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

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?

• In other terms, you are asking for a closed form of $$\sum_{n\geq 0}\frac{27\cdot 16^n}{(2n+3)^3 (2n+1)^2 \binom{2n}{n}^2}.$$ Commented Feb 1, 2017 at 1:57
• It is interesting to point out that if the factor $(2n+3)^3$ was $(2n+3)^2$, it would not be too hard to tackle such series through Parseval's theorem and known Taylor series for $\arcsin(x)$ and $\arcsin(x)^2$. Commented Feb 1, 2017 at 2:04
• Namely $$\sum_{n\geq 0}\frac{16^n}{(2n+3)^2 (2n+1)^2 \binom{2n}{n}^2} = \pi-3.$$ Commented Feb 1, 2017 at 2:41
• Wow, that was unexpected. The question has an affirmative answer, such value of a hypergeometric function has a manageable closed form! Commented Feb 5, 2017 at 20:29
• @TitoPiezasIII I encountered the given hypergeometric series while in the process of researching an integral transform which may be used to evaluate new classes of infinite series involving harmonic numbers. Using this integral transform, I discovered that $$\sum_{n=1}^{\infty} \left( \frac{4^{n}}{n(2n+1)\binom{2n}{n}} \right)^{2}H_{2n}'$$ may be expressed in a natural way in terms of $${}_{4}F_{3}\left(1,1,1,\frac{3}{2};\frac{5}{2};\frac{5}{2};\frac{5}{2};1\right),$$ letting $H_{2n}'=H_{2n}-H_{n}$ for $n \in \mathbb{N}$. Commented Feb 8, 2017 at 20:19

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 $\int_{0}^{\pi/2}\theta\cot(\theta)\,d\theta=\frac{\pi}{2}\log(2).$

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.

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.

• You knocked-out all math software (Mathematica, Maple, and Mathcad) by computing this closed form!!! BRAVO - Jack - BRAVO. And the method is very instructive!!! Please make sure to publish it. (+1) Commented Feb 6, 2017 at 8:57
• @HazemOrabi: I am quite fond of this proof, and ready to buy a shirt saying better than Mathematica. I do not know if this proof deserves publishing as-it-is, I actually just used Parseval's theorem in a non conventional form. But if the method leads to an explicit evaluation of other interesting series, that might be the case. Commented Feb 6, 2017 at 9:08
• Hi Jack, I am a professional software developer with 17 years of experience and a member of British Computer Society. Up to my knowledge, all CAD systems had a special library to compute/simplify hyper-geometric functions considering the special values (at least I am sure about this in Wolfram from its technical documentation). With a development of an algorithm, the method you introduced is applicable to be generalized for calculating a closed form for a class of hyper-geometric functions. Generalize-it and Publish-it, I know what I see, it is an achievement. BRAVO again and good luck. Hazem Commented Feb 6, 2017 at 12:02
• @JackD'Aurizio "Please, do not ask me to do the same for other values of ${_4F_3}$"... I've actually been curious for some time now if somebody somewhere has compiled a list of closed forms for ${_4F_3}$'s at low half-integer parameters, and classified by complexity. At the very least, it would be nice to see the individual results scattered about this site gathered in one place... (BTW, speaking as a person who is impressed by your work on a regular basis, I agree this one belongs on the highlights reel ;) ) Commented Feb 9, 2017 at 14:34
• @HazemOrabi: I have to thank you again for the encouragement, since I am really going to publish it, together with another interesting identity. Commented Aug 16, 2017 at 0:46

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

$$_4F_3$$ 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

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.