Area under inverse gamma function branch from minimum to 1. Equivalent to $1-\int\limits_0^1x! dx=.077254…$ in closed form. I have an interesting question about an integral which follows from this question:

Here.

This is about the integral of the inverse gamma function, which is the solution of y to x=y!. This function will be denoted by $y=Π^{-1}(x)$ which is simply x! reflected around y=x. Here is the integral in question when we integrate over its domain. This constant will be denoted by Γ:
$$\mathrm{Γ=\int_{\mathrm{min}(x!),x>0}^1 Π^{-1}(x)dx=1-\int_0^1 x!dx⇔1-Γ=\int_I x! dx}$$
Here is the plot of the wanted figure. The minimum value in the lower bound is very close to $\frac{\sqrt{\pi}}{2}=.886...\ $:

$$\mathrm{Link!}$$

Of course we can just do $$\mathrm{Γ=1-\lim_{x\to \infty}\frac 1x \sum_{n=0}^x\frac nx!=.077254...}$$ for the area of the black figure, which is obvious of how to derive it. Here is what the unmodified area looks like which obviously is unrealistic to integrate:


Graph and sum

This will also give the area under the factorial/gamma function.
I will list out some other attempts soon enough. What is an alternate form of this boring Riemann Sum and Γ constant? This is simply based on the factorial function, the Desmos type and not the boring integer defined one, integrated over the unit interval denoted I. You could try doing the unmodified area if you want, but this probably is undefined or is infinite.As always, please correct me and give me feedback!
 A: Here is a possible answer for the integral of the gamma function. This then can be used to find “Γ” constant:
From source 1 with the nth derivative and source 2. Lets assume that the n=0 terms are by convention or limit defined. Here is a demo of the first series here:
1.$$\mathrm{Γ(x)=\sum_{n=0}^\infty \frac{Γ^{(n)}(1)x^{n-1}}{n!},|x|<1}$$
2.$$\mathrm{Γ^{(n)}(x)=\lim_{t\to 0} t^{-n}\sum_{m=0}^n(-1)^{n- m}\binom nm Γ(x+kt),n\in \Bbb N}$$
3.$$\mathrm {\int Γ(x)dx=\int \sum_{n=0}^\infty \frac{{\lim_{t\to 0} t^{-n}\sum_{k=0}^n(-1)^{n-k}\binom nk Γ(1+kt)}x^{n-1}}{n!}dx= \quad \lim_{t\to 0}\sum_{n=0}^\infty  \sum_{k=0}^n \binom nk \frac{{
(-1)^{n-k}Γ(1+kt)}x^n}{t^nnn!}= \sum_{n=0}^\infty \frac{Γ^{(n)}(1)x^n}{nn!},|x|<1}$$
Also, you can use the fact that $$\mathrm{Γ(x+y)=Γ(y)\sum_{n=0}^\infty \frac {Γ^{(n)}(y)x^n}{Γ(y)n!},y\not \in \Bbb Z\ and \ y\not \le 0}$$
Therefore:
$$\mathrm{\int Γ(x+y) dx=\int Γ(y)\sum_{n=0}^\infty \frac {Γ^{(n)}(y)x^n}{Γ(y)n!} dx= \sum_{n=0}^\infty \frac {Γ^{(n)}(y)x^{n+1}}{(n+1)!}= \quad \lim_{t\to 0} \sum_{n=0}^\infty \sum_{k=0}^n \binom nk\frac {(-1)^{n-k}Γ(y+kt)x^{n+1}}{t^n(n+1)!}, y\not \in \Bbb Z \ and \ y\not \le 0}$$
You can also use the classic Riemann Sum Definition:
$$\int_a^b f(x) dx=\lim_{n\to \infty} \sum_{k=1}^n f(x_k=a+kΔx)Δx, Δx=\frac {b-a}{n}$$
to find that:
$$\mathrm{\int_a^b Γ(x) dx=\lim_{n\to \infty} \frac{b-a}{n} \sum_{k=1}^n Γ\left(a+k\frac{b-a}{n}\right)}$$
There also exists this expansion for the reciprocal gamma function which can be used to find the Fransén Robinson Constant, aka F, as there does not appear to be a restriction on the series. I will take Wolfram’s Website’s word that it converges. $\mathrm x_0$ is any real value to produce a vertical shift.
$$\mathrm{F=\int_{\Bbb R^+}\frac {dx}{Γ(x)}= \frac1\pi \sum_{k=0}^\infty \left(\sum_{n=0}^k \frac{\pi^{k-n}(-1)^nΓ^{(n)}(1-x_0)}{n!(k-n)!} sin\left(\pi x_0+\frac{\pi(k-n)}{2}\right)\right)\frac{(x-x_0)^{k+1}}{k+1}  \mathop=^?\frac1\pi \sum_{k=0}^\infty \left(\sum_{n=0}^k \frac{\pi^{k-n}(-1)^n \lim_{t\to 0} t^{-n}\sum_{m=0}^n(-1)^{n- m}\binom nm Γ(1-x_0+kt)}{n!(k-n)!} sin\left(\pi x_0+\frac{\pi(k-n)}{2}\right)\right)\frac{(x-x_0)^{k+1}}{k+1}}$$
Here is another Beta Function based answer with the Pochhammer Symbol. Here is proof this really works
$$\mathrm{\int\frac{Γ(x)}{Γ(x+y)}dx=\frac1{Γ(y)}\left(ln(x)+\sum_{n=1}^\infty \frac{(1-y)_n}{n!}ln\left(1+\frac{x}{n}\right)\right)= \frac{lnx}{Γ(y)}+\sum_{n=1}^\infty\frac{sin(\pi y)(n-y)!}{\pi n!}ln\left(1+\frac{x}{n}\right),Re(y)>0}$$
Browsing Wolfram Mathworld made me find some generalized reciprocal factorial integral and generalized factorial integral functions. These are nicely named the Lamda function, Mu functions, and Nu functions. These may seem like they have the sole purpose of finding a closed form of an integral, but they should have some nice properties. Note I only list out a few:
$$\mathrm{λ(x,y)\mathop=^{def}\int_0^y \frac{t!}{x^t}dt, \frac{μ(x,y,z)\,y!}{x^z} \mathop=^{def}\int_0^\infty\frac{x^t t^y}{(t+z)!}dt,\frac{ν(x,z)}{x^z}=\int_0^\infty\frac{x^t}{(t+z)!}dt= \frac{μ(x,0,z)}{x^z},μ(x,y,0) \mathop=^{def}μ(x,y) ,ν(x,0) \mathop=^{def}ν(x)=μ(x,0,0)=μ(x,0)}$$
This means we can put our original problem and the Fransén-Robinson constant in closed form. Note the conventions for Γ in the question:
$$\mathrm{Γ \mathop=^{def}\int_{min\ Γ(x), x>0}^1 Γ^{-1}(x)dx=1-\int_0^1 t! dt=1-λ(1,1)= 0.0772540493193693948561195176…}$$
$$\mathrm{F \mathop=^{def} \int_{\Bbb R^+} \frac{dx}{Γ(x)}=\int_0^\infty \frac{1^t t^0dt}{Γ(t-1+1)}=μ(1,0,-1) =ν(1,-1)= 2.8077702420285193652215011865577…}$$
Please correct me and give me feedback!
