Determination of special values of Eisenstein series 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 $\sum_n \frac 1 {x+n} = \pi \cot (\pi z)$
 A: 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.
A: *

*$$\wp(z) = \frac1{z^2}+\sum_{(n,m)\ne (0,0)} \frac1{(z+ni+m)^2}-\frac1{(ni+m)^2}$$
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 $$\wp'(z)^2=  4 \wp(z)^3-60G_4(i) \wp(z)+O(z^2)$$ The $O(z^2)$ term vanishes because it is analytic doubly periodic with a zero at $0$.
$\wp(\frac{1-i}2)= -\wp(\frac{1+i}2)=0$

*$$\frac{1+i}2=\int_0^{\frac{1+i}2} dz = \int_0^{\frac{1+i}2} \frac{d\wp(z)}{\wp'(z)}=\int_0^{\frac{1+i}2} \frac{d\wp(z)}
{\sqrt{4\wp(z)^3-60G_4(i)\wp(z)}}$$ $$=\int_\infty^0 \frac{dx}{2\sqrt{x^3-15 G_4(i) x}}
=\frac{i}{2}( 15 G_4(i))^{-1/4}\int_0^1+\int_1^\infty \frac{dX}{\sqrt{X-X^3}}$$
$$=\frac{i+1}{2}(15G_4(i))^{-1/4} \int_0^1 \frac{dt^{1/2}}{\sqrt{t^{1/2}(1-t)}}=
\frac{i+1}{4}( 15G_4(i))^{-1/4}\beta(1/4,1/2)$$ $$=\frac{i+1}{4}( 15 G_4(i))^{-1/4}\frac{\Gamma(1/4)\Gamma(1/2)}{\Gamma(3/4)} =\frac{i+1}{4}( 15G_4(i))^{-1/4}\Gamma(1/4)^2 \sqrt{\pi}  \frac{\sin(\pi/4)}{\pi}$$
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} (1-e^{2i \pi nz})^{24} $ of weight $12$ , since it has only one simple zero at $i\infty$ then $\frac{E_4(z)^3-E_6(z)^2}{\Delta(z)}$ is a modular form of weight $0$ thus it is constant,  thus for $f$ of weight $2k=4a+6b\ge 12$  then $\frac{f-f(i\infty) E_4(z)^a E_6(z)^b}{\Delta(z)}$ is of weight $2k-12$ 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 $E_{4k},E_4,E_6$.
A: It is well known that $$G_{4k}(i)=\left(4\int_{0}^1\frac{1}{\sqrt{1-x^4}}dx\right)^{4k}\frac{H_{4k}}{(4k)!}, $$ 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.
A: Im writing a paper about negative weights Eisenstein series evaluated at $z=i$. Set
$$
E_{\nu}(z)=1+\frac{1}{\zeta(1-\nu)}\sum_{n=1}^{\infty}\sigma_{\nu-1}(n)q^n\textrm{, }q=e(z)\textrm{, }Im(z)>0.
$$
I have evaluated among other values ($C$ is Catalan constant):
$$
E_{-2}(i)=\frac{7\pi^3}{\zeta(3)}\textrm{, }E_{-2}'(i)=\frac{i(240C\pi-29\pi^3)}{720 \zeta(3)}\textrm{, }E_{-2}''(i)=\frac{80C\pi-11\pi^3}{240 \zeta(3)}
$$
$$
E_{-2}^{(3)}(i)=\frac{i\left(1-E_{4}(i)\right)\pi^3}{15\zeta(3)}\textrm{, }E_{-2}^{(4)}(i)=\frac{2E_{4}(i)\pi^3}{15\zeta(3)}
$$
$$
E_{-4}^{(5)}(i)=\frac{8i\pi^5}{63\zeta(5)}
$$
In general holds the trivial formula
$$
E_{2n+2}(z)=1+\frac{\zeta(2n+1)}{\zeta(-2n-1)(2\pi i)^{2n+1}}\frac{d^{2n+1}}{dz^{2n+1}}E_{-2n}(z)
$$
Also
$$
E'_{-4}(i)=-2iE_{-4}(i)+\frac{13i\pi^5}{1890\zeta(5)}
$$
$$
E_{-4}''(i)=-3E_{-4}(i)+\frac{11\pi^5}{378\zeta(5)}-\frac{\pi^2\zeta(3)}{4\zeta(5)}
$$
$$
E_{-4}^{(3)}(i)=3iE_{-4}(i)-\frac{i\pi^5}{14\zeta(5)}+\frac{3i\pi^2\zeta(3)}{2\zeta(5)}
$$
