$\Large{\text{Introduction:}}$
Here is a link to the Inverse Regularized Incomplete Gamma function used in this problem.
For simplicity, let the unit interval be expressed as $I=[0,1]$: $$\mathfrak{Q}=\int_I Q^{-1}(x,x)dx=.1984234858….$$ Here is a visual representation of this constant. Note that the constant is the area over the domain for $$x\in\Bbb R,(0,0):$$
$\large{\text{Identities}}$:
\begin{gather*} x=Q\bigl(x,Q^{-1}(x,x)\bigr)=Q^{-1}(x,Q(x,x)) \\ \Gamma\bigl(x,Q^{-1}(x)\bigr)=x!=\Gamma(x+1)=x\Gamma(x)=\int_{Q^{-1}(x,x)}^{\infty}t^{x-1}e^{-t}dt=\int_0^{\infty}t^xe^{-t}dt \\ \Gamma\left(x,f(x)\right)=\Gamma(x+1), f(x)=Q^{-1}(x,x). \end{gather*}
$\large{\text{Sum Representation:}}$
Here is the function’s sum representation up to 10 terms. Please look here for patterns as denominator looks easy to find.
$$ Q^{-1}(x,x)=((1-x)x!)^\frac1x+\frac {((1-x)x!)^\frac2x}{x+1}+ \frac {(3x+5)((1-x)x!)^\frac3x}{2(x+1)^2(x+2)}+O\left((x-1)^\frac4x \right)$$
Therefore:$$\int_0^1 Q^{-1}(x,x)dx=\sum_{n=1}^\infty\int_0^1C_{n,x}(1-x)^{n \over x} x!^{n\over x}dx$$
Here is graphical analytic continuation.
I need to find a formula for $C_x$ to solve this problem. $$C_{1,x}=1,C_{2,x}=\frac{1}{x+1},C_{3,x}=\frac{3x+5}{2(x+1)^2(x+2)}$$
An exact solution is preferred. Please correct me with feedback!
I think I found an exact form for the first term…
$\frak Q$:
Note this answer needs help from you to find the full solution.
Amazingly there is another identity found based on this similar question about
which shows it can actually solved for. Here are some gamma functions for context.
It actually is possible to integrate $Q^{-1}(x,x)$ many ways, but one way which does not use a general definite integration formula is harder to find. I suspect a closed form is impossible using currently accepted functions. Let’s start with the series representation of $Q^{-1}(x,x)$. This is the notation I will use called $C_{n,x}$ and its definition:
$$Q^{-1}(x,x)=Q^{-1}(x,x,Q(x,x)-x)=((1-x)x!)^\frac1x+\frac {((1-x)x!)^\frac2x}{x+1}+ \frac {(3x+5)((1-x)x!)^\frac3x}{2(x+1)^2(x+2)}+O\left((x-1)^\frac4x \right)= C_{1,x}(1-x)^{1\over x} x!^{1\over x} + C_{2,x}(1-x)^{2 \over x} x!^{2\over x} + C_{3,x}(1-x)^{3 \over x} x!^{3\over x}+…= \sum_{n=1}^\infty C_{n,x}(1-x)^{n \over x} x!^{n\over x} $$
Therefore:$$\int_0^1 Q^{-1}(x,x)dx=\sum_{n=1}^\infty\int_0^1C_{n,x}((1-x)x!)^{\frac nx} dx$$
The next step is to use a Taylor series of $((1-x)x!)^{\frac nx}$ at x=0 which has the amazing form of the following with Apéry’s constant, the Euler-Mascheroni constant, π, and e. Feel free to simplify it as it appears to have a higher degree polynomial with each term. Notice how $\pi$ and ζ($3$) also exhibit this polynomial behavior with each increasing m:
$$\sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac nx}\right]_{x=0}x^m}{m!}= e^{-(1 + γ ) n}\left[1 + \frac{π^2 - 6} {12} n x +\frac { nx^2 ((π^2 - 6)^2 n - 96 (ζ(3) +1)) } {288} + \frac{n x^3 (5 (π^2 - 6)^3 n^2 -1440 (π^2 - 6) n (ζ(3) +1)+ 144 (π^4 - 90))}{51840} +\frac{ n x^4 (5 (π^2 - 6)^4 n^3 -2880 (π^2 - 6)^2 n^2 (ζ(3) +1) + 576 n (-90 π^2 - 6 π^4 + π^6 + 60 (13 +8ζ(3) + 4ζ^2(3))) -497664 (ζ(5) ))}{2488320} + O(x^5)\right] $$
This series converges to $((1-x)x!)^{\frac nx},0\le x\le 1$, so here is the constant so far:
$$\int_0^1 Q^{-1}(x,x)dx= \sum_{n=1}^\infty\int_0^1C_{n,x} \sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac nx}\right]_{x=0}x^m}{m!} dx = \sum_{n=1}^\infty\sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac nx}\right]_{x=0}}{m!} \int_0^1 C_{n,x} x^m dx$$
Here is where I need to find an expression for $C_{n,x}$. I already know the first few terms using this generalized power series. Here is the constant which should hopefully converge using the n=1,2,3 cases of $C_{n,x}$. Note I used Wolfram Functions and Wolfram Alpha to evaluate and simplify. Note that $H_y$ are the Harmonic numbers and Ф(a,b,c) is the Lerch Transcendent/Zeta function.
Note that I could have integrated up to 4 or more terms, but this becomes tedious and has more interesting complicated functions:
$${\frak Q}=\int_0^1 Q^{-1}(x,x)dx=\int_0^1 P^{-1}(x,x)=\int_0^1 Q^{-1}(x,x,Q(x,x)-x)dx=\int \sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac 1x}\right]_{x=0}}{m!} \int_0^1 x^m +\sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac 2x}\right]_{x=0}}{m!} \int_0^1\frac{1}{x+1} x^m +\sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac 3x}\right]_{x=0}}{m!} \int_0^1\frac {(3x+5)}{2(x+1)^2(x+2)} x^m + \sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac 4x}\right]_{x=0}}{m!} \int_0^1\frac{8x^2+33x+31}{3(x+1)^3(x+2)(x+3)}x^m+…=\sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac 1x}\right]_{x=0}}{(m+1)!} +\frac12 \sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac 2x}\right]_{x=0}}{m!} \left(H_\frac m2-H_\frac{m-1}{2}\right)+ \frac14 \sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac 3x}\right]_{x=0}}{m!}\left((2 m - 1) H_\frac m2 - (2 m - 1) H_\frac{m -1}2 -Ф\left(-\frac 12,1,m+1\right)+2\right)+…= \sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac 1x}\right]_{x=0}+ \left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac 2x}\right]_{x=0}+ \left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac 3x}\right]_{x=0}+…}{m!} \left[ \frac{1}{m+1}+\frac12 \left(H_\frac m2-H_\frac{m-1}{2}\right)+ \frac14\left((2 m - 1) H_\frac m2 - (2 m - 1) H_\frac{m -1}2 -Ф\left(-\frac 12,1,m+1\right)+2\right)+…\right]=.1984234858… $$
If this formula works, then what is an explicit form for $C_{n,x}$? An explicit formula for $\sum_{m=0}^\infty\frac{\left[\frac{d^m}{dx^m} ((1-x)x!)^{\frac nx}\right]_{x=0}x^m}{m!}$ is also appreciated as its expansion seems to be a double sum of n and x. Any other simplifications are also appreciated.
This leads to the fact that an amazing approximation for $\frak Q$ is the following and proves the series representation of the constant. Note that the rest of the $n\ge 1$ terms will add on smaller values making the approximation better each time:
$${\frak Q}= .19842…≈ \frac{588 + 24 π^2 + π^4 - 96 ζ(3)}{864 e^{γ+1}}=.19289…$$
Note there may be a typo. I hope this shows you that a general Taylor series is all you need to integrate such problems with inverse gamma type functions. Please correct me and give me feedback!
$\mathcal I$:
The real purpose of this question was to find a convergent interval found by @qifeng618:
$$\mathcal I=\int_0^1 \text I^{-1}_x(x,x)=.45697517...$$
which does graphically converge on $0<x\le 2$ from the question.
The area is the area over the function’s domain for $x\in\Bbb R$ Here is the graph of this function:
Here is the definition of the function and its properties along with some various beta functions:
$$x=\text I_{I^{-1}_x(y,z)}(y,z)= \text I_{\text I^{-1}_x(x,x)}(x,x)=\frac{\int_0^{\text I^{-1}_x(x,x)}\left(t-t^2\right)^{x-1}dt}{\text B(x,x)}\implies x \text B(x)= \text B_{\text I^{-1}_x(x,x)}(x,x)= \int_0^{\text I^{-1}_x(x,x)}\left(t-t^2\right)^{x-1}dt$$
See this link to derive more results from the definition of the function. I did find this possible form.
Let’s actually begin solving the integral with this series expansion for the Inverse of the Regularized Incomplete Beta function. See the question where @qifeng618 answers well for the first term approximation. Here appears the definition of our $B_{n,x}$ coefficients unrelated to other functions:
$$\text I^{-1}_x(x,x)\mathop=^{x>0}(x^2 \text B(x,x))^\frac 1 x+\frac{x-1}{x+1}(x^2 \text B(x,x))^\frac2x+\frac{2(x-1)(x^2+x-1)}{(x+1)^2(x+2)} (x^2 \text B(x,x))^\frac 3 x+…=B_{1,x} (x^2 \text B(x,x))^\frac 1 x +B_{2,x} (x^2 \text B(x,x))^\frac 2 x +B_{3,x} (x^2 \text B(x,x))^\frac 1 x +…$$
Therefore:
$$\mathcal I=\int_0^1 \text I^{-1}_x(x,x) =\int_0^1 \sum_{n=0}^\infty B_{n,x} (x^2 \text B(x,x))^\frac n x dx= \int_0^1 \sum_{n=0}^\infty B_{n,x} \frac{x!^\frac{2n}{x}}{Γ(2x)^\frac nx} dx $$
Let’s now use the Taylor series for $x!^\frac{2n}{x}$ at $x=0$ for easier integration. Our choice of integration will have consequences for future steps. I could easily have used a more complex Taylor Series, but it would be too complicated to find a formula for. It would also be impractical. Here is the series converging almost certainly for $0\le x\le 2$: …
Other constant:
Note we also have $$\int_0^.489 \text{erf}^{-1}(x,x)dx$$ as the Inverse of the Generalized Error function. This will be the place for the other main “constants”.
This is a work in progress for fun