Entry 185

For the third McKay-Thompson series of the Monster, or \(j_3(\tau)\) 

$$j_3(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3\left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2$$

I haven't yet found a general formula for the complete elliptic integral of the first kind \(K(k_d)\) except for the special case \(K(k_3)\). Note that,

$$\pi = \beta\big(\tfrac12,\tfrac12\big) = 2\int_0^1\frac1{(1-x^2)^{1/2}}dx = 3.14159 $$

$$\pi_3 = \beta\big(\tfrac13,\tfrac13\big) = 3\int_0^1\frac1{(1-x^3)^{1/3}}dx = 5.29992 $$

where \(\pi_3 = 2^{4/3}\,3^{1/4}K(k_3) = \frac{\sqrt3}{2\pi}\Gamma^3\big(\tfrac13\big)\), the real period of the Dixon elliptic functions, can be considered as a cubic analogue of \(\pi\). Let 

$$\begin{align}j_3\big(\tfrac{1+3\sqrt{-1/3}}2\big) &= -3\cdot 4^3\\ j_3\big(\tfrac{1+5\sqrt{-1/3}}2\big) &= -5\cdot 12^3\\ j_3\big(\tfrac{1+7\sqrt{-1/3}}2\big) &= -7\cdot 36^3\\ j_3\big(\tfrac{1+11\sqrt{-1/3}}2\big) &= -\sqrt{11}\,(150\sqrt3+78\sqrt{11})^3\end{align}$$

Hence we propose

$$2^{1/3}K(k_{3}) = 3^{1/4}\,\frac{\pi}2 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-3\cdot 4^3}\big)^n}}$$

$$2^{1/3}K(k_{3}) = \frac{5^{2/3}}{3^{3/4}}\,\frac{\pi}2 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-5\cdot 12^3}\big)^n}}$$

$$2^{1/3}K(k_{3}) = \frac{7^{5/6}}{3^{5/4}}\,\frac{\pi}2 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-7\cdot 36^3}\big)^n}}$$

$$\frac{2^{1/3}K(k_{3})}{\sqrt{11}-\sqrt3\,} = \frac{11^{5/6}}{3^{3/4}}\,\frac{\pi}8 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-\sqrt{11}\,(150\sqrt3+78\sqrt{11})^3}\big)^n}}$$