Ridiculous hypergeometric closed form for $\sum_{k=0}^{\infty} (-1)^k \frac{\binom{2k}{k}}{4^k}\frac{H_{2k}}{4k-1}$

Question

In this question I asked about a closed form for a very simple-looking integral.

$$\int _0^1\int _0^1 \frac{\sqrt{x^2+y^2}}{1-xy} dxdy$$

This problem was boiled down to evaluating the following two series involving the Harmonic Numbers \begin{eqnarray}S_1 &=& \sum_{k=0}^{\infty}(-1)^k \frac{\binom{2k}{k}}{4^k}\frac{H_{2k}}{2k-1}\\ S_2 &=& \sum_{k=0}^{\infty} (-1)^k \frac{\binom{2k}{k}}{4^k}\frac{H_{2k}}{4k-1} = -0.179653 \cdots \end{eqnarray}

The first sum turned out to be

$$S_1 = \frac12 \bigl( 2 - 2\sqrt{2} + \sqrt{2} \log \left( 48-32\sqrt{2} \right) + \log \left( 3-2\sqrt{2} \right) \bigr)$$

But finding a polylogarithmic closed form for $S_2$ proved difficult. Using a generating function technique, I managed to express $S_2$ as

$$S_2 = -\frac12 \int_1 ^{\sqrt{2}} \frac{1}{(1-y^2)^{5/4}} \log \left( \frac{1+y}{2y^2} \right) \mathrm{d}y$$

After tinkering around with this in Wolfram Mathematica, it seems like the expression for $S_2$ is quite involved:

$$ S_2 = \left(\frac{1}{2}+\frac{i}{2}\right)\frac{\pi^{3/2}\Bigl(-8-\pi+\ln16\Bigr)} {\Gamma\Bigl(\frac{1}{4}\Bigr)^2}\, -\operatorname{arcsinh}(1) + 2^{3/4}(1-\sqrt{2})^{1/4} \Biggl[ \Bigl(-2+\operatorname{arcsinh}(1)\Bigr) \, {}_2F_{1}\!\Bigl(-\frac{1}{4},\frac{1}{4};\frac{3}{4};\frac{1}{2}+\frac{1}{\sqrt{2}}\Bigr) + 4\, {}_3F_{2}\!\Bigl(-\frac{1}{4},-\frac{1}{4},\frac{1}{4};\frac{3}{4},\frac{3}{4};\frac{1}{2}+\frac{1}{\sqrt{2}}\Bigr) \Biggr] - (1+i)\, {}_3F_{2}\!\Bigl(\frac{1}{2},\frac{1}{2},\frac{5}{4};\frac{3}{2},\frac{3}{2};2\Bigr) + \sqrt{2}\,\ln2 + \left(\frac{1}{2}+\frac{i}{2}\right) \Biggl[ -5\, {}_2F_{1}\!\Bigl(-\frac{3}{4},\frac{1}{2};\frac{3}{2};2\Bigr) + {}_2F_{1}\!\Bigl(\frac{1}{4},\frac{1}{2};\frac{3}{2};2\Bigr) \Biggr]\ln2 $$

My question is, how would one prove and simplify something like this?

Addendum. As Efim Mazhnik showed below, the answer seems to involve two hypergeometric functions. I suppose I will ask two separate questions to see if they could be simplified further:

This one will be for $\mathcal{H}_1$:

$$ \mathcal{H}_1 = {}_3F_2 \left(-\frac14,-\frac14,\frac14;\,\frac34,\frac34;\,\frac12\right) $$

This one will be for $\mathcal{H}_2$:

$$ \mathcal{H}_2 = {}_3F_2 \left(\frac14,\frac12,\frac12;\,\frac32,\frac32;2(\sqrt{2}-1)\right) $$

Answer 1

(A partial answer.)

I. Part 1

The OP gives,

$$S_1 = \frac12 \bigl( 2 - 2\sqrt{2} + \sqrt{2} \log \left( 48-32\sqrt{2} \right) + \log \left( 3-2\sqrt{2} \right) \bigr)$$

We can simplify this to,

$$S_1 = 1-\sqrt2+2\sqrt2\ln2+(1+\sqrt2)\ln(-1+\sqrt2)$$

---

II. Part 2

First, define the second lemniscate constant $L_2$,

$$L_2 = \int_0^1\frac{x^2}{\sqrt{1-x^4}}dx = \frac{\sqrt2\,\pi^{3/2}}{\Gamma\big(\tfrac14\big)^2} = 0.599070117\dots$$

then the second part,

$$S_2 = \sum_{k=0}^{\infty} (-1)^k \frac{\binom{2k}{k}}{4^k}\frac{H_{2k}}{4k-1} = -0.17965340854\dots$$

can be divided into five portions with almost closed-forms.

---

$$A = \ln(-1+\sqrt2)+\sqrt{2}\,\ln2$$

---

$$B=\left(\frac{1}{2}+\frac{i}{2}\right)\frac{\pi^{3/2}\Bigl(-8-\pi+\ln16\Bigr)} {\Gamma\Bigl(\frac{1}{4}\Bigr)^2}=\left(\frac{1}{2}+\frac{i}{2}\right)\frac{\Bigl(-8-\pi+4\ln2\Bigr)\,L_2} {\sqrt2}$$

---

$$C=\left(\frac{1}{2}+\frac{i}{2}\right) \Biggl[ -5\, {}_2F_{1}\!\Bigl(-\frac{3}{4},\frac{1}{2};\frac{3}{2};2\Bigr)+{}_2F_{1}\!\Bigl(\frac{1}{4},\frac{1}{2};\frac{3}{2};2\Bigr) \Biggr]\ln2 = -\frac{(2\ln2)\,L_2\, \sqrt{-1}}{\sqrt2}$$

---

$$D=- (1+i)\, {}_3F_{2}\!\Bigl(\frac{1}{2},\frac{1}{2},\frac{5}{4};\frac{3}{2},\frac{3}{2};2\Bigr)=\color{red}{a_1}-\frac{\pi\,L_2\sqrt{-1}}{2\sqrt2}$$

---

$$E=2^{3/4}(1-\sqrt{2})^{1/4} \Biggl[ \Bigl(-2+\operatorname{arcsinh}(1)\Bigr) \, {}_2F_{1}\!\Bigl(-\frac{1}{4},\frac{1}{4};\frac{3}{4};\frac{1}{2}+\frac{1}{\sqrt{2}}\Bigr) + 4\, {}_3F_{2}\!\Bigl(-\frac{1}{4},-\frac{1}{4},\frac{1}{4};\frac{3}{4},\frac{3}{4};\frac{1}{2}+\frac{1}{\sqrt{2}}\Bigr) \Biggr]=\color{red}{a_2}+\frac{(4+\pi)L_2\sqrt{-1}}{\sqrt2}$$

---

Then,

$$S_2 = A+B+C+D+E = −0.17965340854580453658796807817043609357\dots$$

where the imaginary values cancel out and only two real values from $(D,E)$ have unknown closed-forms, namely,

\begin{align} \color{red}{a_1} &=-1.691614532217260425137039402197102725918878\dots\\ \color{red}{a_2} &=+3.185658976555447234030013042707392415451352\dots \end{align}

Do these have closed-forms?

---

III. Addendum

The 5th portion $E$ is the most complicated, but we can split it up. Let,

$$\beta=2^{3/4}(1-\sqrt{2})^{1/4}=\sqrt[4]{\frac2{1+\sqrt2}}+\sqrt[4]{\frac2{1+\sqrt2}}\,\sqrt{-1}$$

and,

$$F = \small{-2\,\beta\times{}_2F_{1}\!\Bigl(-\frac{1}{4},\frac{1}{4};\frac{3}{4};\frac{1}{2}+\frac{1}{\sqrt{2}}\Bigr) = \color{blue}{b_1}-\sqrt8\,L_2\sqrt{-1}}$$

$$G = \small{\operatorname{arcsinh}(1)\,\beta\times{}_2F_{1}\!\Bigl(-\frac{1}{4},\frac{1}{4};\frac{3}{4};\frac{1}{2}+\frac{1}{\sqrt{2}}\Bigr) = \color{blue}{b_2}+\sqrt2\,\ln(1+\sqrt2)\,L_2\sqrt{-1}}$$

$$H = \small{4\,\beta\times {}_3F_{2}\!\Bigl(-\frac{1}{4},-\frac{1}{4},\frac{1}{4};\frac{3}{4},\frac{3}{4};\frac{1}{2}+\frac{1}{\sqrt{2}}\Bigr) = \color{blue}{b_3}+\frac{8+\pi-2\ln(1+\sqrt2)}{\sqrt2}\,L_2\sqrt{-1}}$$

then $E = F+G+H.$ The real parts are,

\begin{align} \color{blue}{b_1}&=-1.4329995224208840378048103992724935579117740\dots\\ \color{blue}{b_2}&=+0.6315039646366933169680288981706533913747556\dots\\ \color{blue}{b_3}&=+3.9871545343396379548667945438092325819883707\dots \end{align}

and $a_2 = b_1+b_2+b_3.\,$ Do these have closed-forms?

---

IV. Update (A day later):

Courtesy of Nikitan's comment and Efim Mazhnik's answer, then the $a_n$ and $b_m$ above do have closed-forms in terms of real hypergeometrics. Define,

\begin{align} a&=\frac{2^{5/4}}9\times\,_3F_2\left(\tfrac34,\tfrac34,\tfrac54;\,\tfrac74,\tfrac74;\,\tfrac12\right)\\[5pt] b&=2^{9/4}\times\,_3F_2\left(-\tfrac14,-\tfrac14,\tfrac14;\,\tfrac34,\tfrac34;\,\tfrac12\right)\\[5pt] c&=2\times\,_3F_2\left(\tfrac14,\tfrac12,\tfrac12;\,\tfrac32,\tfrac32;\,2(-1+\sqrt2)\right)\quad \end{align}

Then,

\begin{align} b_1 &= (-2+\sqrt2-\sqrt2\,L_2)\\[5pt] b_2 &= (-2+\sqrt2-\sqrt2\,L_2)\ln(-1+\sqrt2)/2\\[5pt] b_3 &= -2\sqrt2+\sqrt2\,L_2(\pi+2)+\left(-a+b+c\sqrt{-1+\sqrt2\,}\right)\quad \end{align}

and

\begin{align} \quad a_1 &= - \frac{3\pi\,L_2}{2\sqrt2}+a\\[5pt] \quad\, a_2 &=\; b_1+b_2+b_3\\ &= -(2+\sqrt2)-\left(\frac{-1+\sqrt2}{\sqrt2}\right)\ln(-1+\sqrt2)+\left(\frac{-6+2\pi-\ln(-1+\sqrt2)}{\sqrt2}\right)L_2+\left(-a+b+c\sqrt{-1+\sqrt2}\right)\end{align}

P.S. The first two hypergeometrics $(a,b)$ are related by $\,-a+b=2\sqrt2\,(1+L_2)$.

Answer 2

Here is an answer which includes only ${}_3F_2$ real hypergeometric functions $$S_2 = 4\times2^{1/4} {}_3F_2 \left(-\frac14,-\frac14,\frac14;\,\frac34,\frac34;\,\frac12\right) + 2\sqrt{\sqrt{2}-1} {}_3F_2 \left(\frac14,\frac12,\frac12;\,\frac32,\frac32;2(\sqrt{2}-1)\right) + \frac{\ln 4 - \operatorname{arcsinh}(1)}{\sqrt{2}} + \frac{\pi^{3/2}(-10+\operatorname{arcsinh}(1)+\ln 4)}{\Gamma \left(\frac14\right)^2} - 2-\sqrt{2}$$

---

Using the following generating function $$\sum_{k=0}^\infty H_{2k} \begin{pmatrix}2k\\k\end{pmatrix} x^k = \frac{\ln \left(\frac{1+\sqrt{1-4x}}{2}\right) - \ln (1-4x)}{\sqrt{1-4x}}$$ we obtain $$S_2 = \sum_{k=0}^\infty \frac{(-1)^{k}}{4^k} \begin{pmatrix}2k\\k\end{pmatrix} \frac{H_{2k}}{4k-1} =\int_0^1 \frac{1}{x^2} \sum_{k=0}^\infty \frac{(-1)^{k}}{4^k} \begin{pmatrix}2k\\k\end{pmatrix} H_{2k} x^{4k} = -\underbrace{\int_0^1 \frac{\ln (1+x^4)dx}{x^2\sqrt{1+x^4}}}_{\mathcal{G}_1} + \underbrace{\int_0^1 \frac{\ln \left(\frac{1+\sqrt{1+x^4}}{2}\right)dx}{x^2\sqrt{1+x^4}}}_{\mathcal{G}_2}$$ $$\mathcal{G}_1 = \int_0^1 \frac{\ln (1+x^4)dx}{x^2\sqrt{1+x^4}} \stackrel{x \, \mapsto 1/(1+x^4)}{=} \frac14\int_1^{1/2} x^{-1/4}(1-x)^{-5/4} \ln x dx$$ We can prove the following indefinite integral (see Appendix A1) $$\frac14\int x^{-1/4}(1-x)^{-5/4} \ln x dx = (1-x)^{-1/4}x^{-1/4} \ln x + x^{-1/4}(4-\ln x) {}_2F_1 \left(-\frac14,\frac14;\frac34;x\right) - 4x^{-1/4} {}_3F_2 \left(-\frac14,-\frac14,\frac14;\,\frac34,\frac34;\,x\right)$$ It is straightforward to evaluate $${}_2F_1 \left(-\frac14,\frac14;\frac34;1\right) = \frac{2\pi^{3/2}}{\Gamma \left(\frac14\right)^2}, \quad {}_2F_1 \left(-\frac14,\frac14;\frac34;\frac12\right) = 2^{-3/4} + 2^{-1/4} \frac{\pi^{3/2}}{\Gamma \left(\frac14\right)^2}$$ $${}_3F_2 \left(-\frac14,-\frac14,\frac14;\,\frac34,\frac34;\,1\right) = \frac{\pi^{3/2}(8+\pi-\ln 4)}{4\Gamma \left(\frac14\right)^2}$$ Therefore $$\mathcal{G}_1 = -2^{5/4} {}_3F_2 \left(-\frac14,-\frac14,\frac14;\,\frac34,\frac34;\,\frac12\right) + \frac{\pi^{3/2}(4+\pi-\ln 2)}{\Gamma \left(\frac14\right)^2} + \frac{4-\ln 2}{\sqrt{2}}$$ Next $$\mathcal{G}_2 = \int_0^1 \frac{\ln \left(\frac{1+\sqrt{1+x^4}}{2}\right)dx}{x^2\sqrt{1+x^4}} = \stackrel{x \, \mapsto \frac{2}{1+\sqrt{1+x^4}}}{=} \frac{1}{4\sqrt{2}}\int_1^{2(\sqrt{2}-1)} x^{1/2}(1-x)^{-5/4} \ln x dx$$ We can prove the following indefinite integral (see Appendix A2) $$\frac14 \int x^{1/2} (1-x)^{-5/4} \ln x dx = x^{1/2}(1-x)^{-1/4} \ln x - \sqrt{x}\,{}_2F_1 \left(\frac14,\frac12;\frac32;\,x\right) (2+\ln x) + 2\sqrt{x}\,{}_3F_2 \left(\frac14,\frac12,\frac12;\,\frac32,\frac32;\,x\right)$$ It is also straightforward to evaluate $${}_2F_1 \left(\frac14,\frac12;\frac32;1\right) = \frac{2\sqrt{2}\pi^{3/2}}{\Gamma \left(\frac14\right)^2}, \quad {}_3F_2 \left(\frac14,\frac12,\frac12;\,\frac32,\frac32;\,1\right) = \frac{\sqrt{2}\pi^{3/2}(8-\pi-2\ln 2)}{2\Gamma \left(\frac14\right)^2}$$ It is also possible to evaluate $${}_2F_1 \left(\frac14,\frac12;\frac32;2(\sqrt{2}-1)\right)$$ using elliptic methods. Apply the transformation of hypergeometric function that changes $x$ to $1-x$: $${}_2F_1 \left( \frac14,\frac12;\,\frac32;\,2(\sqrt{2}-1)\right) = \frac{\Gamma \left(\frac32 \right)\Gamma \left(\frac34 \right)}{\Gamma \left(\frac54 \right)}{}_2F_1 \left( \frac14,\frac12;\,\frac14;\,3-2\sqrt{2}\right) + \frac{\Gamma \left(\frac32 \right)\Gamma \left(-\frac34 \right)}{\Gamma \left(\frac14 \right)\Gamma \left(\frac12 \right)} (3-2\sqrt{2})^{3/4}{}_2F_1 \left( \frac54,1;\,\frac74;\,3-2\sqrt{2}\right) =$$ $$=\frac{2\sqrt{2}\pi^{3/2}}{\Gamma \left(\frac14\right)^2} {}_1F_0 \left(\frac12;\,;\,3-2\sqrt{2}\right) - \frac23 (3-2\sqrt{2})^{3/4} {}_2F_1 \left( \frac54,1;\,\frac74;\,3-2\sqrt{2}\right)$$ $$=\frac{2\sqrt{2}\pi^{3/2}}{\Gamma \left(\frac14\right)^2} \frac{1}{\sqrt{2(\sqrt{2}-1)}} - \frac23 (3-2\sqrt{2})^{3/4} {}_2F_1 \left( \frac54,1;\,\frac74;\,3-2\sqrt{2}\right)$$ Let $\wp(x)$, $\zeta(x)$ and $\sigma(x)$ be the Weierstrass elliptic, zeta and sigma functions associated with the differential equation $\wp'(x)^2 = 4\wp(x)^3 - \wp(x)$. We also let $\omega = \frac{\Gamma \left( \frac14\right)^2}{4\sqrt{\pi}}$ to be the half period for $\wp(x)$ on the real axis and $\omega' = i\omega$. It was proven in this answer that $$\alpha^2 {}_3F_2 \left(1,1,\frac54;2,\frac74;\,\alpha^2\right) = 12\ln \left(\frac{\sigma(\rho)}{\sqrt{2\alpha}}\right)$$ where $\rho$ is defined such as $\wp(\omega' + \rho) = \frac12 (\alpha-1)/(\alpha+1)$.

Differentiate both parts with respect to $\alpha^2$: $$\,{}_2F_1 \left( 1,\frac54;\,\frac74;\,\alpha^2\right) = 12 \frac{d}{d(\alpha^2)} \ln \left( \frac{\sigma (\rho)}{\sqrt{2\alpha}}\right) = 12 \frac{\sqrt{2\alpha}}{\sigma (\rho)} \left( \frac{1}{\sqrt{2\alpha}} \frac{d \sigma(\rho)}{d (\alpha^2)} - \frac{\sigma(\rho)}{4\sqrt{2\alpha}\alpha^2}\right) = $$ $$=\frac{12}{\sigma(\rho)} \left( \sigma (\rho) \zeta(\rho) \frac{d \rho}{d (\alpha^2)} - \frac{\sigma (\rho)}{4\alpha^2}\right) = 12 \left( \zeta(\rho) \frac{d \rho}{d (\alpha^2)} - \frac{1}{4\alpha^2} \right) = 12 \left( \zeta(\rho) \frac{[\wp'(\omega'+\rho)]^{-1}}{2\alpha (1+\alpha)^2} - \frac{1}{4\alpha^2} \right)$$ For $\alpha = \sqrt{2}-1$ we have $\rho = \omega/2$ and $\wp' \left(\omega'+\frac{\omega}{2}\right) = \sqrt{2}-1$. Using the duplication formula $$\zeta(2z) = 2\zeta(z) + \frac12 \frac{\zeta'''(z)}{\zeta''(z)} = 2\zeta(z) + \frac12 \frac{\wp''(z)}{\wp'(z)}$$ and $\zeta \left(\omega \right) = \frac{\pi}{4\omega}$, $\wp' \left(\frac{\omega}{2}\right) = -\sqrt{2}-1$, $\wp'' \left(\frac{\omega}{2}\right) = 4+3\sqrt{2}$ we find $\zeta \left(\frac{\omega}{2}\right) = \frac{\pi}{8\omega} + \frac{2+\sqrt{2}}{4}$ and finally $$\,{}_2F_1 \left( 1,\frac54;\,\frac74;\,3-2\sqrt{2}\right) = \frac{3}{3-2\sqrt{2}} \left( \frac{\pi}{8\omega} + \frac{2+\sqrt{2}}{4} - 1\right) = \frac34 \left( \frac{2\pi^{3/2}}{(3-2\sqrt{2})\Gamma \left(\frac14\right)^2} - 2-\sqrt{2}\right)$$ Therefore $${}_2F_1 \left(\frac14,\frac12;\frac32;2(\sqrt{2}-1)\right) = \frac{1}{\sqrt{\sqrt{2}-1}} \left( \frac{\pi^{3/2}}{\Gamma \left(\frac14\right)^2} + 1 - \frac{1}{\sqrt{2}}\right)$$ It gives $$\mathcal{G}_2 = 2\sqrt{\sqrt{2}-1}\,{}_3F_2 \left(\frac14,\frac12,\frac12;\,\frac32,\frac32;\,2(\sqrt{2}-1)\right) + \frac{\ln(2(\sqrt{2}-1))}{\sqrt{2}} + \sqrt{2}-2 + \frac{\pi^{3/2}(-6+\ln (2(\sqrt{2}+1)))}{\Gamma \left(\frac14\right)^2}+\frac{\pi^{5/2}}{\Gamma \left(\frac14\right)^2}$$ The result can be then obtained by algebraic manipulations.

---

Appendix

A1 $$\frac14\int x^{-1/4}(1-x)^{-5/4} \ln x dx =-\frac12 \int x^{-5/4}(1-x)^{-1/4} \ln x \left(\frac{1/2-x}{1-x}-1/2\right) dx = $$ $$=-\frac12 \int x^{-5/4}(1-x)^{-5/4}(1/2-x) \ln x dx + \frac14 \int x^{-5/4}(1-x)^{-1/4} \ln x dx$$ We can evaluate both of these integrals using integration by parts $$-\frac12 \int x^{-5/4}(1-x)^{-5/4}(1/2-x) \ln x dx \stackrel{\text{IBP}}{=} (1-x)^{-1/4} x^{-1/4} \ln x - \int x^{-5/4} (1-x)^{-1/4} dx$$ $$=(1-x)^{-1/4} x^{-1/4} \ln x + 4x^{-1/4}\,{}_2F_1 \left(-\frac14,\frac14;\,\frac34;\,x\right)$$ $$\frac14 \int x^{-5/4}(1-x)^{-1/4} \ln x dx \stackrel{\text{IBP}}{=} -x^{-1/4}\ln x \,{}_2F_1 \left(-\frac14,\frac14;\,\frac34;\,x\right) + \int x^{-5/4} {}_2F_1 \left(-\frac14,\frac14;\,\frac34;\,x\right) dx = $$ $$=-x^{-1/4}\ln x \,{}_2F_1 \left(-\frac14,\frac14;\,\frac34;\,x\right) - 4x^{-1/4} {}_3F_2 \left(-\frac14,-\frac14,\frac14;\,\frac34,\frac34;\,x\right)$$

A2 $$\frac14\int x^{1/2}(1-x)^{-5/4} \ln x dx =\frac14 \int x^{-1/2}(1-x)^{-1/4} \ln x \left(\frac{2-x}{1-x}-2\right) dx = $$ $$=\frac14 \int x^{-1/2}(1-x)^{-5/4}(2-x) \ln x dx - \frac12 \int x^{-1/2}(1-x)^{-1/4} \ln x dx$$ We can evaluate both of these integrals using integration by parts $$\frac14 \int x^{-1/2}(1-x)^{-5/4}(2-x) \ln x dx \stackrel{\text{IBP}}{=} (1-x)^{-1/4} x^{1/2} \ln x - \int x^{-1/2} (1-x)^{-1/4} dx$$ $$=(1-x)^{-1/4} x^{1/2} \ln x - 2x^{1/2}\,{}_2F_1 \left(\frac14,\frac12;\,\frac32;\,x\right)$$ $$\frac12 \int x^{-1/2}(1-x)^{-1/4} \ln x dx \stackrel{\text{IBP}}{=} x^{1/2}\ln x \,{}_2F_1 \left(\frac14,\frac12;\,\frac32;\,x\right) - \int x^{-1/2} {}_2F_1 \left(\frac14,\frac12;\,\frac32;\,x\right) dx = $$ $$=x^{1/2}\ln x \,{}_2F_1 \left(\frac14,\frac12;\,\frac32;\,x\right) - 2x^{1/2} {}_3F_2 \left(\frac14,\frac12,\frac12;\,\frac32,\frac32;\,x\right)$$