Wednesday, June 11, 2025

Entry 160

We continue with closed-forms for the Dedekind eta function \(\eta^2(\sqrt{-d})\) for \(d=11,19,43,67,163\). As discussed in Entry 159, the closed-form for the complete elliptic integral of the first kind \(K(k_d)\) then necessarily follows. 

$$\eta^2(\sqrt{-11}) = \frac1{x_{11}^2}\frac{\Gamma\big(\tfrac1{11}\big)\,\Gamma\big(\tfrac3{11}\big)\,\Gamma\big(\tfrac4{11}\big)\,\Gamma\big(\tfrac5{11}\big)\,\Gamma\big(\tfrac9{11}\big)}{11^{1/4}\,(2\pi)^3}$$ $$\eta^2(\sqrt{-19}) = \frac1{x_{19}^2}\frac{\Gamma\big(\tfrac1{19}\big)\,\Gamma\big(\tfrac4{19}\big)\,\Gamma\big(\tfrac5{19}\big)\,\Gamma\big(\tfrac6{19}\big)\dots\Gamma\big(\tfrac{17}{19}\big)}{19^{1/4}\,(2\pi)^5}$$

$$\eta^2(\sqrt{-43}) = \frac1{x_{43}^2}\frac{\Gamma\big(\tfrac1{43}\big)\,\Gamma\big(\tfrac4{43}\big)\,\Gamma\big(\tfrac6{43}\big)\,\Gamma\big(\tfrac9{43}\big)\dots\Gamma\big(\tfrac{41}{43}\big)}{43^{1/4}\,(2\pi)^{11}}$$

$$\eta^2(\sqrt{-67}) = \frac1{x_{67}^2}\frac{\Gamma\big(\tfrac1{67}\big)\,\Gamma\big(\tfrac4{67}\big)\,\Gamma\big(\tfrac6{67}\big)\,\Gamma\big(\tfrac9{67}\big)\dots\Gamma\big(\tfrac{65}{67}\big)}{67^{1/4}\,(2\pi)^{17}}$$

$$\eta^2(\sqrt{-163}) = \frac1{x_{163}^2}\frac{\Gamma\big(\tfrac1{163}\big)\,\Gamma\big(\tfrac4{163}\big)\,\Gamma\big(\tfrac6{163}\big)\,\Gamma\big(\tfrac9{163}\big)\dots\Gamma\big(\tfrac{161}{163}\big)}{163^{1/4}\,(2\pi)^{41}}$$

where the \(x_d\) are the real roots of the following simple cubics, respectively

$$\begin{align}& x^3-2x^2+2x-2 = 0\\ & x^3-2x-2 = 0 \\ & x^3-2x^2-2 = 0 \\ & x^3-2x^2-2x-2 = 0 \\ & x^3-6x^2+4x-2=0\end{align}$$

The numerators, for example, of \(\Gamma\big(\frac{n}{19}\big)\) can be found using the Kronecker symbol and this Wolfram command which yields the \(\frac{19-1}2 = 9\) numerators as \(n = 1, 4, 5, 6, 7, 9, 11, 16, 17\).

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