$\infty$-multifactorial: $\displaystyle z!_{(\infty)}:=\lim_{\alpha\to\infty}z!_{(\alpha)}$

Question

Introduction

Let $z!_{(\alpha)}$ the $\alpha$-multifactorial.

$$z!_{(\alpha)}=\alpha^{\frac{z}{\alpha}}\Gamma\left(1+\frac{z}{\alpha}\right)\prod_{j=1}^{\alpha-1}\left(\frac{\alpha^{\frac{\alpha-j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right)^{C_{\alpha}(z-j)}$$ Where $$C_{\alpha}(z)=\frac{1}{\alpha}\left(1+2\sum_{k=1}^{\left\lfloor\frac{\alpha-1}{2}\right\rfloor}\cos\left(\frac{2k\pi}{\alpha}z\right)+\text{mod}(\alpha-1,2)\cos(\pi z)\right)$$

My intent

I would like to calculate the limit of the multifactorial formula for $\alpha$ which approaches infinity.

Leaving aside the various steps that are long and irrelevant to the question

I arrived at this formula:

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

Where

• $\text{sinc}(x):=\begin{cases}\dfrac{\sin(\pi x)}{\pi x}&\text{if }x\neq0\\1&\text{if }x=0\end{cases}$

Unfortunately I can't move forward with this last sum :/, does anyone have any suggestions?
(It would also be acceptable to write it as an integral)

At the end of everything, a function with the following characteristics must emerge:

• $n!_{(\infty)}=n$ for $n\in\mathbb{N}^{+}$
• $n!_{(\infty)}=1$ for $n=-1,-2,-3,...$
• $\displaystyle\prod_{i=0}^{\infty}(z-i)!_{(\infty)}=z!$

This is the graph:

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I'll also leave this link which leads to the desmos graph I made in case anyone wants to try their hand at trying to solve it.

Update

The series is equal to $$z!_{(\infty)}=\exp\left(\frac{\sin(\pi z)}{\pi}\sum_{n=1}^{\infty}(-1)^n\frac{\ln(n)}{z-n}\right)\qquad \text{for }z\not\in\mathbb{N}^{+}$$

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I take inspiration from an answer to make a clarification:

It would be useful to be able to use a result like this:

$$\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{z-k}f(k)=\pi\frac{f(z)}{\sin(\pi z)}$$

Applying this result with $f(n)=\begin{cases}\ln(n)&\text{if }n>1\\0&\text{if }n\leq 1\end{cases}$ the result would be: $$z!_{(\infty)}=\begin{cases}z&\text{if }z>1\\1&\text{if }z\leq 1\end{cases}$$ Unfortunately:

• In this case it is not applicable because one of the hypotheses requires that $f(n)$ is entire and the logarithm is not (and it can not be continued analytically to an entire function).
• $z!_{(\infty)}$ is a complex variable function, $\mathbb{C}$ is not ordered so it's not possible to say things like "$z>1$" or "$z\leq 1$"
• According to this solution for "$z\leq 1$" the series should ALWAYS be $0$.

Answer 1

This is a partial answer.

If I understand correctly from your comment, you extended the Wolfram alpha definition of $z!! = z!_{(2)}$ to all the multifactorials $z!_{(\alpha)}$ instead of using the Wikipedia definition, so the formula $$ z! = \prod_{i = 0}^{\alpha - 1} (z - i)!_{(\alpha)}, $$ holds for all $z \in \mathbb{C}\backslash\mathbb{Z}_-$, right ? We clearly have convergence of $n!_{(\alpha)}$ toward some ramp function when $n$ is an integer. Let $z \in \mathbb{C}\backslash\mathbb{Z}$.

Step 1 : simplifications and convergence of $C_\alpha$. First of all, $\alpha^{\frac{z}{\alpha}}\Gamma\left(1 + \frac{z}{\alpha}\right) \rightarrow 1$ so we can ignore this part. The only limit that remains to compute is the limit of, $$ \prod_{j = 1}^{\alpha - 1} \left(\frac{\alpha^{\frac{\alpha - j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right)^{C_\alpha(z - j)} = \exp\left(\sum_{j = 1}^{\alpha - 1} C_\alpha(z - j)\ln\left(\frac{\alpha^{\frac{\alpha - j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right)\right). $$ Let us simplify the $C_\alpha$ now. For all real number $x \in \mathbb{R}\backslash\mathbb{Z}$ and all integer $n$, we have, \begin{align*} \sum_{k = 0}^n \cos(k\pi x) & = \Re\left(\sum_{k = 0}^n e^{ik\pi x}\right)\\ & = \Re\left(\frac{1 - e^{i(n + 1)\pi x}}{1 - e^{i\pi x}}\right)\\ & = \Re\left(\frac{e^{i\frac{(n + 1)\pi}{2}x}}{e^{i\frac{\pi}{2}x}}\frac{e^{-i\frac{(n + 1)\pi}{2}x} - e^{i\frac{(n + 1)\pi}{2}x}}{e^{-i\frac{\pi}{2}x} - e^{i\frac{\pi}{2}x}}\right)\\ & = \cos\left(\frac{n\pi}{2}x\right)\frac{\sin\left(\frac{(n + 1)\pi}{2}x\right)}{\sin\left(\frac{\pi}{2}x\right)}\\ & = (n + 1)\cos\left(\frac{n\pi}{2}x\right)\frac{\mathrm{sinc}\left(\frac{n + 1}{2}x\right)}{\mathrm{sinc}\left(\frac{1}{2}x\right)}. \end{align*} As $\mathbb{R}\backslash\mathbb{Z}$ is not discrete in $\mathbb{C}$ and by uniqueness of the analytic continuation, this equality remains true for any $x \in \mathbb{C}\backslash 2\mathbb{Z} \cup \{0\}$. In particular, if is true for any point in $z + \mathbb{Z}$.

If $\alpha = 2\beta$ is even, we have, \begin{align*} C_\alpha(z) & = \frac{1}{2\beta}\left(1 + 2\sum_{k = 0}^{\beta - 1} \cos\left(\frac{k\pi}{\beta}z\right) + \cos(\pi z)\right)\\ & = \frac{1}{2\beta}\left(1 + \cos(\pi z) + 2\beta\cos\left(\frac{(\beta - 1)\pi}{2\beta}z\right)\frac{\mathrm{sinc}\left(\frac{1}{2}z\right)}{\mathrm{sinc}\left(\frac{1}{2\beta}z\right)}\right)\\ & = \cos\left(\frac{(\beta - 1)\pi}{2\beta}z\right)\frac{\mathrm{sinc}\left(\frac{1}{2}z\right)}{\mathrm{sinc}\left(\frac{1}{2\beta}z\right)} + \mathrm{o}(1) \textrm{ when } \beta \rightarrow +\infty,\\ & \rightarrow \cos\left(\frac{\pi}{2}z\right)\mathrm{sinc}\left(\frac{1}{2}z\right)\\ & = \frac{2}{\pi z}\cos\left(\frac{\pi}{2}z\right)\sin\left(\frac{\pi}{2}z\right)\\ & = \mathrm{sinc}(z). \end{align*} And when $\alpha = 2\beta - 1$ is odd, similarly, \begin{align*} C_\alpha(z) & = \frac{1}{2\beta - 1}\left(1 + 2\sum_{k = 0}^{\beta - 1} \cos\left(\frac{2k\pi}{2\beta - 1}z\right)\right)\\ & = \frac{1}{2\beta - 1}\left(1 + 2\beta\cos\left(\frac{(\beta - 1)\pi}{2\beta - 1}z\right)\frac{\mathrm{sinc}\left(\frac{\beta}{2\beta - 1}z\right)}{\mathrm{sinc}\left(\frac{1}{2\beta - 1}z\right)}\right)\\ & \rightarrow \cos\left(\frac{\pi}{2}z\right)\mathrm{sinc}\left(\frac{1}{2}z\right)\\ & = \mathrm{sinc}(z). \end{align*}

Step 2 : convergence of the other part of the product. Fix some integer $j$. $\Gamma(\varepsilon) \sim \frac{1}{\varepsilon}$ when $\varepsilon \rightarrow 0$ thus, $$ \frac{\alpha^{\frac{\alpha - j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)} \sim \frac{j}{\alpha}\alpha^{\frac{\alpha - j}{\alpha}} = j\alpha^{-\frac{j}{\alpha}} \rightarrow j. $$ We deduce that, $$ C_\alpha(z - j)\ln\left(\frac{\alpha^{\frac{\alpha - j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right) \rightarrow \mathrm{sinc}(z - j)\ln(j). $$

Step 3 : convergence of the final sum. The hard part is that the sum of the $\mathrm{sinc}(z - j)\ln(j)$ doesn't converge absolutely. However, we do have convergence. Indeed, we have for all $j \in \mathbb{N}$, $$ \mathrm{sinc}(z - j) = \frac{\sin(\pi z - \pi j)}{\pi z - \pi j} = (-1)^j\frac{\sin(\pi z)}{\pi z}\frac{z}{z - j} = (-1)^j\mathrm{sinc}(z)\frac{z}{z - j}. $$ Therefore, $$ \left|\mathrm{sinc}(z - j)\ln(j) - (-1)^j\mathrm{sinc}(z)\frac{\ln(j)z}{j}\right| = \left|\mathrm{sinc}(z)z\right|\left|\frac{1}{z - j} - \frac{1}{j}\right|\ln(j) \leqslant C\frac{\ln(j)}{j^2}, $$ for some constant $C$ depending on $z$. The series of the $\frac{\ln(j)}{j^2}$ is absolutely convergent and the series of the $(-1)^j\frac{\ln(j)}{j}$ is convergent by an alternate sum argument. We deduce that the series of the $\mathrm{sinc}(z - j)\ln(j)$ converges. Moreover, we verify easily that the constant $C$ can be chosen locally uniformly with respect to $z$. It implies the local uniform convergence. In particular, $z \mapsto \sum_{j \geqslant 1} \mathrm{sinc}(z - j)\ln(j)$ is holomorphic on $\mathbb{C}$ i.e. it is an entire function.

Now, to conclude, we would need to bound the difference $|C_\alpha(z - j)\ln(\cdots) - \mathrm{sinc}(z - j)\ln(j)|$ by some $K(j,\alpha) \geqslant 0$ (ideally, $K$ would be defined in a neighborhood of each $z \in \mathbb{C}\backslash\mathbb{Z}$ to make sure we have local uniform convergence) such that $\sum_{j = 1}^{\alpha - 1} K(j,\alpha) \rightarrow 0$ when $\alpha \rightarrow +\infty$, but it seems very hard, especially for the $j$ close to $\alpha$.