Integral ∫10lnln3F2(14,12,34;23,43;x)dx\int_0^1\ln\ln\,_3F_2\left(\frac{1}{4},\frac{1}{2},\frac{3}{4};\frac{2}{3},\frac{4}{3};x\right)\,dx

I encountered this scary integral
10lnln3F2(14,12,34;23,43;x)dx
where 3F2 is a generalized hypergeometric function
3F2(14,12,34;23,43;x)=1+8π33n=128n Γ(4n)Γ(n) Γ(n+1) Γ(n+23) Γ(n+43)xn.
I do not hope much that anything can be done with it, but maybe somebody got an idea how to find a closed form for this integral.

Answer

Let us denote
F(x)=3F2(14,12,34;23,43;x),
and the integral in question
I=10lnlnF(x)dx.
I am still working on some details in my proof, but quite amazingly, yes, there is a closed form for this integral, albeit a non-elementary one:
I=25627(Ei(4ln34)Ei(3ln34)ln43)+lnln43,
where Ei(x) is the exponential integral:
Ei(x)=xettdt.


Proof sketch:

The crucial fact is that on the interval x(0,1) the function xF(x) is an inverse of the function z256(z1)27z4, so that the following equalities hold:
F(256(z1)27z4)=z,
27x256F(x)4+1=F(x).
This can be seen from comparing coefficients in corresponding series expansions (for details please refer to the paper M.L. Glasser, Hypergeometric functions and the trinomial equation).

It means the function F(x) on the interval x(0,1) can be expressed in radicals, as I mentioned in my comment above. But, actually, I do not need that expansion.

Note that the integrand lnlnF(x) is negative on the integration interval, and monotonically increasing from to lnln431.245899…, so we can think of an absolute value of I as the area above the function graph. Now, if we transpose this picture, we will see that it equals to the integral of the inverse function plus the rectangular area to the right of it:
|I|=lnln43256(eet1)27e4etdt+0lnln43dt.
Using the identity
eet1e4etdt=Ei(3et)Ei(4et)+C
(that can be checked by taking derivatives from both parts), we get the final result.

Attribution
Source : Link , Question Author : Nik Z. , Answer Author : Vladimir Reshetnikov

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