We have the Eisenstein series of weight $k$: $G_k(z)=\frac 1 2 \sum_{m,n} \frac 1 {(mz+n)^k}$. Can we evaluate it in closed form for some special values of $z$, eg. $z=i$ or $z=\omega$?
It is clear by symmetry that $G_k(i)=0$ unless $k$ is a multiple of 4, but is there a closed form for $G_4(i)$, for example?
The problem is very similar to determining the values of the $\zeta$ function at even integers, so I guess that the Weierstrass elliptic function could be of use here (it is the “equivalent” of the cotangent function for lattices in $\mathbb C$, as $\sum_n \frac 1 {x+n} = \pi \cot (\pi z)$

$\begingroup$ Perhaps this is not going in a direction of interest to you, but...: suitable linear combinations at CM/Heegner points gives special values of Hecke Lfunctions. Suitable of these special values were investigated by Dammerell and by Shimura c. 1970s, and in terms of Deligne's conjectures on special values, are "knowable" in a certain finite range (of your "k"), with a common (presumed) transcendental (in addition to $\pi$) consisting of a Petersson inner product of the associated modular form. Is this of interest? $\endgroup$– paul garrettSep 8, 2019 at 22:33

2$\begingroup$ These: $1+240\sum_{n=1}^{\infty}\frac{n^{3}e^{2\pi n}}{1e^{2\pi n}}=\frac{3\pi^{2}}{4\Gamma(3/4)^{8}}$, $1+240\sum_{n=1}^{\infty}\frac{n^{3}e^{4\pi n}}{1e^{4\pi n}}=\frac{33\pi^{2}}{64\Gamma(3/4)^{8}}$, $1+480\sum_{n=1}^{\infty}\frac{n^{7}e^{2\pi n}}{1e^{2\pi n}}=\frac{9\pi^{4}}{16\Gamma(3/4)^{16}}$ were produced by looking at special values of Jacobian theta functions. I only hazily remember how I arrived at them. $\endgroup$– graveolensaSep 9, 2019 at 13:48
4 Answers
It is well known that $$G_{4k}(i)=\left(4\int_{0}^1\frac{1}{\sqrt{1x^4}}dx\right)^{4k}\frac{H_{4k}}{(4k)!}, $$ where $H_{4k}$ are called Hurwitz numbers. $H_4=\frac{1}{10}$ and $H_{8}=\frac{3}{10}$ are the first two such numbers, but in general these come as coefficients in the Laurent series of Weierstrass's elliptic function. The introduction of Tsumura's paper On certain analogues of Eisenstein series and their evaluation formulas of Hurwitz type gives more details, and the relevant paper by Hurwitz is cited there.

$\begingroup$ Am I correct in assuming there's no known closed form for that integral? $\endgroup$ Sep 8, 2019 at 22:42

5$\begingroup$ @GerryMyerson No, let $t= x^4$ you obtain $$\int_0^1 (1x^4)^{1/2}dx=\frac14\int_0^1 (1t)^{1/2} t^{3/4} dt= \beta(1/2,1/4) =\frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)}=\sqrt{\pi}\Gamma(1/4)^2\frac{\sin(\pi/4)}{\pi}$$ 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 $\Bbb{C}[E_4,E_6]$ $\endgroup$– reunsSep 8, 2019 at 23:16

$\begingroup$ @reuns There is a factor of four missing in the denominator of the last expression (the final expression should be $\Gamma(1/4)^2\frac{\sqrt{2}}{8\sqrt{\pi}}$). $\endgroup$ Sep 8, 2019 at 23:22

5$\begingroup$ 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 Lseries. $\endgroup$ Sep 9, 2019 at 11:35

1$\begingroup$ 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 $\omega_1=2\int\limits_{0}^1 \frac{1}{\sqrt{1x^6}} dx$, $\omega_2=i\omega_1(1/2+\sqrt{3}/2i)$, $G_6=4$. There are more formulas for imaginary quadratics $\tau=\omega_2/\omega_1$ in Katayama's paper "On the Values of Eisenstein Series". $\endgroup$ Sep 9, 2019 at 11:56
For real Eisenstein series $$ \sum_{(m,n)\ne(0,0)}\frac{1}{m\tau+n^s}, $$ the Kronecker limit formula gives the value at $s=1$ in terms of the Dedekind eta function. See https://en.wikipedia.org/wiki/Kronecker_limit_formula For CM values of $\tau$, one gets a product of $\Gamma$ values.
$$\wp(z) = \frac1{z^2}+\sum_{(n,m)\ne (0,0)} \frac1{(z+ni+m)^2}\frac1{(ni+m)^2}$$ is the unique even $\Bbb{Z}+i\Bbb{Z}$ periodic meromorphic function with only one double pole at $0$ where $\wp(z) = \frac1{z^2}+O( z^2)$. We obtain $\wp(z)=\frac1{z^2}+3 G_4(i)z^2 + 5 G_6(i)+O(z^6)$ where $G_6(i)=0$ so that $$\wp'(z)^2= 4 \wp(z)^360G_4(i) \wp(z)+O(z^2)$$ The $O(z^2)$ term vanishes because it is analytic doubly periodic with a zero at $0$.
$\wp(\frac{1i}2)= \wp(\frac{1+i}2)=0$
$$\frac{1+i}2=\int_0^{\frac{1+i}2} dz = \int_0^{\frac{1+i}2} \frac{d\wp(z)}{\wp'(z)}=\int_0^{\frac{1+i}2} \frac{d\wp(z)} {\sqrt{4\wp(z)^360G_4(i)\wp(z)}}$$ $$=\int_\infty^0 \frac{dx}{2\sqrt{x^315 G_4(i) x}} =\frac{i}{2}( 15 G_4(i))^{1/4}\int_0^1+\int_1^\infty \frac{dX}{\sqrt{XX^3}}$$ $$=\frac{i+1}{2}(15G_4(i))^{1/4} \int_0^1 \frac{dt^{1/2}}{\sqrt{t^{1/2}(1t)}}= \frac{i+1}{4}( 15G_4(i))^{1/4}\beta(1/4,1/2)$$ $$=\frac{i+1}{4}( 15 G_4(i))^{1/4}\frac{\Gamma(1/4)\Gamma(1/2)}{\Gamma(3/4)} =\frac{i+1}{4}( 15G_4(i))^{1/4}\Gamma(1/4)^2 \sqrt{\pi} \frac{\sin(\pi/4)}{\pi}$$
and hence $$G_4(i)= (\frac{1}{2}15^{1/4}\Gamma(1/4)^2 (2\pi)^{1/2})^4$$
If $k$ is odd $G_{2k}(i)=0$. To find $G_{4k}(i)$ we'll need to show that the first cusp form for the full modular group is $\Delta(z) = (2\pi)^{12}e^{2i\pi z} \prod_{n \ge 1} (1e^{2i \pi nz})^{24} $ of weight $12$ , since it has only one simple zero at $i\infty$ then $\frac{E_4(z)^3E_6(z)^2}{\Delta(z)}$ is a modular form of weight $0$ thus it is constant, thus for $f$ of weight $2k=4a+6b\ge 12$ then $\frac{ff(i\infty) E_4(z)^a E_6(z)^b}{\Delta(z)}$ is of weight $2k12$ and by induction $f$ is a polynomial in $E_4,E_6$.
Whence $$E_{4k}(z) = \sum_{4a+6b=4k} c_{a,b} E_4(z)^aE_6(z)^b, \qquad G_{4k}(i)= c_{k,0} 2 \zeta(4k) E_4(i)^k=c_{k,0} 2 \zeta(4k)\frac{G_4(i)^k}{(2\zeta(4))^k}$$ where the $c_{a,b} \in \Bbb{Q}$ are found from the first few coefficients of the $q$expansion of $E_{4k},E_4,E_6$.

$\begingroup$ Your formula for $G_4(i)$ is off by a factor of $\frac1 2$ compared to the literature. That is probably because you define $G_k$ as $\sum \frac 1 {(mz+n)^k}$, whereas I defined it as $\frac 1 2 \sum \frac 1 {(mz+n)^k}$. Also, the limits on the second and third integrals should be $0$ and $\infty$, not the corresponding values of $z$. $\endgroup$– FusRoDahSep 10, 2019 at 14:13

1$\begingroup$ The standard definition is $G_k(z)=\sum_{(m,n) \ne (0,0)} \frac{1}{(mz+n)^k},E_k(z)= \frac{G_k(z)}{G_k(i\infty)}$, and they are riemann stieltjes integrals over $z$ $\endgroup$– reunsSep 10, 2019 at 23:27
Im writing a paper about negative weights Eisenstein series evaluated at $z=i$. Set $$ E_{\nu}(z)=1+\frac{1}{\zeta(1\nu)}\sum_{n=1}^{\infty}\sigma_{\nu1}(n)q^n\textrm{, }q=e(z)\textrm{, }Im(z)>0. $$ I have evaluated among other values ($C$ is Catalan constant): $$ E_{2}(i)=\frac{7\pi^3}{\zeta(3)}\textrm{, }E_{2}'(i)=\frac{i(240C\pi29\pi^3)}{720 \zeta(3)}\textrm{, }E_{2}''(i)=\frac{80C\pi11\pi^3}{240 \zeta(3)} $$ $$ E_{2}^{(3)}(i)=\frac{i\left(1E_{4}(i)\right)\pi^3}{15\zeta(3)}\textrm{, }E_{2}^{(4)}(i)=\frac{2E_{4}(i)\pi^3}{15\zeta(3)} $$ $$ E_{4}^{(5)}(i)=\frac{8i\pi^5}{63\zeta(5)} $$ In general holds the trivial formula $$ E_{2n+2}(z)=1+\frac{\zeta(2n+1)}{\zeta(2n1)(2\pi i)^{2n+1}}\frac{d^{2n+1}}{dz^{2n+1}}E_{2n}(z) $$ Also $$ E'_{4}(i)=2iE_{4}(i)+\frac{13i\pi^5}{1890\zeta(5)} $$ $$ E_{4}''(i)=3E_{4}(i)+\frac{11\pi^5}{378\zeta(5)}\frac{\pi^2\zeta(3)}{4\zeta(5)} $$ $$ E_{4}^{(3)}(i)=3iE_{4}(i)\frac{i\pi^5}{14\zeta(5)}+\frac{3i\pi^2\zeta(3)}{2\zeta(5)} $$