# A tough series related with a hypergeometric function with quarter integer parameters

Is it possible to express $$\sum_{n\geq 0}\frac{\binom{4n}{2n}\binom{2n}{n}}{64^n(4n+1)} = \phantom{}_3 F_2\left(\frac{1}{4},\frac{1}{4},\frac{3}{4}; 1,\frac{5}{4}; 1\right)$$ in terms of standard mathematical constants given by Euler sums and values of the $$\Gamma$$ function?

This problem arise from studying the interplay between elliptic integrals, hypergeometric functions and Fourier-Legendre expansions. According to Mathematica's notation we have $$\sum_{n\geq 0}\frac{\binom{4n}{2n}\binom{2n}{n}}{64^n}y^{2n}=\frac{2}{\pi\sqrt{1+y}}\,K\left(\frac{2y}{1+y}\right)$$ for any $$y\in[0,1)$$, where the complete elliptic integral of the first kind fulfills the functional identity $$\forall x\in[0,1),\qquad K(x) = \frac{\pi}{2\cdot\text{AGM}\left(1,\sqrt{1-x}\right)}$$ hence the computation of the above series boils down to the computation of $$\int_{0}^{1}K\left(\frac{2y^2}{1+y^2}\right)\frac{2\,dy}{\pi\sqrt{1+y^2}}\stackrel{y\mapsto\sqrt{\frac{x}{2-x}}}{=}\frac{\sqrt{2}}{\pi}\int_{0}^{1}\frac{K(x)}{\sqrt{x}(2-x)}\,dx\\=\frac{1}{\pi}\int_{-1/2}^{+\infty}\frac{\arctan\sqrt{u}}{\sqrt{u(1+u)(1+2u)}}\,du$$ where $$K(x),\sqrt{2-x},\frac{1}{\sqrt{x}},\frac{1}{\sqrt{2-x}}$$ all have a pretty simple FL expansion, allowing an easy explicit evaluation of similar integrals. This one, however, is a tougher nut to crack, since $$\frac{1}{2-x}$$ does not have a nice FL expansion. There are good reasons for believing $$\Gamma\left(\frac{1}{4}\right)$$ is involved, since a related series fulfills the following identity: $$\sum_{n\geq 0}\frac{\binom{2n}{n}^2}{16^n(4n+1)}=\frac{1}{2\pi}\int_{0}^{1}K(x)\,x^{-3/4}\,dx = \frac{1}{16\pi^2}\,\Gamma\left(\frac{1}{4}\right)^4$$ which ultimately is a consequence of Clausen's formula, stating that in some particular circumstances the square of a $$\phantom{}_2 F_1$$ function is a $$\phantom{}_3 F_2$$ function.

March 2019 Update: after some manipulations, it turns out that the computation of the original $$\phantom{}_3 F_2$$ is equivalent to the computation of the integral $$\int_{0}^{1}\frac{-\log x}{\sqrt{x(1+6x+x^2)}}\,dx = \frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{z\,dz}{\sqrt{3+\cosh z}}$$ which is way less scary. Additionally, nospoon has already tackled similar integrals, so I guess he might have something interesting to share.

• The hypergeometric term in question can also be expressed as a double integral of a relatively "nice" rational function: ${_3F_2}{\left(\frac14,\frac14,\frac34;1,\frac54;1\right)}=\frac{2\sqrt{2}}{\pi}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{x^{2}}{1-x^{4}+x^{4}y^{4}}$. – David H Sep 15 '17 at 10:27
• Which essentially boils down to $$\int_{0}^{1}\frac{\arctan(z)+\text{arctanh}(z)}{(1-z^4)^{3/4}}\,dz.$$ Maybe using some change of variables $\varphi:(0,1)^2\to (0,1)^2$ the integral $$\iint_{(0,1)^2}\frac{x^2\,dx\,dy}{1-x^4+x^4 y^4}$$ becomes something with a nicer structure. – Jack D'Aurizio Sep 15 '17 at 15:11
• I've mulled over the integral some more, and I think it may actually be a special case of an integral I asked about here. Some connections connections are made there with Appell functions that may prove useful somewhere, but for most part every way I try to attack this integral seems to be sending me in circles... – David H Sep 25 '17 at 16:44

This is not going to be a full answer but I think I am able to contribute a little more compared to what has been presented above therefore I present my approach. Denote : \begin{equation} S(y):= \sum\limits_{n=0}^\infty \binom{4 n}{2 n} \binom{2 n}{n} \frac{y^{2 n}}{64^n} \end{equation} We use the good old identity: \begin{equation} \binom{2 n}{n}=(-4)^n \cdot \binom{-\frac{1}{2}}{n} \end{equation} and we get \begin{eqnarray} S(y)&=& \sum\limits_{n=0}^\infty \binom{-\frac{1}{2}}{n} \cdot \underbrace{\binom{-\frac{1}{2}}{2n}}_{\frac{1}{\pi} \int\limits_0^1 t^{-1/2} (1-t)^{2 n-1/2} dt} \cdot (-y^2)^n\\ &=& \frac{1}{\pi} \int\limits_0^1 \frac{1}{\sqrt{t(1-t)(1-y^2 t^2)}} dt &=& \end{eqnarray} Now in the last equation above we replace $y$ by $y^2$ and integrate over $y$ from zero to unity. This gives: \begin{eqnarray} \int\limits_0^1 S(y^2) dy&=& \frac{1}{\pi} \int\limits_0^1 \frac{F_{2,1}[\frac{1}{4},\frac{1}{2},\frac{5}{4},t^2]}{\sqrt{t(1-t)}}dt\\ &=&\frac{1}{\pi} \int\limits_0^1 \frac{F[\arcsin(\sqrt{t}),-1]}{t \sqrt{1-t}} dt\\ &=&-\frac{1}{\pi}\int\limits_0^1 \frac{2\log(1-\sqrt{1-t})-\log(t)}{2\sqrt{t(1-t^2)}}dt\\ &=&-\frac{1}{\pi} \left(2\int\limits_0^1 \frac{\log(t)}{\sqrt{t(t-2)(t-1+\sqrt{2})(t-1-\sqrt{2})}}dt+\pi^{3/2} \frac{\Gamma(5/4)}{\Gamma(3/4)}\right)\\ &=&-\frac{1}{\pi} \left( 2 \sqrt{2} \int\limits_0^{\pi} \frac{\log[\sin(u/4)]}{\sqrt{3-\cos(u)}}- \frac{2 \sqrt{\pi} \Gamma(5/4) (\pi+4 \log(2))}{\Gamma(-1/4)} \right)\\ &=&-\frac{1}{\pi} \left( \sqrt{2} \int\limits_0^\pi \frac{\log[1-\cos(u/2)]}{\sqrt{3-\cos(u)}}du- \frac{2 \pi^{3/2} \Gamma(5/4)}{\Gamma(-1/4)} \right)\\ &=& -\frac{1}{\pi} \left( \sqrt{2} \int\limits_0^{\pi/2} \frac{\log[1-\cos(u)]}{\sqrt{1-1/2\cos(u)^2}}du- \frac{2 \pi^{3/2} \Gamma(5/4)}{\Gamma(-1/4)} \right) \end{eqnarray} In the first line above I expanded the square root in a series and integrated over $y$ term by term. In the second line above I went to Wolframs site http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/09/19/02/ and looked up the closed form for the particular hypergeometric function. Here $F[\phi,m]$ is the elliptic function of the first kind. In the third line I integrated by parts once and finally in the forth line I integrated the second term by using appropriate Euler sums and in the first term we substituted for $1-\sqrt{1-t}$.In the fifthe line I substituted for $t:=2 \sin(u)^2$ and in the sixth line I simplified the result using trigonometric half-angle identities.
The remaining integral can be actually simplified by expanding the integrand in a double series and then doing one of the sums. Here we just state the result: \begin{eqnarray} &&\int\limits_0^{\pi/2} \frac{\log[1-\cos(u)]}{\sqrt{1-1/2\cos(u)^2}}du = -\frac{\pi}{2}\sum\limits_{\lambda=0}^\infty \binom{-1/2}{\lambda} (-\frac{1}{2})^\lambda \cdot \\ && \left(\frac{1+\lambda}{1+2\lambda} \binom{1/2+\lambda}{-1/2}(2 \log(2)+H_\lambda) + \binom{\lambda}{-1/2} F_{3,2}[\begin{array}{rrr} \frac{1}{2}& 1&1+\lambda\\ \frac{3}{2}& \frac{3}{2}+\lambda & \end{array};1] \right) \end{eqnarray} Now, the remaining hypergeometric functions $F_{3,2}$ can also be simplified and expressed through Catalan numbers. We lack time to complete the whole calculations and therefore we have to leave this apart for the time being and complete this later.
Update: By using the following identity: \begin{equation} \frac{(1+\lambda)^{(l)}}{(3/2+\lambda)^{(l)}} = \frac{(3/2)^{(\lambda)}}{1^{(\lambda)}} \cdot \frac{1^{(l)}}{(3/2)^{(l)}} \cdot \frac{(l+1)^{(\lambda)}}{(l+3/2)^{(\lambda)}} \end{equation} and then by decomposing the last term on the righthand side into partial fractions in $l$ we easily get the following identity: \begin{eqnarray} &&F_{3,2}[\begin{array}{rrr} \frac{1}{2}& 1&1+\lambda\\ \frac{3}{2}& \frac{3}{2}+\lambda & \end{array};1]= \frac{(3/2)^{(\lambda)}}{1^{(\lambda)}} \cdot \left(\right.\\ &&\left. 2 C- \sum\limits_{j=1}^\lambda j \binom{-1/2+j}{\lambda} \binom{\lambda}{j} (-1)^{\lambda-j} \cdot \int\limits_0^1 \theta^{j-1/2} \cdot F_{3,2}[\begin{array}{rrr} \frac{1}{2}& 1&1\\ \frac{3}{2}& \frac{3}{2} & \end{array};\theta]d\theta\right.\\ &&\left. \right)=\\ &&\frac{(3/2)^{(\lambda)}}{1^{(\lambda)}} \cdot \left(\right.\\ &&\left. 2 C-(1-\frac{(-1)^l \sqrt{\pi}}{\Gamma[1/2-l]l!}) \frac{\imath \pi^2}{4}-\right.\\ &&\left. (-1)^l\sum\limits_{j=1}^l j \binom{-1/2}{j} \binom{-1/2}{l-j} \cdot \right.\\ &&\left.\int\limits_0^{\pi/2} [\sin(u)]^{2j-1} 2 \cos(u)\left( i \text{Li}_2\left(-e^{i u}\right)-i \text{Li}_2\left(e^{i u}\right)+u \log \left(\frac{1-e^{i u}}{1+e^{i u}}\right)\right)du\right.\\ &&\left. \right)=\\ &&\frac{(3/2)^{(\lambda)}}{1^{(\lambda)}} \cdot \left(\right.\\ &&\left. 2 C-(1-\frac{(-1)^l \sqrt{\pi}}{\Gamma[1/2-l]l!}) \frac{\imath \pi^2}{4}-\right.\\ &&\left. (-1)^l\sum\limits_{j=1}^l\sum\limits_{p=-j-1}^{j-1} j \binom{-1/2}{j} \binom{-1/2}{l-j} \cdot \frac{(-1)^{p+l}}{2^{2j-1}}[\binom{2j-1}{p+1} - \binom{2j-1}{p+j+1}]\cdot\right.\\ &&\left.\int\limits_1^{\imath} z^{p+1} \left(\log(z) \log(\frac{1-z}{1+z})+Li_2(z)-Li_2(-z)\right)dz\right.\\ &&\left. \right)=\\ &&\frac{(3/2)^{(\lambda)}}{1^{(\lambda)}} \cdot \left(\right.\\ &&\left. 2 C+\right.\\ &&\left. \frac{1}{2} \sum\limits_{j=0}^{l-1}\sum\limits_{p=-l-1,p\neq -1}^{l-1} \binom{-3/2}{j} \binom{-1/2}{l-1-j} \frac{1}{2^{2j+1}} [\binom{2j+1}{p+j+1}-\binom{2j+1}{p+j+2}]\cdot\right.\\ &&\left. \frac{(-1)^{p+l}}{p+1} \left((1+(-1)^p) \cdot C - \sum\limits_{k=0}^{p \cdot 1_{p\ge 0}-(p+2) 1_{p<0}} \frac{(-1)^k}{(2k+1)^2}\right)\right.\\ &&\left. \right) \end{eqnarray} where in the second line we used the closed form expression from http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/08/05/02/01/07/0001/ and we changed the variables appropriately. In the third line we substituted for $z:=\exp(\imath u)$. In the fourth line we calculated the integrals using integration by parts and then simplified the result.