Entry 166

The case $d=8$ is special since it is even but has odd class number $h(-d)=1$. Given the Kronecker symbol $\big(\tfrac{-d}{m}\big)$, we propose a symmetrical closed-form 

$$\begin{align}K(k_8) &=\frac{\sqrt{2\pi}}{16}\frac1{\sqrt{U_2}}\left(\sqrt{\frac{U_2-1}2}+\sqrt{\frac{U_2+1}2}\right)^2 \left(\prod_{m=1}^{8}\Big[\Gamma\big(\tfrac{m}{8}\big)\Big]^{\big(\tfrac{-8}{m}\big)}\right)^{1/2}\\ &=\frac{\sqrt{2\pi}}{16}\frac1{\sqrt{U_2}}\left(\sqrt{\frac{U_2-1}2}+\sqrt{\frac{U_2+1}2}\right)^2 \left(\frac{\Gamma\big(\frac18\big)\,\Gamma\big(\frac38\big)}{\Gamma\big(\frac58\big)\,\Gamma\big(\frac78\big)}\right)^{1/2}\end{align}$$ with fundamental unit $U_2=1+\sqrt2$ and form reminiscent of the odd $d$ with class number $1$.