I. Some results
Let $T$ be the tribonacci constant, the real root of $x^3-x^2-x-1=0$. Then the exact value of the complete elliptic integral of the first kind $K(k_d)$ for $d=11$ is,
$$K(k_{11}) = \frac{(2T)^{2/3}}{11^{1/4}(4\pi)^2} \Gamma\big(\tfrac1{11}\big) \Gamma\big(\tfrac3{11}\big) \Gamma\big(\tfrac4{11}\big) \Gamma\big(\tfrac5{11}\big) \Gamma\big(\tfrac9{11}\big)$$
There are similar $K(k_d)$ formulas for $d=11,19,43,67,163$ with class number $1$ below.
II. Definitions
Given class number $n = h(-d).$ For primes $d=4m+3$, we separate cases $d=8m+7$ and $d=8m+3$ for matters which will be clear shortly. In a previous post, we considered $K(k_d)$ for the first form, while this new post will be for the second form.
Let $\eta(\tau)$ be the Dedekind eta function where $\tau = \tfrac{1+\sqrt{-d}}2$ and gamma functions,
$$\prod_{m=1}^{d}\Gamma\left(\frac{m}{d}\right)^{\big(\frac{-d}{\; m}\big)}$$
with exponent $\big(\frac{-d}{\; m}\big)$ as the Kronecker symbol.
A. Form $d=8m+7$
$$w_d = e^{\pi\,i/24}\frac{\;\eta(\tau)}{\color{red}{\sqrt2}\,\eta(2\tau)}$$
$$\prod_{m=1}^{d}\Gamma\left(\frac{m}{d}\right)^{\big(\frac{-d}{\; m}\big)} = x_{d}^{-4}\times\left(K(k_{d})\sqrt{\frac{2d}{\pi}}\right)^{2n}$$
then $w_d$ and $x_d$ are algebraic numbers of degree $n$, with $n$ being the class number.
B. Form $d=8m+3$
$$u_d = e^{\pi\,i/24}\frac{\;\eta(\tau)}{\eta(2\tau)}$$
$$\prod_{m=1}^{d}\Gamma\left(\frac{m}{d}\right)^{\big(\frac{-d}{\; m}\big)} = x_{d}^{-4}\times\left(\color{red}2K(k_{d})\sqrt{\frac{2d}{\pi}}\right)^{2n}$$
then $u_d$ and $x_d$ are now algebraic numbers of degree $3n$ (which is one reason to distinguish between the forms). The obvious question is to determine the relationship between $u_d$ and $x_d$.
III. Class Number n = 1
There are six, namely $d=3,11,19,43,67,163$. For this class, then $u_d = x_d$,
$$\prod_{m=1}^{3}\Gamma\left(\frac{m}{3}\right)^{\big(\frac{-3}{\; m}\big)} = x_{3}^{-4}\times\left(2K(k_{3})\sqrt{\frac{2\cdot3}{\pi}}\right)^\color{red}2$$
$$\prod_{m=1}^{11}\Gamma\left(\frac{m}{11}\right)^{\big(\frac{-11}{\; m}\big)} = x_{11}^{-4}\times\left(2K(k_{11})\sqrt{\frac{2\cdot11}{\pi}}\right)^\color{red}2$$
and so on. So $(x_3, x_{11},\dots)$ are algebraic numbers of degree $3n=3$,
\begin{align} d = 3:&\;x^3-2=0\\ d = 11:&\;x^3-2x^2+2x-2=0\\ d = 19:&\;x^3 - 2x - 2=0\\ d = 43:&\;x^3 - 2x^2 - 2=0\\ d = 67:&\;x^3 - 2x^2 - 2x - 2=0\\ d = 163:&\;x^3 - 6x^2 + 4x - 2=0 \end{align}
The first two are special,
$$x_3 = u_3 = 2^{1/3}\quad\\ x_{11} = u_{11} = (2T)^{1/3}$$
where $2^{1/3}$ is the Delian constant and $T$ is the tribonacci constant. For the next class, unfortunately $u_d \neq x_d$.
IV. Class Number n = 3
There are many, the smallest being $d=59$,
$$\prod_{m=1}^{59}\Gamma\left(\frac{m}{59}\right)^{\big(\frac{-59}{\; m}\big)} = x_{59}^{-4}\times\left(2K(k_{59})\sqrt{\frac{2\cdot59}{\pi}}\right)^\color{red}6$$
There doesn't seem to be a simple relationship between $x_{59}$ and $u_{59}$, as well as for the rest. The number $x_{59}$ is the real root of a degree $3n=9$ equation,
$$-2^9 + 2560x - 3840x^2 - 5056x^3 - 1024x^4 + 512x^5 + 152x^6 - 32x^7 - 8x^8 + x^9 = 0$$
while $u_{59}$ is the real root also of a nonic,
$$-2^3 + 16u - 8u^2 + 4u^3 - 8u^4 + 4u^5 - 2u^6 + 4u^7 - 4u^8 + u^9 = 0$$
They are solvable in radicals.
V. Class Number n = 5
The smallest being $d=131$,
$$\prod_{m=1}^{131}\Gamma\left(\frac{m}{131}\right)^{\big(\frac{-131}{\; m}\big)} = x_{131}^{-4}\times\left(2K(k_{131})\sqrt{\frac{2\cdot131}{\pi}}\right)^\color{red}{10}$$
The number $x_{131}$ is the real root of a degree $3n=15$ equation,
$$ -2^{25} - 2030043136x - 821915811840x^2 + 504210915328x^3 + \dots + x^{15}=0$$
while $u_{131}$ is the real root of,
$$-2^5 - 64u - 128u^2 - 112u^3 - 160u^4 - 128u^5 - 128u^6 - \dots + u^{15}=0$$
They are also solvable in radicals. And so on for all odd class numbers.
VI. Question
We ask a slightly different version from the previous post. Given prime $d=8k+3$ with class number $n=h(-d)$ and the proposed equation,
$$\prod_{m=1}^{d}\Gamma\left(\frac{m}{d}\right)^{\big(\frac{-d}{\; m}\big)} = x_{d}^{-4}\times\left(\color{red}2K(k_{d})\sqrt{\frac{2d}{\pi}}\right)^{2n}$$
The equation of degree $3n$ for the algebraic numbers $x_d$ and $u_d$ can be found using computers. But why is the degree now $3n$, and do all $x_d$ have a closed-form in terms of special functions, just like the $x_d$ for class number $n=1$?
