I know this looks like a silly question, but please see the motivation.
$\large\text{Goal Integral:}$
Let me introduce the Racah/Wigner 6j symbols which are defined discretely because of a $\text{PhysicalQ}(\{j_1,j_2,j_3\},\{j_4,j_5,j_6\})$ function. Let’s use ignore that restriction in the bolded link to analytically extend the domain. The link also has a few definitions in terms of regularized hypergeometric functions and factorials, also seen in the link. What would happen if we tried to make the inputs central and integrate using this new definition?. Here I integrate term by term:
$$\int \left\{ \begin{matrix} x&x&x \\ x&x&x \\ \end{matrix}\right \}dx=\int \frac{(-1)^{4x} x!^6}{(3x+1)!^2}\sum_{n=0}^x\frac{(-1)^n(4x-n+1)!}{n!^3(x-n)!^4}dx=\pi\int\frac{\csc(4\pi x) x!^6}{(-1)^{4x}(3x+1)!^2}\sum_{n=0}^\infty \frac{dx}{(x-n)!^4n!^3(n-4x-2)!}\mathop=\pi\int \frac{\csc(4\pi x)x!^2\,_4\mathrm {\tilde F_3}(-x,-x,-x,-x,-4x-1,1,1,1)}{(3x+1)!^2}dx= \pi\int \frac{\csc(4\pi x)x!^2\,_4\text F_3(-x,-x,-x,-x,-4x-1,1,1,1)}{(3x+1)!^2(-4x-1)!}dx=\pi\sum_{n=0}^\infty\frac1{n!^3}\int\frac{\csc(4\pi x) x!^6}{(-1)^{4x} (3x+1)!^2 (x-n)!^4(n-4x-2)!}dx$$
This is the source for the main series representation ignoring the integer restriction.
$\large\text{Motivaton:}$
I am looking for a series representation solution for the integral for any non-degenerate interval of convergence. I also know that we are integrating a factorial, but you actually can integrate a factorial using the 2 argument Lambda function and Euler-Mascheroni Integrals $I_n$. Here is also a gamma function integral for variety or which I am sure about the interval of convergence
$$\int_0^y x! dx=λ(1,y),\int Γ(x)dx=C-\sum_{n=1}^\infty \frac{(-1)^n I_{n-1} (x-1)^n}{n!}=C+\sum_{n=1}^\infty \frac{\left(\frac {d^{n-1}}{dx^{n-1}} Γ(x)\right)_{x=1}\ (x-1)^n}{n!}$$
Here is the closed form of the central Wigner Coefficients /Racah 3j Symbol assuming an analytic continuation of the function without the $\text{PhysicalQ}(\{j_1,j_2,j_3\},\{j_4,j_5,j_6\})$ function constraint also seen in the link:
$$\int \left( \begin{matrix} x&x&x \\ x&x&x \\ \end{matrix}\right ) dx=\int(-1)^{2x}dx=C+\frac{i(-1)^{2x}}{2\pi}$$
The same goes for the Clebsh-Gordon Coefficients. If we ignore the $\text{PhysicalQ}(\{j_1,j_2,j_3\},\{j_4,j_5,j_6\})$ restriction in the link, then we get an analytic continuation. Let’s make the function central by making all arguments equal to one another using the Incomplete Gamma function for $x\ne \frac12$:
$$\int \langle x\,x\,x\,x|x\,x\,x\,x\rangle dx=\int (-1)^{3x} \sqrt{2x+1}dx=\frac{(1+i)Γ\left(\frac32,-3i\pi \left(x-\frac12\right)\right)}{(3\pi)^\frac32}$$
$\large\text{Conclusion:}$
So as you can see, each integral is possible, but the mystery of the Wigner 6j symbol integral remains unsolved. A closed form is probably impossible, so how can I find a series representation of the integral in the title? A series expansion like a Taylor expansion or Darboux formula or inversion formula work. Please correct me and give me feedback!