Proving $_4F_3\left(\frac{1}{2},\frac{3}{4},\frac{3}{4},1;\frac{5}{4},\frac{5}{4}, \frac{3}{2};1\right)=\frac{\Gamma^8(1/4)}{768\pi^3}$

Question

I have been trying to Prove the following Sum.

$$\frac{\Gamma^2(5/4)}{\Gamma^2(3/4)}\sum_{r=0}^{\infty}\left(\frac{1}{2r+1}\right)\frac{\Gamma^2(r+3/4)}{\Gamma^2(r+5/4)}=\frac{\Gamma^8(1/4)}{768\pi^3}$$

Or in terms of Pochhammer Symbol:

$$\sum_{r=0}^{\infty}\left(\frac{1}{2r+1}\right)\frac{(3/4)_r^2}{(5/4)_r^2}=\frac{\Gamma^8(1/4)}{768\pi^3}$$

One may convert it into a Hypergeometric Representation which gives:

$$_4F_3\left(\frac{1}{2},\frac{3}{4},\frac{3}{4},1;\frac{5}{4},\frac{5}{4}, \frac{3}{2};1\right)=\frac{1}{768\pi^3}\Gamma^8\left(\frac{1}{4}\right)$$

Also note that, $$\sum_{i=1}^3b_i-\sum_{i=1}^{4}a_i=1$$

I just mentioned the above because I heard it's a kind of Property of such Functions.

I don't see how a $8$th Power Gamma Term appears.

The following List at Functions.Wolfram doesn't have any similar ones nor am I able to use any formulas either.

One may use the following to Numerically Test it.

N[HypergeometricPFQ[{1/2,3/4,3/4,1},{5/4,5/4,3/2},1],300]-Gamma[1/4]^8/Pi^3*1/768

EDIT:

I was able to Convert it to an Integral: $$\int_{0}^{1}\int_{0}^{1}\frac{\ln(1+xy)-\ln(1-xy)}{\sqrt{xy(1-x^2)(1-y^2)}}dxdy=\frac{\Gamma^4(1/4)}{48}$$

Answer 1

We have the following formula, which transforms a well-poised $_5F_4$ into two $1$-balanced $_4F_3$, here all $_pF_q$ are assume to be at $z=1$: $$\begin{aligned}&{}_5 F_4\left(\begin{matrix}a,\quad b,\quad c,\quad d,\quad e\\1+a-b,\quad 1+a-c, \quad 1+a-d,\quad 1+a-e\end{matrix}\right) \\ &= \frac{\Gamma (a-b+1) \Gamma (a-c+1) \Gamma (a-d+1) \Gamma (a-b-c-d+1)}{\Gamma (a+1) \Gamma (a-b-c+1) \Gamma (a-b-d+1) \Gamma (a-c-d+1)} \times {}_4F_3\left(\begin{matrix}b, \quad c,\quad d,\quad \frac{a}{2}-e+1 \\ \frac{a}{2}+1, \quad d-a+b+c, \quad 1+a-e\end{matrix}\right) \\ &\quad +\frac{\Gamma \left(\frac{a}{2}+1\right) \Gamma (a-b+1) \Gamma (a-c+1) \Gamma (a-d+1) \Gamma (a-e+1) \Gamma (-a+b+c+d-1) \Gamma \left(\frac{3 a}{2}-b-c-d-e+2\right) }{\Gamma (a+1) \Gamma (b) \Gamma (c) \Gamma (d) \Gamma \left(\frac{a}{2}-e+1\right) \Gamma \left(\frac{3 a}{2}-b-c-d+2\right) \Gamma (2 a-b-c-d-e+2)} \\ &\quad \times {}_4F_3\left(\begin{matrix}a-b-c+1, \quad a-b-d+1,\quad a-c-d+1,\quad \frac{3 a}{2}-b-c-d-e+2 \\ a-b-c-d+2, \quad \frac{3 a}{2}-b-c-d+2, \quad 2 a-b-c-d-e+2\end{matrix}\right) \end{aligned}$$ this is a special case of very-well-poised $_7F_6$ reduction formula, which can be found in Bailey's Generalized hypergeometric series p.29. Another proof using Barne's integral can be found in p.44 of same book.

Let $(a,b,c,d,e) = (1,1,1/2,3/4,3/4)$, LHS actually reduces to $_4F_3$, one term on RHS reduces to $_3F_2$ that can be summed using Dixon's theorem. The form will be

$$_4F_3\left(\begin{matrix}\frac{1}{2},\frac{3}{4},\frac{3}{4},1 \\ \frac{5}{4},\frac{5}{4},\frac{3}{2}\end{matrix}\right)=\frac{\Gamma \left(-\frac{1}{4}\right) \Gamma \left(\frac{5}{4}\right) }{2 \Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{3}{4}\right)} {}_4F_3\left(\begin{matrix}\frac{1}{2},\frac{3}{4},\frac{3}{4},1 \\ \frac{5}{4},\frac{5}{4},\frac{3}{2}\end{matrix}\right)+\frac{\pi \Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{5}{4}\right)^3}{2 \Gamma \left(\frac{3}{4}\right)^4}$$

Solving this gives $${}_4F_3\left(\begin{matrix}\frac{1}{2},\frac{3}{4},\frac{3}{4},1 \\ \frac{5}{4},\frac{5}{4},\frac{3}{2}\end{matrix}\right) = \frac{\Gamma \left(\frac{1}{4}\right)^8}{768 \pi ^3}$$ as desired.