What is $\lim_{x\to \infty}f(x)=\lim_{x\to \infty}\int_{0}^{x}\prod_{n=1}^{\infty}\left(1-e^{-t-n}\right)dt-x+1$

Question

Working on the Gamma function I found the following:

Let

$$f(x)=\int_{0}^{x}\prod_{n=1}^{\infty}\left(1-e^{-t-n}\right)dt-x+1.$$

What is the limit

$$\lim_{x\to \infty}f(x)=\,?$$

As possible guess it's near by the absissa of the minimum for the factorial function which is $x\simeq 0.4616$.

Question:

Does it admits a closed form?

Answer 1

$$f(x)=(1-x)+\int_0^x \frac{e^t}{e^t-1} \left(e^{-t};\frac{1}{e}\right){}_{\infty }\,\,dt$$ does not seem to have a closed form.

Numerically, the convergence is quite fast $$\left( \begin{array}{cc} x & f(x) \\ 10 & 0.4620257654385160639655594317 \\ 15 & 0.4619995218587963453456974771 \\ 20 & 0.4619993450303188952646355559 \\ 25 & 0.4619993438388579577552602090 \\ 30 & 0.4619993438308299571055464707 \\ 35 & 0.4619993438307758648626600170 \\ 40 & 0.4619993438307755003919943864 \\ 45 & 0.4619993438307754979362103587 \\ 50 & 0.4619993438307754979196634161 \\ 55 & 0.4619993438307754979195519237 \\ 60 & 0.4619993438307754979195511724 \\ 65 & 0.4619993438307754979195511669 \\ 70 & 0.4619993438307754979195511671 \\ 75 & 0.4619993438307754979195511673 \\ \end{array} \right)$$

Nothing found using inverse symbolic calculators (interesting are the two expressions above and below this value in the $ISC$).

Answer 2

$\prod_{n=1}^{\infty}\left(1-e^{-t-n}\right)$ has its roots at $t=-n, t \in (-\infty, -1]$

Also $$\sin(x) = x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)$$

$$\frac{\sin(\pi x)}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right)$$

$$\frac{\sin(\pi e^{\frac{-x}{2}})}{\pi}e^{\frac{x}{2}} = \prod_{n=1}^\infty \left(1-\frac{e^{-x}}{n^2}\right)$$ The last problem seems to be how to convert $n^{2}$ to $e^n$. This is probably a modification of a infinite product, I found some similarities with the $\sin(x)$ one but it didnot work out in the end.

Answer 3

Not a closed form, but a rapidly (i.e. quadratically) convergent series, worth noting: $$\color{blue}{1-\frac1{e-1}+\frac1{2(e-1)(e^2-1)}-\frac1{3(e-1)(e^2-1)(e^3-1)}+\dots}$$ Indeed, we're evaluating $1-I(e^{-1})$ where, in this notation, $$I(q)=\int_0^\infty\big(1-(qe^{-t};q)_\infty\big)\,dt=\sum_{n=1}^\infty(-1)^{n-1}\frac{q^{n(n+1)/2}}{n(q;q)_n}$$ after using a series for $(x;q)_\infty$ from here, and termwise integration.