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1. Premise

I would like to find the closed form of this function:

I have the infinite-multifactorial function $f(x)=x!_{(\infty)}$ and I want to calculate its Taylor series.

$$f(x)=\prod_{j=1}^{\infty} j^{\text{sinc}(x-j)}\qquad\text{ where}\quad\text{sinc}(z)=\begin{cases}\frac{\sin(\pi z)}{\pi z}&z\neq 0\\ 1&z=0\end{cases}$$

After various calculations that I am not here to write (they are very long and repetitive), I have arrived at this formulation:

$$f'(x)=f(x)\cdot\underbrace{\sum_{j=1}^{\infty}\ln(j)\text{sinc}(x-j)}_{=: g(x)}$$

and applying the Leibnitz rule

$$f^{(n)}(x)=\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}f(x)g(x)=\sum_{k=0}^{n-1}\binom{n-1}{k}f^{(k)}(x)g^{(n-k-1)}(x)$$

I get that:

$$f^{(n)}(0)=\sum_{k=0}^{n-1}\binom{n-1}{k}f^{(k)}(0)g^{(n-k-1)}(0)$$

$f^{(n)}$ is obtained iteratively, while $$g^{(n)}(0)=\sum_{j=1}^{\infty}\ln(j)\text{sinc}^{(n)}(-j)=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\frac{(-1)^k\pi^{2k}}{2k+1}\binom{n}{2k} b_{n-2k}$$ $$g^{(n)}(0)=b_n+\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\frac{(-1)^k\pi^{2k}}{2k+1}\binom{n}{2k} b_{n-2k}$$ Where: $$b_n=-n!\cdot \eta'(n)=\begin{cases} \dfrac{\ln\left(2\right)^{2}}{2}-\ln\left(2\right)\gamma&n=1\\ -n!\cdot \dfrac{(2^{n-1}-1)\zeta'\left(n\right)+\zeta\left(n\right)\ln\left(2\right)}{2^{n-1}}&n\geq 2 \end{cases}$$ Where $\eta(z)$ is the Dirichlet Eta function

By writing the series up to grade 5 I obtained this sum: $$\begin{align} f(x)\approx& 1+\frac{b_{1}}{1!}x+\frac{b_{1}^{2}+b_{2}}{2!}x^{2}+\frac{b_{1}^{3}+3b_{1}b_{2}+b_{3}-\pi^{2}b_{1}}{3!}x^{3}\\ +&\frac{\left(b_{1}^{4}+6b_{1}^{2}b_{2}+4b_{1}b_{3}+3b_{2}^{2}+b_{4}\right)-\pi^{2}\left(4b_{1}^{2}+2b_{2}\right)}{4!}x^{4}\\ +&\frac{{\left(b_{1}^{5}+10b_{1}^{3}b_{2}+15b_{1}b_{2}^{2}+10b_{1}^{2}b_{3}+5b_{1}b_{4}+10b_{2}b_{3}+b_{5}\right)+\left(\pi^{4}b_{1}-20b_{1}b_{2}\pi^{2}-10b_{1}^{3}\pi^{2}-20\pi^{2}\frac{b_{3}}{6}\right)}}{5!}x^{5} \end{align}$$

It may seem complicated but the part of the coefficients not containing powers of $\pi$ can be written as complete Bell polynomials: $$\begin{align} f(x)\approx&B_0(\{b_i\})+\frac{B_1(\{b_i\})}{1!}x+\frac{B_2(\{b_i\})}{2!}x^{2}+\frac{B_3(\{b_i\})-\pi^{2}b_{1}}{3!}x^{3}\\ +&\frac{B_4(\{b_i\})-\pi^{2}\left(4b_{1}^{2}+2b_{2}\right)}{4!}x^{4}\\ +&\frac{B_5(\{b_i\})+\left(\pi^{4}b_{1}-20b_{1}b_{2}\pi^{2}-10b_{1}^{3}\pi^{2}-20\pi^{2}\frac{b_{3}}{6}\right)}{5!}x^{5} \end{align}$$

Where: $B_n(\{b_i\})=B_n(b_1,...,b_n)$ is the $n$-th complete Bell Polynomial

So I separated the first part from the rest:

$$f(x)=\sum_{k=0}^{\infty}\frac{B_k(\{b_i\})}{k!}x^k+\frac{-\pi^{2}b_{1}}{3!}x^{3}+\frac{-\pi^{2}\left(4b_{1}^{2}+2b_{2}\right)}{4!}x^{4}+\frac{\pi^{4}b_{1}-20b_{1}b_{2}\pi^{2}-10b_{1}^{3}\pi^{2}-20\pi^{2}\frac{b_{3}}{6}}{5!}x^{5}+...$$

My problem now is that I can't understand what the closed formulation of the remaining piece of coefficients could be, can anyone help me?

2. Motivation of interest

Using the formula

$$\sum _{n=0}^{\infty }{B_{n}(b_{1},\dots ,b_{n}) \over n!}x^{n}=\exp \left(\sum _{i=1}^{\infty }{b_{i} \over i!}x^{i}\right)$$

We have that the part of $f$ in which the first series appears can be written as: $$\sum _{n=0}^{\infty }{B_{n}(\{b_i\}) \over n!}x^{n}=\exp\left[\left(\dfrac{\ln\left(2\right)^{2}}{2}-\ln\left(2\right)\gamma\right)x-\sum_{n=2}^{\infty} \dfrac{(2^{n-1}-1)\zeta'\left(n\right)+\zeta\left(n\right)\ln\left(2\right)}{2^{n-1}}x^n\right]$$ $$=\exp\left[\left(\frac{\ln\left(2\right)^{2}}{2}-\ln\left(2\right)\gamma\right)x+\sum_{n=2}^{\infty}\frac{\zeta'\left(n\right)}{2^{n-1}}x^{n}-\sum_{n=2}^{\infty}\zeta'\left(n\right)x^{n}-\ln\left(2\right)\sum_{n=2}^{\infty}\frac{\zeta\left(n\right)}{2^{n-1}}x^{n}\right]$$ $$=\exp\left[\frac{\ln\left(2\right)^{2}}{2}x+2\sum_{n=2}^{\infty}\zeta'\left(n\right)\left(\frac{x}{2}\right)^{n}-\sum_{n=2}^{\infty}\zeta'\left(n\right)x^{n}+\ln\left(2\right)x\psi\left(1-\frac{x}{2}\right)\right]$$

Here a series involving the derivative of the zeta function appears twice.
I would like to find some way to express the series in this way $$\sum_{n=2}^{\infty}\zeta'(n)x^{n-1}=K-\ln(2)f(1-x)$$ so the sum becomes $$\sum _{n=0}^{\infty }{B_{n}(\{b_i\}) \over n!}x^{n}=\exp\left[\ln\left(2\right)x\left(\frac{\ln\left(2\right)}{2}+f\left(1-\frac{x}{2}\right)-f\left(1-x\right)+\psi\left(1-\frac{x}{2}\right)\right)\right]$$ I think there may be connections with the gamma function being $$\sum_{n=2}^{\infty}\zeta(n)x^{n-1}=-\gamma-\psi(1-x)$$

I think it could be interesting that the infinte-multifactorial could be connected with some function deriving from the gamma function.

3. Update

I realized I can write the infinite-multifactor function as

$$x!_{(\infty)}=\exp\left(\sum_{n=1}^{\infty}\left(-1\right)^{n}\frac{\ln\left(n\right)x}{x-n}\right)^{\text{sinc}(x)}$$

In this way I only study the function inside the exponential so as not to have that uncomfortable presence of $\pi$

$$\begin{align}\sum_{n=1}^{\infty}\left(-1\right)^{n}\frac{\ln\left(n\right)x}{x-n}=&\sum_{s=1}^{\infty}x^{s}\left(-\sum_{n=1}^{\infty}\left(-1\right)^{n}\frac{\ln\left(n\right)}{n^{s}}\right)\\ =&\sum_{s=1}^{\infty}x^{s}\left(\frac{\mathrm{d}}{\mathrm{d}s}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n^{s}}\right)\\ =&-\sum_{s=1}^{\infty}\eta'(s)x^s\\ =&-\sum_{s=1}^{\infty}\frac{(2^{s-1}-1)\zeta'(s)+\ln(2)\zeta(s)}{2^{s-1}}x^s \end{align}$$ So there is a strong correlation between the infinite-multifactorial and the series $$\begin{align}x!_{(\infty)}=&\exp\left(-\sum_{s=1}^{\infty}\eta'(s)x^s\right)^{\text{sinc}(x)}\\ =&\exp\left(-\sum_{s=1}^{\infty}\dfrac{\ln(2)\zeta(s)+(2^{s-1}-1)\zeta'(s)}{2^{s-1}}x^s\right)^{\text{sinc}(x)}\\ =&\exp\left(\ln(2)\psi\left(1-\frac{x}{2}\right)x+\sum_{s=1}^{\infty}(2^{1-s}-1)\zeta'(s)x^{s}\right)^{\text{sinc}(x)}\end{align}$$ We considered the regularized zeta where $\zeta(1)=\gamma$ and $\eta'(1)=\gamma\ln(2)-\dfrac{\ln(2)^2}{2}$

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    $\begingroup$ Impressive work so far. Interesting to see the generating function of $\zeta'(n)$ appearing in the calculations. $\endgroup$ Nov 16, 2023 at 21:48
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    $\begingroup$ @C-RAM I wrote an update that I think is very interesting $\endgroup$ Nov 16, 2023 at 22:14

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