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S Oct 11, 2022 at 18:14 history suggested user682141 CC BY-SA 4.0
Fix a mistake
Oct 11, 2022 at 15:50 review Suggested edits
S Oct 11, 2022 at 18:14
Sep 9, 2019 at 12:13 comment added Josiah Park The same observation of @reuns applies to writing the second integral in terms of Beta/Gamma function values.
Sep 9, 2019 at 11:56 comment added Josiah Park Another simple formula holds for $G_6(\omega_1,\omega_2)=120 \sum\limits_{(m,n)\in\mathbb{Z}^2} \frac{1}{(m\omega_1+n\omega_2)^6}$. When $\omega_1=2\int\limits_{0}^1 \frac{1}{\sqrt{1-x^6}} dx$, $\omega_2=i\omega_1(-1/2+\sqrt{3}/2i)$, $G_6=4$. There are more formulas for imaginary quadratics $\tau=\omega_2/\omega_1$ in Katayama's paper "On the Values of Eisenstein Series".
Sep 9, 2019 at 11:35 comment added David Loeffler There is nothing special about $i$ (or $\omega$) here; in principle you can do this for any $z$-value of the form $a + b \sqrt{-D}$ with $b^2 D > 0$, but the explicit formulae will get messy quite fast. The role of the Hurwitz numbers will be played by special values of Dirichlet L-series.
Sep 9, 2019 at 11:29 comment added FusRoDah Is there a similar expression for other values of $z$? I would guess that $z=\omega$ could be another “nice” value.
Sep 8, 2019 at 23:22 comment added Josiah Park @reuns There is a factor of four missing in the denominator of the last expression (the final expression should be $\Gamma(1/4)^2\frac{\sqrt{2}}{8\sqrt{\pi}}$).
Sep 8, 2019 at 23:16 comment added reuns @GerryMyerson No, let $t= x^4$ you obtain $$\int_0^1 (1-x^4)^{-1/2}dx=\frac14\int_0^1 (1-t)^{-1/2} t^{-3/4} dt= \beta(1/2,1/4) =\frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)}=\sqrt{\pi}\Gamma(1/4)^2\frac{\sin(\pi/4)}{\pi}$$ also the LHS is an elliptic integral which comes from integrating $\frac{dx}{y} =dz$ on a basis of the homology of $y^2=x^3+x \overset{\wp}\cong \Bbb{C/r(Z+iZ)}$ and relating it with $\Delta(i)$ then extending to all $SL_2(\Bbb{Z})$ modular forms using they are all in $\Bbb{C}[E_4,E_6]$
Sep 8, 2019 at 22:42 comment added Gerry Myerson Am I correct in assuming there's no known closed form for that integral?
Sep 8, 2019 at 20:44 history edited Josiah Park CC BY-SA 4.0
Added another detail on relation with Weierstrass's p-function
Sep 8, 2019 at 20:29 history edited Josiah Park CC BY-SA 4.0
added missing factor of two needed for alternative definition used by OP
Sep 8, 2019 at 20:07 history edited Josiah Park CC BY-SA 4.0
Changed notation to explicit elliptic integral
Sep 8, 2019 at 19:59 history answered Josiah Park CC BY-SA 4.0