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So in the process of trying to find a derivation for this answer, the following interesting equalities arose (one can check with Wolfram Alpha/Mathematica):

$$\frac{8\sqrt{2}G^4}{5\pi^2} \left(\left(7 \sqrt{2}-10\right) \beta +5 \left(\sqrt{2}-2\right)\right) = -\pi/2,\tag1$$

$$-\frac{4}{3} \left(\alpha\left(\sqrt{2}-1\right)^2 +6 \ln \left(\sqrt{2}-1\right)\right) = 7\ln 2 - \ln(17-12\sqrt{2})-\pi/2,\tag2$$

where $G = \Gamma\left(\frac{3}{4}\right)$, $\alpha = {}_3F_2\left(1,1,\frac{5}{4};\frac{7}{4},2;3-2\sqrt{2}\right)$ and $\beta = {}_3F_2\left(1,\frac{3}{2},\frac{7}{4};\frac{9}{4},\frac{5}{2};3-2\sqrt{2}\right)$.

Doing some simplification and solving will tell you:

$${}_3F_2\left(1,1,\frac{5}{4};\frac{7}{4},2;3-2\sqrt{2}\right) = \frac{3}{4}\cdot\frac{\pi/2-7\ln{2}-4\ln(\sqrt{2}-1)}{(\sqrt{2}-1)^2},\tag4$$

$${}_3F_2\left(1,\frac{3}{2},\frac{7}{4};\frac{9}{4},\frac{5}{2};3-2\sqrt{2}\right) = \frac{5}{4}\cdot\frac{(\pi/2)^3-4(\sqrt{2}-1)G^4}{G^4(\sqrt{2}-1)^3}.\tag5$$

It's very bizarre that aside from the Gamma function valued at 3/4, these come out to (relatively) nice closed forms. My guess is that they are a result of integrals, but I have no idea what those integrals could be. Mathematica doesn't get anywhere with the hypergeometric functions, so I'm a bit stuck.

Note: $(\sqrt{2}-1)^2 = 3 - 2\sqrt{2}$. That last number seems to come out, sort of, in the result of the hypergeometric functions' closed form, but I can't place why.

EDIT: Here are some ways of describing both functions simultaneously, which may help in some way:

${}_3F_2\left(a,b,c;c+\frac{1}{2},b+1;z\right)$, ${}_3F_2\left(a,b,c;c+\frac{1}{2},a+b;z\right)$, ${}_3F_2\left(a,b,b+\frac{1}{4};a+b-\frac{1}{4},a+b;z\right)$, ${}_3F_2\left(1,a,a+\frac{1}{4};a+\frac{3}{4},a+1;z\right)$

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  • $\begingroup$ Do I understand correctly that so far these are only conjectures supported by numerical calculations? How did they arise? $\endgroup$ – Vladimir Reshetnikov Mar 14 '14 at 2:52
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    $\begingroup$ Nice Question! I have been trying to solve the same integral for a lot of time but without success. Perhaps the following quadratic transformation may prove useful: $${\;}_3F_2 \left(\begin{matrix}a,b,c\\ 1+a-b,1+a-c \end{matrix}\Big| z \right)= (1-z)^{-a}{\;}_3F_2 \left(\begin{matrix}\frac{a}{2},\frac{1+a}{2},1+a-b-c\\ 1+a-b,1+a-c \end{matrix} \Big|- \frac{4z}{(1-z)^2}\right)$$ $\endgroup$ – Shobhit Bhatnagar Mar 14 '14 at 3:51
  • $\begingroup$ @VladimirReshetnikov I don't know what it is about the hypergeometric functions that results in these closed forms. If you read the question I linked to and (I believe) Lucian's answer to it you'll see that it stems from a complicated definite integral. I simply equated the Mathematica-derived solution and the closed form found in my link and simplified. $\endgroup$ – Eugene Bulkin Mar 14 '14 at 6:09
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    $\begingroup$ @EugeneBulkin OK, but you should clearly state that it is a conjecture supported by numeric evidence, rather than a rigorously proven fact (unless we take Cleo's answers as new axioms, as she suggests to do). $\endgroup$ – Vladimir Reshetnikov Mar 15 '14 at 2:28
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    $\begingroup$ I have finally learned enough Weierstrass elliptic functions to derive Cleo's answer in the other question. All the strange closed form of hypergeometric functions mentioned here probably comes from some magic identities among the underlying elliptic functions. $\endgroup$ – achille hui Mar 15 '14 at 9:59
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I'm going to only cover conjecture $(4)$. We can use elliptic functions to show it is true.

To simplify setup, we will adopt all conventions and notation as in this answer.
For $x \in (0, 1)$, let $F(x) = x _3F_2(1,1,\frac54;2,\frac74;x)$, we have

$$ F(x) = \sum_{k=0}^\infty \frac{x^{k+1}}{k+1}\frac{(\frac54)_k}{(\frac74)_k} = \frac{\Gamma(\frac74)}{\Gamma(\frac12)\Gamma(\frac54)} \sum_{k=0}^\infty \frac{x^{k+1}}{k+1}\int_0^1 t^{\frac14+k}(1-t)^{-\frac12}dt\\ = -\frac{3}{2\sqrt{2}\omega}\int_0^1 \log(1-xt) t^{-\frac34}(1-t)^{-\frac12} dt $$ Let $x = \alpha^2$, $\beta = \frac{1-\alpha}{1+\alpha}$ and substitute $t$ by $\left(\frac{p-\frac12}{p + \frac12}\right)^2$ and then $p$ by $\wp(z)$, we have:

$$\begin{align} F(\alpha^2) &= -\frac{3}{\omega}\int_{\infty}^{\frac12} \log\left[(1-\alpha^2)\frac{(p+\frac12\beta)(p+\frac12\beta^{-1})}{(p+\frac12)^2}\right] \frac{dp}{\sqrt{4p^3-p}}\\ &= -\frac{3}{2\omega}\int_{-\omega}^\omega \log\left[(1-\alpha^2)\frac{(\wp(z)+\frac12\beta)(\wp(z)+\frac12\beta^{-1})}{(\wp(z)+\frac12)^2}\right] dz \end{align}\tag{*1} $$ Notice $1 - (\sqrt{2}-1)^2 = 2 (\sqrt{2}-1)$, conjecture $(4)$ can be rewritten as

$$F((\sqrt{2}-1)^2) \stackrel{?}{=}\frac34\left[ \frac{\pi}{2} - 3\log 2\right] - 3 \log\left[1 - (\sqrt{2}-1)^2\right]$$ Compare this with $(*1)$, we find conjecture $(4)$ is equivalent to

$$\frac{1}{2\omega}\int_{-\omega}^\omega \log\left[\frac{(\wp(z)+\frac12\beta)(\wp(z)+\frac12\beta^{-1})}{(\wp(z)+\frac12)^2}\right] dz \stackrel{?}{=}\frac{1}{4}(3\log 2 - \frac{\pi}{2})\tag{*2} $$ when $\alpha = \beta = \sqrt{2}-1$.

Like the other answer, we can express the RHS of $(*2)$ using Weierstrass sigma function. Before we do that, I will like to point out

$$\left(\frac{d}{dz}\wp(z)\right) ^2 = 4 \wp(z)^3 - \wp(z)^3 \;\;\implies\;\; \left(\frac{d}{dz}\frac{1}{4\wp(iz)}\right)^2 = 4 \left(\frac{1}{4\wp(iz)}\right)^3 - \left(\frac{1}{4\wp(iz)}\right) $$ One consequence of this is if we pick a $\rho$ such that $\wp( \pm ( \omega' + \rho) ) = -\frac12\beta$, then $$\wp(\pm ( \omega' + i\rho)) = \wp(\pm( \omega' - i\rho)) = -\frac12\beta^{-1} \quad\text{ and }\quad\wp( \pm( \omega' - \rho)) = -\frac12\beta$$

When $\alpha = \beta = \sqrt{2}-1$, we can pick $\rho = \frac{\omega}{2}$. The RHS of $(*2)$ becomes

$$\frac{1}{2\omega}\int_{-\omega}^{\omega} \left\{ \sum_{k=0}^3\log\left[ \frac{\sigma(z + \omega' + i^k\rho)}{\sigma(\omega' +i^k\rho))} \right] - 4 \log\left[ \frac{\sigma(z + \omega')}{\sigma(\omega')} \right] \right\} dz$$

One can perform same analysis as in my other answer by playing with $\varphi_{\pm}(\tau)$ functions. Because of the symmetry of $\wp(z)$ around the point $z = \omega' = i\omega$, their contributions will cancel out with each other. The net result is

$$\text{RHS}[*2] = \log\left[\frac{\sigma(\omega')^4}{\prod_{k=0}^3\sigma(\omega' + i^k\rho))}\right]\tag{*3}$$

For Weierstrass elliptic functions with general $g_2, g_3$, we know

$$\begin{cases} \wp'(z) &= -\frac{\sigma(2z)}{\sigma(z)^4}\\ \sigma(z+2\omega) &= -e^{2\eta(z+\omega)}\sigma(z)\\ \sigma(z+2\omega') &= -e^{2\eta'(z+\omega')}\sigma(z) \end{cases}$$

This implies $$\sigma(z+\omega)^4 = -\frac{\sigma(2z+2\omega)}{\wp'(z+\omega)} = e^{2\eta(2z+\omega)}\frac{\sigma(2z)}{\wp'(z+\omega)} \quad\implies\quad \sigma(\omega)^4 = e^{2\eta\omega}\frac{2}{\wp''(\omega)}$$ For our case where $(g_2,g_3) = (1,0)$, we know $\eta = \frac{\pi}{4\omega}$, this leads to $\sigma(\omega) = e^{\frac{\pi}{8}}\sqrt[4]{2}$. Let me call this number $\Omega$.

When $g_3 = 0$, the double poles of $\wp(z)$ forms a square lattice. This 4-fold symmetry around the origin give us $\sigma(i\omega) = i\sigma(\omega) = i\Omega$.

The value of sigma functions at other points in $(*3)$ can be deduced in similar manner:

$$\begin{align} \left|\sigma\left(\frac{\omega'}{2}\right)\right| &= \left|\frac{\sigma(\omega')}{\wp'(\frac{\omega'}{2})}\right|^{1/4} = \left(\frac{\Omega}{\sqrt{2}+1}\right)^{1/4}\\ \left|\sigma\left(\frac{3\omega'}{2}\right)\right| &= \left|e^{\eta'\omega'}\sigma\left(-\frac{\omega'}{2}\right)\right| = e^{\pi/4}\left(\frac{\Omega}{\sqrt{2}+1}\right)^{1/4}\\ \left|\sigma\left(\omega'\pm\frac{\omega}{2}\right)\right| &= \left|\frac{\sigma\left(2\omega'\pm\omega\right)}{\wp'(\omega'\pm\frac{\omega}{2})}\right|^{1/4} = \left|e^{2\eta'(\omega'\pm\omega)}\frac{\Omega}{\sqrt{2}-1}\right|^{1/4} = e^{\pi/8}\left(\frac{\Omega}{\sqrt{2}-1}\right)^{1/4} \end{align}$$

Combine all this, we find

$$\text{RHS}[*2] = \log\left(\frac{\Omega^4}{e^{\pi/2}\Omega}\right) = 3\log\Omega - \frac{\pi}{2} = 3\left(\frac14\log 2 + \frac{\pi}{8}\right) - \frac{\pi}{2} = \frac14\left(3\log 2 - \frac{\pi}{2}\right)$$

i.e Conjecture $(4)$ is true.

Update

It turns out there is a cleaner algebraic relation between the hypergeometric function in conjecture $(4)$ and elliptic functions. Using the addition formula for sigma function:

$$\wp(z)-\wp(u) = - \frac{\sigma(z+u)\sigma(z-u)}{\sigma(z)^2\sigma(u)^2}$$

We find $$\frac{\sigma(\omega')^4}{\prod_{k=0}^3\sigma(\omega'+ i^k\rho)} =\left(\frac{1}{\sigma(\rho)^2(\wp(\omega')-\wp(\rho))}\right) \left(\frac{1}{\sigma(i\rho)^2(\wp(\omega')-\wp(i\rho))}\right) $$

Notice $$\begin{cases} \wp(\omega'\pm\rho) &= -\frac12\beta\\ \wp(\omega'\pm i\rho) &= -\frac12\beta^{-1} \end{cases} \quad\implies\quad \begin{cases} \wp(\pm \rho) &= \frac{1}{2\alpha}\\ \wp(\pm i\rho) &= -\frac{1}{2\alpha}\\ \end{cases} $$ We get $$\frac{\sigma(\omega')^4}{\prod_{k=0}^3\sigma(\omega'+ i^k\rho)} = \frac{4\alpha^2}{\sigma(\rho)^4(1-\alpha^2)} $$ Which in terms implies a simple relation between $F(\alpha^2)$ and $\sigma(\rho)$:

$$_3F_2\left(1,1,\frac54; 2,\frac74; \alpha^2\right) = \frac{F(\alpha^2)}{\alpha^2} = \frac{12}{\alpha^2}\log\left(\frac{\sigma(\rho)}{\sqrt{2\alpha}}\right)$$

When $\alpha = \sqrt{2}-1$, $\rho = \frac{\omega}{2}$ and since $\sigma(\frac{\omega}{2}) = \left(\frac{\Omega}{\sqrt{2}+1}\right)^{1/4}$, we again obtain

$$_3F_2\left(1,1,\frac54; 2,\frac74; (\sqrt{2}-1)^2\right) = \frac{3}{(\sqrt{2}-1)^2}\log\left(\frac{e^{\pi/8}\sqrt[4]{2}}{(\sqrt{2}+1)4(\sqrt{2}-1)^2}\right) = \frac{3}{4(\sqrt{2}-1)^2}\left[\frac{\pi}{2} - 7\log 2 - 4\log(\sqrt{2}-1)\right] $$

Setting $\rho$ to other rational multiples of $\omega$, we can deduce a bunch of similar identities. For example,

  1. When $\rho = \omega$, it is clear $\alpha = 1$ and $\sigma(\omega) = \Omega$, this leads to $$ _3F_2\left(1,1,\frac54; 2,\frac74; 1\right) = 12\log\left(\frac{\Omega}{\sqrt{2}}\right) = \frac32\left(\pi- 2\log 2\right) $$
  2. When $\rho = \frac{2\omega}{3}$, one can use the triplication formula of sigma function $$\frac{\sigma(3z)}{\sigma(z)^9} = 3\wp(z)\wp'(z)^2 - \frac14 \wp''(z)^2$$ to show $$\alpha = \sqrt{2\sqrt{3}-3} \quad\text{ and }\quad \sigma\left(\frac{2\omega}{3}\right) = e^{\pi/18} 3^{1/8} (2-\sqrt{3})^{1/12}$$ This leads to the identity $$_3F_2\left(1,1,\frac54; 2,\frac74; 2\sqrt{3}-3\right) = \frac{1}{2\sqrt{3}-3}\left(\frac{2\pi}{3} - 6\log 2 - 2\log(2-\sqrt{3})\right) $$
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  • $\begingroup$ And it's basically the same to calculate the second one, because it's the same idea. Great! $\endgroup$ – Eugene Bulkin Mar 16 '14 at 18:30
  • $\begingroup$ @achille hui: Thank you very much for this answer.I would really love to understand deeper that method. From what I can see your approach boils down to three steps.1. Reduce the hypergeometric function to an integral (easy),2. Change variables in the integral to map it to the elliptic Weierstrass function (hard ?), 3. Reduce the quantity to residues of the Weieirstrass fct. In (2.) you somehow figured out that the equation $t^{-3/4} (1-t)^{-1/2} dt\simeq (4p^3-p)^{-1/2} dp$ is solved by $t=(p-1/2)^2/(p+1/2)^2$. Can you disclose how you knew that? Can we do that for different exponents as well? $\endgroup$ – Przemo Sep 13 '17 at 15:14
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    $\begingroup$ @Przemo I answered this 3 years ago, I forgot what I'm thinking at that point. If I redo this now, I will probably first change variable to $s = t^{1/4}$ and get something like $(1-s^4)^{-1/2} ds$. This one has 4 branch points at $\pm 1, \pm i$. The next thing I will try is find a suitable Mobius transform to move the the 4 branch points to the right places (the 3 roots of $4p^3-p$ and $\infty$). I'm not sure this will get to above change of variable but this is the general step. $\endgroup$ – achille hui Sep 13 '17 at 16:17
  • $\begingroup$ I checked your 2-\sqrt(3) example. You had a small typo as the pi should be in the numerator, like in the 1-sqrt(2) example. $\endgroup$ – Tito Piezas III Jan 29 at 11:30
  • $\begingroup$ @TitoPiezasIII thanks for the fix. $\endgroup$ – achille hui Jan 29 at 14:24

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