fit a normal curve very well for more terms, which made me wonder if it had the form of an exponential. The problem, as you can see from @marty cohen’s answer, is that the product diverges.

But what about the partial product?

$$\prod_{n=1}^m\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\Gamma\left(n-x+1\right)}$$

Can we find an exact or asymptotic formula to shed some light on the shape of this function?

Revised version after @tyobrien key remark.

As I wrote in my first answer to your previous question, there is a closed form expressionfor$$f_p(x)=\prod_{n=1}^p\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)}$$ It is given in terms of the Barnes G-function (sometimes named the double gamma function) and write $$f_p(x)=\frac{ G(p+2)^2 \, G(2-x)\, G(2+x) }{G(p+2-x)\, G(p+2+x)}$$

This $G(.)$ is pretty documented (google for it) and seems to be related to many other functions (have a look here).

Concerning the asymptotics, as I wrote in my second answer to your previous question, it seems that, building the Taylor expansion around $x=0$, we have very good approximations looking like $$f_p(x)=\exp\left(-\sum_{i=1}^\infty c_k^{(p)} x^{2k} \right)$$ where all coefficients have an explicit formulation in terms of polygamma and zeta functions (all of them are positive). To make them clearer and to see some possible patters, writing $$c_k^{(p)}=d_k^{(p)}+e_k^{(p)}$$ the first are given in the table below $$\left( \begin{array}{ccc} k & d_k^{(p)} & e_k^{(p)} \\ 1 & \psi ^{(0)}(p+2)+(p+1) \psi ^{(1)}(p+2) & -\frac{\pi ^2}{6}+\gamma \\ 2 & \frac{3 \psi ^{(2)}(p+2)+(p+1) \psi ^{(3)}(p+2)}{12} & -\frac{\pi ^4}{180}+\frac{\zeta (3)}{2} \\ 3 & \frac{5 \psi ^{(4)}(p+2)+(p+1) \psi ^{(5)}(p+2)}{360} & -\frac{\pi ^6}{2835}+\frac{\zeta (5)}{3} \\ 4 & \frac{7 \psi ^{(6)}(p+2)+(p+1) \psi ^{(7)}(p+2)}{20160} & -\frac{\pi ^8}{37800}+\frac{\zeta (7)}{4} \\ 5 & \frac{297 \psi ^{(8)}(p+2)+33 (p+1) \psi ^{(9)}(p+2)}{59875200} & -\frac{\pi ^{10}}{467775}+\frac{\zeta (9)}{5} \\ 6 & \frac{11 \psi ^{(10)}(p+2)+(p+1) \psi ^{(11)}(p+2)}{239500800} & -\frac{691 \pi^{12}}{3831077250}+\frac{\zeta (11)}{6}\\ 7 & \frac{39 \psi ^{(12)}(p+2)+3 (p+1) \psi ^{(13)}(p+2)}{130767436800} &-\frac{2 \pi ^{14}}{127702575}+\frac{\zeta (13)}{7} \end{array} \right)$$

This does confirm your interesting observation.

By the way, you could be interested by this paper.

Numerical aspects

As said earlier, the first term is, from far away, the most significant. In order to check, I computed $$\Phi(a)=\int_{-3}^3 \left(f_{100}(x)-e^{-a x^2}\right)^2\,dx$$ which I minimized with respect to $a$. The optimum is found for $a=4.5645$ to be compared with $c_1^{(100)}=4.5474$.

This is an extension of the work by @Claude Leibovici.

Here we derive the explicit formula for the coefficients in the power series equal to the product.

Begin with the Taylor expansion:

$$\log\left(\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)}\right)=-2\sum_{k=1}^\infty \frac{\psi ^{(2 k-1)}(n+1)}{(2 k)!}x^{2k}$$

The goal then is to evaluate $$\sum_{n=1}^p\psi^{(2k-1)}(n+1).$$

Using the recurrence relation $$\psi^{(m)}(x+1)=\psi^{(m)}(x)+\frac{(-1)^m m!}{x^{m+1}}$$

we can derive

$$\sum_{n=1}^p\psi^{(2k-1)}(n+1)=p\psi^{(2k-1)}(p+2)+(2k-1)!\sum_{i=1}^p\frac{i}{(i+1)^{2k}}.$$

The final sum is

\begin{align} \sum_{i=1}^p\frac{i}{(i+1)^{2k}} &=\sum_{i=1}^p\frac{1}{(i+1)^{2k-1}}-\sum_{i=1}^p\frac{1}{(i+1)^{2k}} \\&=\zeta(2k-1)-1-\sum_{i=0}^\infty\frac{1}{(i+p+2)^{2k-1}} \\&\space\space\space\space\space-\left[\zeta(2k)-1-\sum_{i=0}^\infty\frac{1}{(i+p+2)^{2k}}\right] \\&=\zeta(2k-1)-\zeta(2k)+\frac{\psi^{(2k-2)}(p+2)}{(2k-2)!}+\frac{\psi^{(2k-1)}(p+2)}{(2k-1)!}. \end{align}

Therefore \begin{align} \sum_{n=1}^p\psi^{(2k-1)}(n+1)&=p\psi^{(2k-1)}(p+2)+(2k-1)!(\zeta(2k-1)-\zeta(2k))+(2k-1)\psi^{(2k-2)}(p+2)+\psi^{(2k-1)}(p+2) \\&=(2k-1)\psi^{(2k-2)}(p+2)+(p+1)\psi^{(2k-1)}(p+2)+(2k-1)!(\zeta(2k-1)-\zeta(2k)). \end{align}

So now we have $$\sum_{n=1}^p\log\left(\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)}\right)=-\sum_{k=1}^\infty c_k^{(p)} x^{2k}$$

where $$c_k^{(p)}=\frac{2(2k-1)\psi^{(2k-2)}(p+2)+2(p+1)\psi^{(2k-1)}(p+2)}{(2k)!}+\frac{\zeta(2k-1)-\zeta(2k)}{k}.$$

Note that by using the definition of $\psi^{(m)}(z)$ for positive $m$, the divergent effects of $\psi^{(0)}(z)$ and $\zeta(1)$ cancel each other out.

Update: We can easily simplify this to $$c_k^{(p)}=\frac{p\zeta(2k)+H_{p+1,2k-1}-(p+1)H_{p+1,2k}}{k}$$

Summing over these gives us

\begin{align} \sum_{k=1}^\infty c_k^{(p)} x^{2k} &= p\sum_{k=1}^\infty\frac{\zeta(2k)}{k}x^{2k}+\sum_{k=1}^\infty\frac{H_{p+1,2k-1}}{k}x^{2k}-(p+1)\sum_{k=1}^\infty\frac{H_{p+1,2k}}{k}x^{2k} \\&=p\log\left(\frac{\pi x}{\sin(\pi x)}\right)-\sum_{n=1}^{p+1}n\log\left(1-\left(\frac{x}{n}\right)^2\right)+(p+1)\sum_{n=1}^{p+1}\log\left(1-\left(\frac{x}{n}\right)^2\right) \\&=p\log\left(\frac{\pi x}{\sin(\pi x)}\right)+\sum_{n=1}^{p+1}(p+1-n)\log\left(1-\left(\frac{x}{n}\right)^2\right) \end{align}

Which finally gives us the very nice result $$\prod_{n=1}^p\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)}=\left(\frac{\sin(\pi x)}{\pi x}\right)^p\,\,\prod_{n=1}^{p+1}\left(1-\left(\frac{x}{n}\right)^2\right)^{n-p-1}$$

Or just for completeness, starting at $n=0$, \begin{align} \prod_{n=0}^p\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)} &=\left(\frac{\sin(\pi x)}{\pi x}\right)^{p+1}\,\,\prod_{n=1}^{p+1}\left(1-\left(\frac{x}{n}\right)^2\right)^{n-p-1} \\ &=\left(\frac{\sin(\pi x)}{\pi x \prod_{n=1}^{p+1}\left(1-\left(\frac{x}{n}\right)^2\right)}\right)^{p+1}\,\,\prod_{n=1}^{p+1}\left(1-\left(\frac{x}{n}\right)^2\right)^n \\ &=\left(\frac{\Gamma(p+2)^2}{\Gamma(p+x+2)\,\Gamma(p-x+2)}\right)^{p+1}\,\,\prod_{n=1}^{p+1}\left(1-\left(\frac{x}{n}\right)^2\right)^n \\ &=\left(\frac{\Gamma(p+1)^2}{\Gamma(p+x+1)\,\Gamma(p-x+1)}\right)^{p+1}\,\,\prod_{n=1}^{p}\left(1-\left(\frac{x}{n}\right)^2\right)^n \end{align}

Notice that numerically the first term appears to tend to $e^{-x^2}$ $$\left(\frac{\sin(\pi x)}{\pi x \prod_{n=1}^{p+1}\left(1-\left(\frac{x}{n}\right)^2\right)}\right)^{p+1}\to e^{-x^2}$$ But I'm not sure how to prove it at the moment.

Thus the divergence is all in the product $\prod_{n=1}^{p+1}\left(1-\left(\frac{x}{n}\right)^2\right)^n$.

This is too long for a comment.

Let us consider $$a_p=\left(\frac{\sin(\pi x)}{(\pi x) \prod_{n=1}^{p+1}\left(1-\left(\frac{x}{n}\right)^2\right)}\right)^{p+1}$$ $$\prod_{n=1}^{p+1}\left(1-\left(\frac{x}{n}\right)^2\right)=\frac{(1-x)_{p+1} (x+1)_{p+1}}{((p+1)!)^2}$$ $$\log(a_p)=(p+1) \log \left(\frac{((p+1)!)^2 \sin (\pi x)}{(\pi x) (1-x)_{p+1} (x+1)_{p+1}}\right)$$ Now, expand as Taylor series around $x=0$ to get $$\log(a_p)=-(p+1) \psi ^{(1)}(p+2)x^2+O\left(x^4\right)$$ Now, using asymtotics $$-(p+1) \psi ^{(1)}(p+2)=-1+\frac{1}{2 p}-\frac{2}{3 p^2}+O\left(\frac{1}{p^3}\right)$$