What is $\, _4F_3\left(1,1,1,\frac{3}{2};\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right)$?

Question

I have been trying to evaluate the series $$\, _4F_3\left(1,1,1,\frac{3}{2};\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right) = 1.133928715547935...$$ using integration techniques, and I was wondering if there is any simple way of finding a closed-form evaluation of this hypergeometric series. What is a closed-form expression for the above series?

Answer 1

A complete answer now.

If we exploit the identities $$\frac{4^n}{(2n+1)\binom{2n}{n}}=\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx \tag{1}$$ $$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\frac{1}{2}\sum_{n\geq 1}\frac{4^n x^{2n-1}}{n\binom{2n}{n}},\qquad \arcsin^2(x)=\frac{1}{2}\sum_{n\geq 1}\frac{(4x^2)^n}{n^2\binom{2n}{n}}\tag{2}$$ we get: $$(\pi-2)=\int_{0}^{\pi/2}\theta^2\sin(\theta)\,d\theta = \frac{1}{2}\sum_{n\geq 1}\frac{16^n}{(2n+1)n^2 \binom{2n}{n}^2}=\frac{1}{2}\sum_{n\geq 0}\frac{16^n}{(2n+3)(2n+1)^2\binom{2n}{n}^2} $$ and in a similar way: $$\begin{eqnarray*}\frac{7\pi}{9}-\frac{40}{27}=\int_{0}^{\pi/2}\theta^2\sin^3(\theta)\,d\theta=\frac{1}{2}\sum_{n\geq 1}\frac{4^n 4^{n+1}}{n^2 (2n+3)\binom{2n}{n}\binom{2n+2}{n+1}}\end{eqnarray*}$$ If we integrate $\arcsin^2(x)$ and exploit $(1)$, we get: $$ \sum_{n\geq 1}\frac{16^n}{(2n+1)^2 n^2 \binom{2n}{n}^2} = 4(\pi-3) $$ and maybe it is enough to integrate $\arcsin^2(x)$ once more to get a closed expression for the series of interest: $$ \sum_{n\geq 0}\frac{16^n}{(2n+3)^3(2n+1)^2\binom{2n}{n}^2}. $$ In such a case it appears a dependence on a dilogarithm, arising from the primitive of $\frac{\arcsin x}{x}\sqrt{1-x^2}$. At the moment I do not know if that is manageable or not, I have to carry out further experiments. Probably a logarithm appears from $\int_{0}^{\pi/2}\theta\cot(\theta)\,d\theta=\frac{\pi}{2}\log(2).$

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Now that the path to the answer is a bit more clear, let us put $(1)$ and $(2)$ in a slightly more convenient way: $$ \int_{0}^{\pi/2}\sin(x)^{2n+3}\,dx = \frac{4^{n}(2n+2)}{(2n+3)(2n+1)\binom{2n}{n}}\tag{1bis}$$ $$\arcsin^2(x)=\frac{1}{2}\sum_{n\geq 0}\frac{4^{n+1} x^{2n+2}}{(2n+2)(2n+1)\binom{2n}{n}}\tag{2bis}$$ If we integrate both sides of $(2\text{bis})$ we get: $$ -2x+2\sqrt{1-x^2}\arcsin(x)+x\arcsin^2(x) = \frac{1}{2}\sum_{n\geq 0}\frac{4^{n+1} x^{2n+3}}{(2n+3)(2n+2)(2n+1)\binom{2n}{n}}\tag{3}$$ We just have to gain an extra $\frac{1}{(2n+3)}$ factor. For such a purpose, we divide both sides of $(3)$ by $x$ and perform termwise integration again, leading to: $$ -4x+2\sqrt{1-x^2}\arcsin(x)+x\arcsin^2(x)+2\int_{0}^{\arcsin(x)}\frac{u\cos^2(u)}{\sin(u)}\,du\\= \frac{1}{2}\sum_{n\geq 0}\frac{4^{n+1} x^{2n+3}}{(2n+3)^2(2n+2)(2n+1)\binom{2n}{n}}\tag{4}$$ Now we evaluate both sides of $(4)$ at $x=\sin\theta$ and exploit $(1\text{bis})$ to perform $\int_{0}^{\pi/2}(\ldots)\, d\theta$.
That leads to: $$ \sum_{n\geq 0}\frac{16^n}{(2n+3)^3(2n+1)^2\binom{2n}{n}^2}=(\pi-4)+\int_{0}^{\pi/2}\int_{0}^{\theta}\frac{u\cos^2(u)}{\sin(u)}\,du\,d\theta\tag{5} $$ and we may start buying beers, since the last integral boils down to $\int_{0}^{\pi/2}\int_{0}^{\theta}\frac{u}{\sin u}\,du\,d\theta$, that is well-known. We get: $$\boxed{\begin{eqnarray*}\phantom{}_4F_3\left(1,1,1,\frac{3}{2};\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right)&=&27\sum_{n\geq 0}\frac{16^n}{(2n+3)^3 (2n+1)^2 \binom{2n}{n}^2}\\&=&\color{red}{\frac{27}{2}\left(7\,\zeta(3)+(3-2K)\,\pi-12\right)}\end{eqnarray*}}\tag{6}$$ where $K$ is Catalan's constant. Please, do not ask me to do the same for other values of $\phantom{}_4 F_3$.
However, this instantly goes in my best of collection.

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Addendum (15/08/2017) This result, together with another interesting identity relating $\phantom{}_4 F_3$ and $\text{Li}_2$, is going to appear on Bollettino UMI. You may have a glance at it on Arxiv.