So far, I have found (p. 5) the following generating functions of the unsigned Stirling numbers of the first kind:

\begin{equation} \tag{1} \label{1} \sum_{l=1}^{n} |S_{1}(n,l)|z^{l} = (z)_{n} = \prod_{k=0}^{n-1} (z+k) =: g(z) , \end{equation} and \begin{equation} \tag{2} \label{2} \sum_{n=l}^{\infty} \frac{|S_{1}(n,l)|}{n!}z^{n} = (-1)^{l} \frac{\ln^{l}(1-z)}{l!} .\end{equation}

Instead of summing the latter expression over $n$, I'm curious whether there is a simpler or different expression of the latter generating sum when summed over the other indices:

\begin{equation} \tag{3} \label{3} f(z) := \sum_{k=1}^{n} \frac{|S_{1}(n,k)|}{k!}z^{k} . \end{equation} (One could also take the sum from $k=1$ to $k=\infty$, as $|S_{1}(n,k)| = 0$ when $k>n$.)

Work so far

Approach 1

We see that $(3)$ is the egf version of the ogf in $(1)$. So one of the ways I've tried to find $f(\cdot)$ is by converting the first equation to the third one by applying the inverse Laplace transform to $g(1/s)/s$.

From $(1)$, we see that $g(1/s)/s = \frac{(1/s)_{n}}{s}$. The tricky part of finding the inverse Laplace transform lies in the numerator. Therefore, I tried to express it in terms of other functions of which the inverse Laplace transform might be known.

For instance, note that the Chu-Vandermonde identity states: \begin{equation} \tag{4} \label{4} \ _2F_1\left(-n,b;c;1\right) = \frac{(c-b)_{n}}{(c)_{n}} . \end{equation}

Now, set $c=1$ and $b = 1 - 1/s$. Then:

\begin{equation} \tag{5} \label{5} \ _2F_1\left(-n,1-1/s;1;1\right) = \frac{(1/s)_{n}}{n!} =: h(s) . \end{equation}

If the inverse Laplace transform of $h(\cdot)$ would be known, then I would only have to multiply by $n!$ and convolve it with $ \{ \mathcal{L}^{-1} (1/s) \} (t) = u(t) $, where $u(t)$ is the unit step function.

However, I did not find an expression of the inverse Laplace transform of the hypergeometric function in $(5)$ in the “Tables of Laplace Transforms” by Oberhettinger and Badii (1973).

Approach 2

Another approach I tried is to note that \begin{equation} \tag{6} \label{6} g(1/s) = (1/s)_{n} = \frac{\Gamma(1/s + n)}{\Gamma(1/s)} . \end{equation}

Unfortunately, the inverse laplace transform of \begin{equation} \tag{7} \label{7} q(s) := \frac{\Gamma(s+a)}{\Gamma(s+b)} \end{equation} is only given by Oberhettinger and Badii (p. 308) in the case when $\Re(b-a) >0$, which is not the case here.

Approach 3

Finally, I tried rewriting $g(1/s)/s$ as follows:

\begin{align} g(1/s)/s &= \frac{1}{s^{2}} \cdot \frac{1+s}{s} \cdot \frac{1+2s}{s} \dots \frac{1+ns}{s} \\ &= \frac{ \prod_{k=1}^{n} (1+ks) }{s^{n+2}} \\ &= \frac{n! \prod_{k=1}^{n}\Big{(}s+\frac{1}{k} \Big{)} }{s^{n+2}}. \end{align}

Observe that $\{ \mathcal{L}^{-1} s^{-(n+2)} \}(t) = [ (n+1)! ]^{-1} t^{n+1} $, so we are left with finding the inverse Laplace transform of the numerator (and convolve afterwards). However, I have not been able to do so thusfar.


  1. Is the inverse Laplace transform of $\frac{(1/s)_{n}}{s}$ known, or can it be calculated somehow?
  2. Are there already any other expressions known for $f(\cdot)$ that can be found by other means than the ones I laid out so far?

N.B. this is a more elaborate version of a question I asked earlier on MSE.


Your question is weird because $\frac{(1/s)_{n}}{s}$ is a rational function vanishing at $\infty$, zero problem to apply the residue theorem to its inverse Laplace transform integral: $$\mathcal{L}^{-1}[\frac{(1/s)_{n}}{s}](t)=\operatorname{Res}(\frac{(1/s)_{n}}{s}e^{st},s=0) 1_{t >0}=\sum_{l=1}^{n} |S_{1}(n,l)| \frac{t^l}{l!} 1_{t >0} $$

  • $\begingroup$ Mathematica 12.2 finds it for concrete values of $n$ by Expand[Table[ InverseLaplaceTransform[Pochhammer[1/s, k]/s, s, t], {k, 1, 5}]] which results in $\left\{t,\frac{t^2}{2}+t,\frac{t^3}{6}+\frac{3 t^2}{2}+2 t,\frac{t^4}{24}+t^3+\frac{11 t^2}{2}+6 t,\frac{t^5}{120}+\frac{5 t^4}{12}+\frac{35 t^3}{6}+25 t^2+24 t\right\}$. $\endgroup$
    – user64494
    Apr 17 at 15:40
  • $\begingroup$ While I disagree with your assessment of the question, I'm grateful for your answer $\endgroup$
    – Max Muller
    Apr 17 at 21:18

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