Strange closed forms for hypergeometric functions

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 $$_3F_2\left(1,1,\frac54; 2,\frac74; 2\sqrt{3}-3\right) = \frac{1}{2\sqrt{3}-3}\left(\frac{2\pi}{3} - 6\log 2 - 2\log(2-\sqrt{3})\right) $$