I'm searching for MZV representations of $_pF_q$. Based on previous computation I conjecture that $$\sum _{n=0}^{\infty } \frac{1}{(2 n+1)^5}\left(\frac{\binom{2 n}{n}}{4^n}\right)^{-2}=\, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1,1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)$$
admits a weight $5$ MZV closed-form. Here is a relevant problem but its method is not directly applicable here due to existence of a square root. So how to find the closed form?

Any help will be appreciated.

  • 3
    $\begingroup$ Perhaps mention the meaning of MZV. $\endgroup$ – GEdgar Jun 27 '20 at 13:16

Solved; I sketch my first solution below.

$1$. By Euler integral the original sum equals $\int_0^1 \frac{\, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};x\right)}{2\sqrt{1-x}} \, dx$, now substitute $x\to x^2$.

$2$. Let $n\to-1, r\to 2$ in formula (proved by termwise integration):

  • $\ _{r+3}F_{r+2}\left(1,1, \{\frac{n+2}{2}\}_{r+1}; \frac32, \{\frac{n+4}{2}\}_{r+1}; x^2\right)=\frac{(-1)^r (n+2)^{r+1}}{x r!} \int_0^1 t^n \frac{ \sin ^{-1}(x t) }{\sqrt{1-x^2 t^2}}\log ^r(t) \, dt$

And use it to substitute $_5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};x^2\right)$, yielding a double integral.

$3$. Substitute $t\to t, x\to \frac zt$, apply Fubini on $(z,t)$, the sum equals $\frac{1}{2} \int _0^1\int _z^1\frac{\log ^2(t) \sin ^{-1}(z)}{t^2 \sqrt{1-z^2} \sqrt{1-\frac{z^2}{t^2}}}dtdz$.

$4$. Integrate w.r.t $t$ by brute force gives

  • $\small f(z)=\frac{1}{z}\left(\frac{1}{2} \pi \log ^2(z)+\frac{1}{24} \pi \left(\pi ^2+12 \log ^2(2)\right)+ \pi \log (2) \log (z)\right)-2 \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2};z^2\right)$

So it boils down to evaluation of $\frac{1}{2} \int _0^1\frac{\sin ^{-1}(z) f(z)}{\sqrt{1-z^2}} dz$, which will be broke into $4$ parts.

$5$. First $3$ parts: By $z\to \frac{2v}{1+v^2}$ one have $\frac{1}{2} \int_0^1 \frac{\sin ^{-1}(z) \log ^k(z)}{z \sqrt{1-z^2}} \, dz=\int_0^1 \frac{\tan ^{-1}(v) \log ^k\left(\frac{2 v}{v^2+1}\right)}{v} \, dv$. In our case $k=0,1,2$, i.e. quadratic log integrals on RHS are of weight $\leq 4$, all of which are calculated here.

$6$. Now we confront the last part i.e.

  • $\int_0^1 \frac{\sin ^{-1}(z) \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2};z^2\right)}{\sqrt{1-z^2}} \, dz=\int_0^1 \frac{\sin ^{-1}(z) \, z _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2};z^2\right)}{z\sqrt{1-z^2}} \, dz$

Due to brute force and $_3F_2$ closed forms (see Y. Brychkov's Handbook of special functions: derivatives, integrals, series and other formulas)

  • $\small \int \frac{\sin ^{-1}(z)}{z \sqrt{1-z^2}} \, dz=i \text{Li}_2\left(-e^{i \sin ^{-1}(z)}\right)-i \text{Li}_2\left(e^{i \sin ^{-1}(z)}\right)+\sin ^{-1}(z) \left(\log \left(1-e^{i \sin ^{-1}(z)}\right)-\log \left(1+e^{i \sin ^{-1}(z)}\right)\right)$

  • $\small \frac{\partial }{\partial z}\left(z \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2};z^2\right)\right)=\, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};z^2\right)=\frac{\log (2 z) \sin ^{-1}(z)}{z}+\frac{\text{Li}_2\left(e^{2 i \sin ^{-1}(z)}\right)-\text{Li}_2\left(e^{-2 i \sin ^{-1}(z)}\right)}{4 i z}$

Thus one may apply IBP, transforming the last part into the following modulo polylog constants:

  • $\scriptsize \int_0^1 \left(i \text{Li}_2\left(-e^{i \sin ^{-1}(z)}\right)-i \text{Li}_2\left(e^{i \sin ^{-1}(z)}\right)+\sin ^{-1}(z) \left(\log \left(1-e^{i \sin ^{-1}(z)}\right)-\log \left(1+e^{i \sin ^{-1}(z)}\right)\right)\right) \left(\frac{\log (2 z) \sin ^{-1}(z)}{z}+\frac{\text{Li}_2\left(e^{2 i \sin ^{-1}(z)}\right)-\text{Li}_2\left(e^{-2 i \sin ^{-1}(z)}\right)}{4 i z}\right) \, dz$

$7$. The final integral: Let $z\to \sin(u), u\to \frac{\log(v)}i$ and deform contour, one arrive at $\int_1^i h(z)dz$ then $\int_0^1 i h(iz)-h(z) dz$. Fortunately the integrand $i h(iz)-h(z)$ admits a $4$-admissible polylog form thus solvable by using special values of numerous level $4$ MZVs (this part is developed by @pisco here, based on rather deep theory).

$8$. Combining all above we conclude

  • $\, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1,1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=-16 \pi \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)+12 \pi \beta(4)+16 \text{Li}_5\left(\frac{1}{2}\right)-\frac{341 \zeta (5)}{32}-\frac{2}{15} \log ^5(2)+\frac{5}{36} \pi ^2 \log ^3(2)-\frac{37}{360} \pi ^4 \log (2)$

Which, unfortunately, does not offer a new representation of irreducible MZVs.

Update : See here for more hypergeometric-MZV relations and a simpler proof of identity above, which is generalizable to prove the MZV-reducibility of case $k>5$ by using iterated integrals. Based on this result (and $7$ other supplementary ones), a general criterion on MZV-reducibility of hypergeometric series is established.

  • 5
    $\begingroup$ (+1) Wow, this technique has a brilliant future. $\endgroup$ – Jack D'Aurizio Jul 6 '20 at 14:28

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