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.

Answer 2

General Principle. Let $A$ (resp. $M, N, B$) be a vector with all components in $\mathbb Z/2$ (resp. $\mathbb N, \mathbb N, \mathbb C$), $A, M$ and $B, N$ are of same length, $S, T$ vectors that met one of five following conditions ($k,m,n,i,j\in\mathbb Z$):

$$\color{blue}{0.\ S=\{k\},\ T=\emptyset}\ \ \ \ \color{green}{1.\ S=\{k+1/2\},\ T=\emptyset}\ \ \ \ \color{purple}{2.\ S=\{k,m\},\ T=\{n+1/2\}}$$ $$\color{red}{3.\ S=\{k+1/2, m+1/2\},\ T=\{n\}}\ \ \color{orange}{4.\ S=\{k,m,n\},\ T=\{i+1/2,j+1/2\}}$$

Then the hypergeometric series $\, _{q+1}F_q(S,A,B;T,A+M,B-N;1)$, whenever convergent and non-terminating, is expressible via level $4$ MZVs. OP's series belongs to case $4$ and is of low weight, thus solved without much difficulty. For the statement's proof and various examples, see Theorem $1$ here. To show its power we illustrate a

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$_4F_3$ table. One may generate an infinitude of $_4F_3$ with half-integer parameters based on principle above. The table below consists of all known $_4F_3$ with $z=1$ and all parameters in $\{1/2,1,3/2,2\}$ that has MZV or Gamma closed-form.

• $\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=\frac{\pi ^3}{48}+\frac{1}{4} \pi \log ^2(2)$
• $\small\pi \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2},\frac{3}{2};1\right)=-16 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{3 \pi ^3}{8}+\frac{1}{2} \pi \log ^2(2)$
• $\small\pi \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},2;1\right)=-8 C-32 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{3 \pi ^3}{4}+4+\pi \log ^2(2)$
• $\small\, _4F_3(1,1,1,1;2,2,2;1)=\zeta (3)$
• $\small\, _4F_3\left(1,1,1,1;\frac{3}{2},\frac{3}{2},2;1\right)=2 \pi C-\frac{7 \zeta (3)}{2}$
• $\small\, _4F_3\left(1,1,1,1;\frac{1}{2},2,2;1\right)=\frac{7 \zeta (3)}{4}+\frac{\pi ^2}{2}-\frac{1}{2} \pi ^2 \log (2)$
• $\small\, _4F_3\left(1,1,1,1;\frac{3}{2},2,2;1\right)=\frac{1}{2} \pi ^2 \log (2)-\frac{7 \zeta (3)}{4}$
• $\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=\frac{7 \zeta (3)}{8}$
• $\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{3}{2},2;1\right)=-\pi +2+\pi \log (2)$
• $\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1;2,2,2;1\right)=8-\frac{16 \Gamma \left(\frac{3}{4}\right) \Gamma \left(\frac{7}{4}\right)}{\pi \Gamma \left(\frac{5}{4}\right)^2}$
• $\small\pi \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1;\frac{3}{2},2,2;1\right)=16 C-24+4 \pi$
• $\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},2;1,\frac{3}{2},\frac{3}{2};1\right)=\frac{\pi }{4}+\frac{1}{4} \pi \log (2)$
• $\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},2;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=\frac{7 \zeta (3)}{16}+\frac{\pi ^2}{16}$
• $\small\, _4F_3\left(1,1,1,\frac{3}{2};2,2,2;1\right)=\frac{\pi ^2}{3}-4 \log ^2(2)$
• $\small\, _4F_3\left(1,1,1,\frac{1}{2};2,2,2;1\right)=-\frac{\pi ^2}{3}+8+4 \log ^2(2)-8 \log (2)$
• $\small\, _4F_3\left(1,1,1,\frac{1}{2};2,2,\frac{3}{2};1\right)=4 \log (2)-\frac{\pi ^2}{6}$
• $\small\, _4F_3\left(1,1,1,\frac{1}{2};2,\frac{3}{2},\frac{3}{2};1\right)=4 C-\frac{\pi ^2}{4}$
• $\small\, _4F_3\left(1,1,1,\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=\frac{7 \zeta (3)}{2}-\pi C$
• $\small\, _4F_3\left(\frac{3}{2},\frac{3}{2},\frac{3}{2},1;2,2,2;1\right)=\frac{8 \pi }{\Gamma \left(\frac{3}{4}\right)^4}-8$
• $\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=4 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{\pi ^3}{32}-\frac{1}{8} \pi \log ^2(2)$
• $\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},2;1\right)=\frac{\pi ^2}{4}-2 \log (2)$
• $\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},2,2;1\right)=2 \pi -8+4 \log (2)$
• $\small\pi \, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;2,2,2;1\right)=-32 C-16 \pi +48+16 \pi \log (2)$
• $\small\pi \, _4F_3\left(1,1,\frac{3}{2},\frac{3}{2};2,2,2;1\right)=16 \pi \log (2)-32 C$
• $\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},1,2;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=C+2 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{\pi ^3}{64}-\frac{1}{16} \pi \log ^2(2)$
• $\small\pi \, _4F_3\left(1,1,\frac{1}{2},\frac{3}{2};2,2,2;1\right)=32 C+8 \pi -16-16 \pi \log (2)$

To elaborate its full power we illustrate more

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Higher weight examples (one for each case).

• $\small \, _7F_6\left(\{1\}_6,\frac{3}{2};\{2\}_3,\{\frac52\}_3;1\right)=1512 \pi C+2592 \pi \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+3456 \pi \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-2592 \text{Li}_4\left(\frac{1}{2}\right)-1728 \text{Li}_5\left(\frac{1}{2}\right)-3024 \zeta (3)+\frac{5859 \zeta (5)}{4}-\frac{81}{8} \pi \zeta \left(4,\frac{1}{4}\right)+\frac{81}{8} \pi \zeta \left(4,\frac{3}{4}\right)-\frac{369 \pi ^4}{10}\\ \scriptsize-1620 \pi +4536+\frac{72 \log ^5(2)}{5}-108 \log ^4(2)-6 \pi ^2 \log ^3(2)+27 \pi ^2 \log ^2(2)+\frac{123}{5} \pi ^4 \log (2)$

• $\small \, _7F_6\left(\frac{1}{2},1,\{\frac54\}_5;\frac{3}{2},\{\frac94\}_5;1\right)=-\frac{3125 C}{81}-\frac{96875 \zeta (5)}{96}-\frac{21875 \zeta (3)}{216}+\frac{756250}{243}-\frac{3125 \pi ^2}{648}-\frac{3125 \pi ^4}{864}-\frac{3125 \pi ^3}{864}-\frac{3125 \pi }{972}-\frac{15625 \pi ^5}{4608}-\frac{3125}{486} \log (2)+\frac{3125 }{2304}\left(\zeta \left(4,\frac{3}{4}\right)-\zeta \left(4,\frac{1}{4}\right)\right)$

• $\small \, _8F_7\left(\{\frac12\}_4,\frac{7}{6},\frac{5}{4},\frac{4}{3},\frac{3}{2};\frac{1}{6},\frac{1}{4},\frac{1}{3},\{\frac52\}_4;1\right)=\frac{2835 \pi \zeta (3)}{32}-\frac{17739 \pi }{128}-\frac{1593 \pi ^3}{512}+\frac{945}{16} \pi \log ^3(2)-\frac{4779}{128} \pi \log ^2(2)+\frac{945}{64} \pi ^3 \log (2)-\frac{3645}{64} \pi \log (2)$

• $\small \, _8F_7\left(\{\frac12\}_4,1,1,\frac{4}{3},\frac{5}{3};\frac{1}{3},\frac{2}{3},\{\frac32\}_4,\frac{5}{2};1\right)=-\frac{3}{8} S+\frac{3}{8} T-\frac{105 C}{64}+\frac{105}{16} \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{3}{4} \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-3 \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{3 \zeta \left(4,\frac{3}{4}\right)}{2048}-\frac{3 \zeta \left(4,\frac{1}{4}\right)}{2048}+\frac{35 \pi ^5}{8192}+\frac{105}{128}-\frac{105 \pi ^3}{2048}+\frac{1}{512} \pi \log ^4(2)+\frac{1}{256} \pi \log ^3(2)+\frac{3 \pi ^3 \log ^2(2)}{1024}-\frac{105}{512} \pi \log ^2(2)+\frac{3 \pi ^3 \log (2)}{1024}$

• $\small \pi \, _7F_6\left(\{-\frac12\}_2,\{1\}_5;\{2\}_6;1\right)=-\frac{2560}{9} S+\frac{9728}{27} T-\frac{47104 C}{243}-\frac{14336}{27} \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{32768}{27} \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{16384}{9} \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{256 \pi \zeta (3)}{27}-\frac{64}{9} \pi \zeta (3) \log (2)+\frac{32 \zeta \left(4,\frac{1}{4}\right)}{9}-\frac{32 \zeta \left(4,\frac{3}{4}\right)}{9}+\frac{4}{27} \zeta \left(4,\frac{1}{4}\right) \log (2)-\frac{4}{27} \zeta \left(4,\frac{3}{4}\right) \log (2)+\frac{25 \pi ^5}{9}+\frac{112 \pi ^3}{9}-\frac{46784 \pi }{729}+\frac{117248}{729}-\frac{1}{9} 32 \pi \log ^4(2)+\frac{512}{27} \pi \log ^3(2)+\frac{16}{3} \pi ^3 \log ^2(2)-\frac{448}{9} \pi \log ^2(2)-\frac{128}{9} \pi ^3 \log (2)+\frac{23552}{243} \pi \log (2)$

Here $S,T$ denotes $\Im \sum_{k>j>0} \frac{i^k}{k^4 j},\ \ \Im \sum_{k>j>0} \frac{i^k (-1)^j}{k^4 j}$ repsectively, which are irreducible level $4$ MZVs. See paper linked above for more.