15
$\begingroup$

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.

$\endgroup$
1
  • 3
    $\begingroup$ Perhaps mention the meaning of MZV. $\endgroup$ – GEdgar Jun 27 '20 at 13:16
18
+100
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.