# What is the inverse Laplace transform of $\frac{(1/s)_{n}}{s}$?

Introduction

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

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

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:

$$$$\tag{3} \label{3} f(z) := \sum_{k=1}^{n} \frac{|S_{1}(n,k)|}{k!}z^{k} .$$$$ (One could also take the sum from $$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: $$$$\tag{4} \label{4} \ _2F_1\left(-n,b;c;1\right) = \frac{(c-b)_{n}}{(c)_{n}} .$$$$

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

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

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 $$$$\tag{6} \label{6} g(1/s) = (1/s)_{n} = \frac{\Gamma(1/s + n)}{\Gamma(1/s)} .$$$$

Unfortunately, the inverse laplace transform of $$$$\tag{7} \label{7} q(s) := \frac{\Gamma(s+a)}{\Gamma(s+b)}$$$$ is only given by Oberhettinger and Badii (p. 308) in the case when $$\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.

Questions

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: $$\mathcal{L}^{-1}[\frac{(1/s)_{n}}{s}](t)=\operatorname{Res}(\frac{(1/s)_{n}}{s}e^{st},s=0) 1_{t >0}=\sum_{l=1}^{n} |S_{1}(n,l)| \frac{t^l}{l!} 1_{t >0}$$
• 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 $\left\{t,\frac{t^2}{2}+t,\frac{t^3}{6}+\frac{3 t^2}{2}+2 t,\frac{t^4}{24}+t^3+\frac{11 t^2}{2}+6 t,\frac{t^5}{120}+\frac{5 t^4}{12}+\frac{35 t^3}{6}+25 t^2+24 t\right\}$. Apr 17, 2021 at 15:40