Is there a closed form for $\int_0^1\,_3F_2(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43;x)dx$?

Question

Is there a closed form evaluation for the integral $$J=\int_0^1 {_3}F_2(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43;x)dx?$$

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Context: I have been investigating integrals of the form $$e_{p,q}^{n,m}\left({\begin{array}ca_1,..., a_p\\b_1,...,b_q\end{array}}\right)=\int_0^1x^n\left[{_p}F_q\left({\begin{array}ca_1,..., a_p\\b_1,...,b_q\end{array}};x\right)\right]^mdx.$$ Obviously there is no reason to expect a general closed form, but I have found the following: $$E_1=e_{2,1}^{1,2}(\tfrac12,1;2)=12-16\ln2,\tag1$$ and $$E_2=e_{2,1}^{1,3}(\tfrac13,\tfrac23;\tfrac32)=\frac{27}{32}.\tag2$$ I found these through applying the Lagrange inversion theorem to the functions $x^2-x$ and $x^3-x$, respectively. The proofs are below.

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Theorem. We have the explicit evaluation $$\int_0^1x\left[{_2}F_1(\tfrac12,1;2;x)\right]^2dx=12-16\ln2.\tag{1'}$$ Proof. Using the Lagrange inversion theorem, the function $g(x)$, satisfying $$g(x)^2-g(x)=x,$$ is given by the hypergeometric series $g(x)=-x\,{_2}F_1(\tfrac12,1;2;-4x)$, for $x\in[-1/4,\infty)$. Thus, the function $F(x)={_2}F_1(\tfrac12,1;2;x)$ satisfies $$xF(x)^2=4(F(x)-1),$$ for $x\in(-\infty,1]$. Thus $$E_1=\int_0^1xF(x)^2dx=-4+4\int_0^1F(x)dx.$$ Then using $$_2F_1(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1\frac{t^{b-1}(1-t)^{c-b-1}}{(1-zt)^a}dt,$$ we have $$F(x)=\frac2\pi\int_0^1\sqrt{\frac{1-t}{t}}\frac{dt}{1-xt}=\frac4\pi\int_0^\infty\frac{t^2dt}{(t^2+1)(t^2+1-x)}=\frac{2}{1+\sqrt{1-x}}.$$ It is then not too difficult to show that $$\int_0^1F(x)dx=\int_0^1\frac{2dx}{1+\sqrt{1-x}}=4-4\ln2,$$ which gives $(1')$ and thus $(1)$. $\square$

Theorem. We have the explicit evaluation $$\int_0^1 x\left[{_2}F_1(\tfrac13,\tfrac23;\tfrac32;x)\right]^3dx=\frac{27}{32}.\tag{2'}$$ Proof. The Lagrange inversion theorem gives $g(x)^3-g(x)=x$, for $$g(x)=-x{_2}F_1(\tfrac13,\tfrac23;\tfrac32;\tfrac{27}{4}x^2),\qquad |x|<\frac{2}{3\sqrt3}.$$ Setting $F(x)={_2}F_1(\tfrac13,\tfrac23;\tfrac32;x)$, we have $$4xF(x)^3=27(F(x)-1),$$ and thus $$E_2=\int_0^1xF(x)^3dx=\frac{27}{4}\left(-1+\int_0^1F(x)dx\right).$$ Then from here and here, we have $$\int_0^1F(x)dx=\left(\frac32\right)^2\left(\frac{\Gamma(\tfrac12)\Gamma(\tfrac32)}{\Gamma(\tfrac56)\Gamma(\tfrac76)}-1\right)=\frac{9}{8},$$ which is equivalent to $(2)$ and $(2')$. $\square$

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Here is my work on the current problem.

As you may have guessed, we use the Lagrange inversion theorem to see that $g(x)^4-g(x)=x$, where $$g(x)=-x{_3}F_2(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43;-\tfrac{4^4}{3^3}x^3).$$ Setting $F(x)={_3}F_2(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43;x)$, we have $$xF(x)^4=\frac{4^4}{3^3}(F(x)-1),$$ so that $$e_{1,4}^{3,2}(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43)=\int_0^1xF(x)^4dx=\frac{4^4}{3^3}(J-1),$$ where $J$ is the integral in the title. It may or may not help, but we can use integral representations of $_pF_q$ to get $$J=\int_0^1 F(x)dx=\frac{\Gamma(\tfrac43)\Gamma(\tfrac23)}{\pi\Gamma(\tfrac7{12})}\int_0^1\int_0^1\int_0^1\frac{dtdxdz}{x^{1/2}(1-x)^{1/2}t^{1/4}(1-t)^{5/12}(1-txz)^{1/4}}.$$ According to Desmos, the value of $J$ is roughly $J\approx 1.08494289471$, but for some reason I can't get wolfram alpha to get me anything better. Is there any way to evaluate $J$? Thanks :)

Answer 1

I did it! We see that $F(x)$ is of the form $$\,{_{k+1}}F_{k}\left(\tfrac{1}{k+1},\tfrac{2}{k+1},...,\tfrac{k}{k+1};\tfrac{2}{k},\tfrac{3}{k},...,\tfrac{k-1}{k},\tfrac{k+1}{k};x\right),$$ so we have $$F\left(-r_3\cdot(m(m^3-1))^3\right)=\frac{1}{1-m^3},$$ where $r_3=4^4/3^3$. The above is from eq. (25) here, with $k=3$. Thus, we may write $x=-r_3m^3(m^3-1)^3$, and we have $$J=\int_0^1F(x)dx=\int_{x=0}^{x=1}\frac{1}{1-m^3}\frac{dx}{dm}dm=3r_3\int_0^\alpha m^2(m^3-1)(4m^3-1)dm,$$ where $\alpha$ is the real root of $r_3\alpha^3(\alpha^3-1)^3+1=0$. Thus, $$J=\frac{2^7}{3^4}\alpha^3(8\alpha^6-15\alpha^3+6).$$ However, we have $f(\alpha)^3=-1/r_3$, so that $\alpha=g(-r_3^{-1/3})=r_3^{-1/3}F(1)$. But according to W|A, we have $F(1)=4/3$, so that $\alpha=\frac{1}{\sqrt[3]{4}}$. Thus

$$J=\frac{88}{81},$$ and thus $$e_{3,2}^{1,4}(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43)=\frac{1792}{2187}.$$