I'm looking for a proof that the following term is an algebraic integer whenever $\tau_N=\frac{N+\sqrt{-N}}{2}$ is a quadratic irrationality with class number $1$:

$$A_N:=\sqrt{-N}\cdot\frac{E_2(\tau_N)-\frac{3}{\pi\cdot Im(\tau_N)}}{\eta^4(\tau_N)}$$

Here $\eta$ denotes the Dedekind $\eta$-Function and $E_2$ is the Eisenstein series of weight $2$.

As @HenriCohen said here: How to compute Coefficients in Chudnovsky's Formula? it follows from theorems of complex multiplication, but I couldn't find such theorems.

I calculated the numerical value of $A_N$ for all discriminants with class number 1. The results are:

  • $A_3 = 0$
  • $A_4 = 0$
  • $A_{7}=3\cdot e^{i\pi/3}$
  • $A_{8}=4\cdot e^{i\pi/2}$
  • $A_{11}=8\cdot e^{i\pi/3}$
  • $A_{12}=6\cdot4^{1/3}\cdot e^{i\pi/2}$
  • $A_{16}=12\cdot2^{1/2}\cdot e^{i\pi/2}$
  • $A_{19}=24\cdot e^{i\pi/3}$
  • $A_{27}=24\cdot9^{1/3}\cdot e^{i\pi/3}$
  • $A_{28}=54\cdot e^{i\pi/2}$
  • $A_{43}=144\cdot e^{i\pi/3}$
  • $A_{67}=456\cdot e^{i\pi/3}$
  • $A_{163}=8688\cdot e^{i\pi/3}$

Thus we get numerically, that these $A_N$ are algebraic integers, but I don't see how I can prove it. Does anyone know how to do that?

EDIT: Thanks to the answer of Nikos Bagis, only the $A_N$ with odd $N$ remain to be proven. I moved the remaining part of the question here, where a complete answer was given by Michael Griffin.

EDIT: Complete Solution:

The answer of Michael Griffin (see here) can now be found in more details in the appendix of this arXiv-preprint.

Edit: Ramanujan-Sato-Series

Tito Piezas III found some Ramanujan-Sato-Series of level 9 which can be expressed with these numbers $A_N$ (see this question).

  • 1
    $\begingroup$ In your first display, what is that funny symbol in 3-over-pi-funnysymbol-tau-sub-N? $\endgroup$ May 29, 2018 at 23:03
  • $\begingroup$ The funny symbol is "imaginary part". I did some calculations to dozens of decimal digits and confirmed half of the values. Unless there is typos in the values of $A_N$ they are wrong for $N=4,8,12,16,27,28$ and these values of N would not be expected to have algebraic values. The others are correct. $\endgroup$
    – Somos
    May 30, 2018 at 1:52
  • $\begingroup$ Thank you @somos! but as the function is periodic with period 1, we get $\tau_N=\frac{1+\sqrt{-N}}{2}$ for odd $N$ and $\tau_N=\frac{\sqrt{-N}}{2}$ for even $N$. $\endgroup$
    – L. Milla
    May 30, 2018 at 3:22
  • $\begingroup$ I made my own typo. $N=27$ value is correct. I corrected your $\tau_N = \frac{1+\sqrt{-N}}2$. $\endgroup$
    – Somos
    May 30, 2018 at 3:27
  • 1
    $\begingroup$ Okay, with $\frac{N+\sqrt{-N}}2$ then your values are correct except for sign. That is, $A_8^6 = -4096$ and you have $4096$. Same for the six values I thought were incorrect. $\endgroup$
    – Somos
    May 30, 2018 at 3:37

3 Answers 3


For $|q|<1$ we consider the null Jacobi theta functions $$ \theta_2(q):=\sum^{\infty}_{n=-\infty}q^{(n+1/2)^2}\textrm{, } \theta_3(q):=\sum^{\infty}_{n=-\infty}q^{n^2}\textrm{, } \theta_4(q):=\sum^{\infty}_{n=-\infty}(-1)^nq^{n^2}. $$ For $q=e^{-\pi \sqrt{r}}$, $r>0$ the elliptic singular modulus $k=k_r$ is given by $$ k_r=\left(\frac{\theta_2(q)}{\theta_3(q)}\right)^2. $$ The complete elliptic integrals of the first and second kind are: $$ K(x)=\frac{\pi}{2}{}_2F_1\left(\frac{1}{2},\frac{1}{2};1;x^2\right) $$ and $$ E(x)=\frac{\pi}{2}{}_2F_1\left(-\frac{1}{2},\frac{1}{2};1;x^2\right). $$ Then


Let $$ f(-q):=\prod^{\infty}_{n=1}(1-q^n)\textrm{, }|q|<1,\tag 1 $$ then if $q=e^{-\pi\sqrt{r}}$, $r>0$, we have $$ P(q^2)=1-24\sum^{\infty}_{n=1}\frac{nq^{2n}}{1-q^{2n}}=\frac{3}{\pi\sqrt{r}}+\left(1+k^2_r-\frac{3\alpha(r)}{\sqrt{r}}\right)\frac{4}{\pi^2}K^2.\tag 2 $$


Let $q=e^{-\pi\sqrt{r}}$, $r>0$. Differentiating with respect to $r$ the relation $$ \log\left(f(-q^2)\right)=\sum^{\infty}_{n=1}\log\left(1-q^{2n}\right)\textrm{, }|q|<1, $$ and using (see [W,W] Chapter 21, Miscellaneous examples 10, pg. 488): $$ f(-q^2)^6=\prod^{\infty}_{n=1}\left(1-q^{2n}\right)^6=\frac{2kk'K(k)^3}{\pi^3q^{1/2}}, $$ we get $$ \frac{1}{6}\frac{d}{dr}\log\left(\frac{2k_rk'_rK^3}{\pi^3q^{1/2}}\right)=-2\sum^{\infty}_{n=1}\frac{nq^{2n-1}}{1-q^{2n}}\frac{dq}{dr}. $$ After some calculations we arrive to (reminder: $P(q^2)=1-24P^{*}(q^2)$): $$ \frac{1}{24}\left(\frac{\pi}{\sqrt{r}}+12\frac{1}{K}\frac{dK}{dk_r}\frac{dk_r}{dr}+\frac{4}{k_r}\frac{dk_r}{dr}+\frac{4}{k'_r}\frac{dk'_r}{dr}\right)=2q^{-1}P^{*}(q^2)\frac{q\pi}{2\sqrt{r}} $$ Using the known relations (see [Ber3] Chapter 17 Entry 9 pg. 120 and [Ber2] Chapter 11 Entry 30 pg. 87-88): $$ \frac{dk_r}{dr}=\frac{-k_r(k'_r)^2K^2}{\pi\sqrt{r}}, $$ $$ \frac{dk_r'}{dr}=\frac{k_r^2k_r'K^2}{\pi\sqrt{r}} $$ and ([Borw,Borw] Chapter 1, Section 1.3, pg. 7-11): $$ \frac{dK}{dk_r}=\frac{E}{k_r(k_r')^2}-\frac{K}{k_r}, $$ $$ \alpha(r)=\frac{\pi}{4K^2}-\sqrt{r}\left(\frac{E}{K}-1\right), $$ we arrive to $$ P^{*}(q^2)=-\frac{1}{24}+\frac{K^2}{6\pi^2}+\frac{K^2k_r^2}{6\pi^2}-\frac{\alpha(r)K^2}{2\pi^2\sqrt{r}}+\frac{1}{8\pi\sqrt{r}}. $$ From this along with $P(q^2)=1-24P^{*}(q^2)$, we get the result. qed

Remark that $\alpha(r)$, $r>0$ is the elliptic alpha function in [Borw,Borw] book. It is proven in [Borw,Borw] that $\alpha(r)$ is algebraic number whenever $r$ is positive rational.


Let $r>0$ and $q=e^{-\pi\sqrt{r}}$, $K=K(k_r)$, then $$ 1-24\sum^{\infty}_{n=1}\frac{nq^n}{1-q^n} =\frac{6}{\pi\sqrt{r}}+\left(1+k^2_r-\frac{6 \alpha(r)}{\sqrt{r}}\right)\frac{4K^2}{\pi^2}= $$ $$ =\frac{6}{\pi\sqrt{r}}+s_1(r)\theta_3(q)^4,\tag 3 $$ where $$ s_1(r):=1-\frac{6\alpha(r)}{\sqrt{r}}+k_r^2.\tag 4 $$


From (1) setting $r\rightarrow r/4$ and using $$ k_{r/4}=\frac{2\sqrt{k_r}}{1+k_r}\textrm{, }M_2(r)=\frac{1+k'_r}{2} $$ and (see [Boew,Borw]) $$ \alpha(4r)=(1+k_{4r})^2\alpha(r)-2\sqrt{r}k_{4r},\tag a $$ we get the result. qed

The relations you are looking for are (3) and (4).

Notes. The upper half plane formulation can constructed if in place of $k_r$ we define for $q=e^{2\pi i \tau}$, $Im(\tau)>0$, the singular modulus: $$ m(q):=\left(\frac{\theta_2(q)}{\theta_3(q)}\right)^2\textrm{, }q=e^{2\pi i \tau}\textrm{, }Im(\tau)>0\textrm{, }-1/2\leq Re(\tau)<1/2 $$ and $m'(q)=\sqrt{1-m(q)^2}$, then $$ \theta_2(q)^2=\frac{2m(q)K(m(q))}{\pi}\textrm{, }\theta_3(q)^2=\frac{2K(m(q))}{\pi}\textrm{, }\theta_4(q)^2=\frac{2m'(q)K(m(q))}{\pi}. $$ Then if also we have $$ i\frac{K\left(m'(q)\right)}{K(m(q))}=2z $$ see [Arm,Eb],[Borw,Borw]$\ldots etc$. You have to find the complex analog of elliptic alpha $\alpha(r)$.


[Arm,Eb]: J.V. Armitage, W.F. Eberlein. 'Elliptic Functions'. Cambridge University Press. (2006)

[Bag]: N.D. Bagis. 'On certain theta functions and modular forms in Ramanujan theories'. arXiv:1511.03716v2 [math.GM] 6 Dec 2017.

[Ber2]: Bruce. C. Berndt. 'Ramanujan`s Notebooks Part II'. Springer-Verlag, New York. 1989.

[Ber3]: B.C. Berndt. 'Ramanujan`s Notebooks Part III'. Springer Verlag, New York (1991).

[Borw,Borw]: J.M. Borwein and P.B. Borwein. 'Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity', Wiley, New York, 1987.

[W,W]: E.T. Whittaker and G.N. Watson. 'A course on Modern Analysis'. Cambridge U.P. 1927.


From relations (3) and (4) we get $q=e^{-\pi\sqrt{r}}$, $r>0$ $$ \left(E_2(q)-\frac{6}{\pi\sqrt{r}}\right)\frac{1}{\theta_3(q)^4}=1-\frac{6\alpha(r)}{\sqrt{r}}+k_r^2\tag 5 $$ If $r$ is positive rational, then $\alpha(r)$ and $k_r$ are algebraic numbers.

Moreover for any integer $N$ we can get easily the evaluation (in each case separate) of $k_r$ usng $$ \left(\frac{256}{r_0^{16}}-r_0^8\right)^3=j\left(\frac{1+\sqrt{-N}}{2}\right), $$
where $r=N-$integer and $$ j\left(z\right)=\left(\left(\frac{\eta(z/2)}{\eta(z)}\right)^{16}+16\left(\frac{\eta(z)}{\eta(z/2)}\right)^8\right)^3 $$ is the known $j-$invariant. Then $$ k_N^2=\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{16}{r_0^{24}}} $$ An interesting article is [Broad] which reduces these calculations very considerably.

Another way is to evaluate it using $$ j_r=j(i\sqrt{r})=\frac{256(k_r^2+(k'_r)^4)^3}{(k_rk'_r)^4}\textrm{, }k'_r=\sqrt{1-k_r^2}.\tag b $$ The definition and evaluations of $\alpha(r)$ function are given in [Borw,Borw] and you must read the related theory (book).

The elliptic alpha function $\alpha(r)$ also has been introduced for evaluation of billions of digits of $\pi$. The Borwein bothers give in their book the underlined theory of $\pi-$formulas and algorithms.

Note also that [W,W] ($q=e^{-\pi\sqrt{r}}$, $r>0$): $$ \prod^{\infty}_{n=1}(1-q^n)=2^{1/3}\pi^{-1/2}q^{-1/24}(k_r)^{1/12}(k'_r)^{1/3}K(k_r)^{1/2}\tag 6 $$ and $$ \theta_3(q)=\sqrt{\frac{2K(k_r)}{\pi}}\tag 7 $$ and $k_r$ is algebraic number when $r-$positive raional.

Note also that the above two formulas have analytic continuations in upper half plane $Im(z)>0$.


[Broad]: David Broadhurst. 'Solutions by radicals at singular values $k_N$ from new class invariants for $N\equiv 3\textrm{mod}8$'. arXiv:0807.2976v3 [math-ph] 31 Jul 2008.


$$ A_r=\sqrt{-r}\left(E_2(q)-\frac{6}{\pi\sqrt{r}}\right)\frac{1}{\eta_1(q)^4}=\sqrt{-r}\frac{1+k_r^2-6\frac{\alpha(r)}{\sqrt{r}}}{2^{-2/3}(k_r)^{1/3}(k'_r)^{4/3}}\textrm{, }q=e^{-\pi\sqrt{r}}\textrm{, }r>0,\tag 8 $$ where $$ \eta_1(q):=q^{1/24}\prod^{\infty}_{n=1}(1-q^n)\textrm{, }|q|<1. $$ Your case is $q=e^{2\pi i z_N}$, where $z_N=\frac{N+i\sqrt{N}}{2}$, $N$ positive integer. Hence if $N$ is even, then $q=e^{2\pi i \left(N+i\sqrt{N}\right)/2}=e^{-\pi\sqrt{N}}$, $N$ positive even integer. Hence I can evaluate all your $N-$even cases.

1) If $N=r=4$, then $$ k_4=3-2\sqrt{2}, $$ $$ k'_4=2\sqrt{-4+3\sqrt{2}}, $$ $$ \alpha(4)=6-4\sqrt{2}. $$ Hence $$ A_4=\ldots=0 $$ 2) If $N=8=r$, then $$ k_8=-2 \sqrt{2 \left(7+5 \sqrt{2}\right)}+4 \sqrt{2}+5, $$ $$ k'_8=2 \sqrt{\sqrt{2 \left(799+565 \sqrt{2}\right)}-4 \left(7+5 \sqrt{2}\right)} $$ $$ \alpha(8)=-8 \sqrt{41+29 \sqrt{2}}+26 \sqrt{2}+36 $$ Hence $$ A_8=4i $$ 3) For $N=r=12$ $$ k_{12}=15-10 \sqrt{2}+8 \sqrt{3}-6 \sqrt{6} $$ $$ k'_{12}=2 \sqrt{3 \sqrt{2 \left(4801-1960 \sqrt{6}\right)}+85 \sqrt{6}-208} $$ $$ \alpha(12)=2 \left(132-94 \sqrt{2}+77 \sqrt{3}-54 \sqrt{6}\right) $$ Hence $$ A_{12}=6i4^{1/3} $$ For $N=r=16$, we have $$ k_{16}=-4 \sqrt{140+99 \sqrt{2}}+24 \sqrt{2}+33, $$ $$ k'_{16}=2 \sqrt{6 \sqrt{69708+49291 \sqrt{2}}-8 \left(140+99 \sqrt{2}\right)} $$ $$ \alpha(16)=4 \left(-4 \sqrt{15900+11243 \sqrt{2}}+252 \sqrt{2}+357\right) $$ Hence $$ A_{16}=12i\sqrt{2} $$ For $N=r=28$, we have $$ k_{28}=255-180 \sqrt{2}+96 \sqrt{7}-68 \sqrt{14} $$ $$ k'_{28}=2 \sqrt{30 \sqrt{9321998-6591648 \sqrt{2}}+45798 \sqrt{2}-64768} $$ $$ \alpha(28)=82464-58312 \sqrt{2}+31170 \sqrt{7}-22040 \sqrt{14} $$ Hence $$ A_{28}=54 i $$

Where for all evaluations I have used [Borw,Borw] pg.172-173 and relation (a).

For odd $N$ we have $q=-e^{-\pi\sqrt{N}}$, which is very hard.


If $k_r=m\left(e^{-\pi\sqrt{r}}\right)$, with $m(q)$ is as below: $$ m(q):=\left(\frac{\vartheta_2(q)}{\vartheta_3(q)}\right)^2, $$ changing $q\rightarrow-q$, we have $$ m(-q)=\left(\frac{i^{1/2}q^{1/4}\sum_{\scriptsize n\in\textbf{Z}}q^{n^2+n}}{\sum_{\scriptsize n\in\textbf{Z}}(-1)^nq^{n^2}}\right)^2=i\left(\frac{\vartheta_2(q)}{\vartheta_4(q)}\right)^2=i\frac{m(q)}{m'(q)}. $$ Hence $$ m(-q)=i\frac{\cdot m(q)}{m'(q)}\textrm{, }m'(q)=\sqrt{1-m(q)^2}. $$ and from $$ \frac{1}{\sqrt{1-x}}K\left(\frac{x}{x-1}\right)=K(x) $$ we get $$ K^{*}=K(m(-q))=m'(q)K(m(q))=m'(q)K. $$ Hence $$ \eta_1(-q)=\frac{e^{i\pi/24}2^{1/3}}{\sqrt{\pi}}\left(m(q)m'(q)\right)^{1/12}K(m(q))^{1/2} $$ Hence if $N=r=$odd, we have $q=-e^{-\pi\sqrt{r}}$ and $$ A_N:=\sqrt{-N}\left(E_2(z_N)-\frac{3}{\pi Im(z_N)}\right)\frac{1}{\eta(z_N)^4}= \sqrt{-r}\left(P(-q)-\frac{6}{\pi\sqrt{r}}\right)\frac{1}{\eta_1(-q)^4}. $$ But as Paramanand Singh noted, we have $$ P(-q)=\left(\frac{2K}{\pi}\right)^2\left(\frac{6E}{K}+4m(q)^2-5\right). $$ Also $$ E=E(m(q))=\frac{\pi}{4\sqrt{r}K}+\left(1-\frac{\alpha(r)}{\sqrt{r}}\right)K. $$ Hence for $N=r=$odd $$ A_N=A_r=\ldots=\frac{e^{i\pi/3}4^{1/3}\left(-6\alpha(r)+\sqrt{r}(1+4m(q)^2)\right)}{\left(m(q)m'(q)\right)^{1/3}}\textrm{, }q=e^{-\pi\sqrt{r}}\tag 9 $$

Hence if $k_r$, $\alpha(r)$, for $r=3,7,11,19,27,43,67,163$ are known, then $A_r=A_N$ are known and hence answer to the evaluations of the case $z_N=\frac{N+\sqrt{-N}}{2}$, with $N-$odd. For all odd cases, since $h(-N)=1$ you can evaluate the $j-$invariant $j(i\sqrt{N})$ which (see [Broad]) have minimal polynomials of degree at most 3. Then using (b) you can find $k_r$.

For the elliptic alpha function, you have to search. With a quick view I find $r=3,7,27$ only (see [Borw,Borw] pg.173)

  • $\begingroup$ Nikos, thanks a lot for your answer! Even if the question is some months old, I'm still very interested in any hints. Unfortunately, I don't see how it is related to proving that $A_N$ is an algebraic integer. Could you please give some details on the connection betwen your answer and my question? $\endgroup$
    – L. Milla
    Sep 27, 2018 at 14:42
  • $\begingroup$ For not using the comments section, I write the details in my answer. $\endgroup$ Sep 27, 2018 at 20:54
  • $\begingroup$ I'm sorry to insist, but I still don't see how this proves that $A_N$ is an algebraic integer... $\endgroup$
    – L. Milla
    Sep 29, 2018 at 5:27
  • 1
    $\begingroup$ For odd $n$ you need to consider $P(-q) $ and the link between $P(q) $ and $P(-q) $ is obtained by the formulas $$P(q) =\left (\frac{2K}{\pi}\right)^2\left(\frac{6E}{K}+k^2-5\right)$$ and $$P(-q) =\left (\frac{2K}{\pi}\right)^2\left(\frac{6E}{K}+4k^2-5\right)$$ and thus $\left(P(-e^{-\pi\sqrt{n}}) - \dfrac{6}{\pi\sqrt{n}}\right) \cdot\dfrac{\pi^2}{4K^2}$ is an algebraic number if $n$ is a positive rational number. $\endgroup$ Nov 17, 2018 at 7:27
  • 1
    $\begingroup$ In particular these posts also show how Chudnovsky series can also be obtained from Ramanujan's ideas and it is rather surprising that Ramanujan himself missed it. $\endgroup$ Nov 17, 2018 at 8:34

This is too long to fit in a comment.

Ramanujan established in his monumental paper Modular Equations and Approximations to $\pi$ that the desired expression is an algebraic number if $\tau_n=\sqrt{-n} $ where $n$ is a positive rational number.

Let $$P(q) =1-24\sum_{j=1}^{\infty}\frac{jq^{j} }{1-q^{j}}\tag{1}$$ and then Ramanujan proved that $$P(e^{-2\pi\sqrt{n}}) =\left( \frac{2K}{\pi}\right)^2A_n+\frac{3}{\pi\sqrt{n}}\tag{2}$$ where $A_n$ is an algebraic number dependent on $n$ provided that $n$ is a positive rational number. Here $K=K(k) $ is the complete elliptic integral of first kind with modulus $k$ and $k$ corresponds to nome $q$ so that $$k=\frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)},q=e^{-\pi\sqrt{n}}\tag{3}$$ and $k$ is an algebraic number if $n$ is a positive rational number. Ramanujan's proof is presented in one of my blog posts.

If $q=\exp(\pi i\tau) $ then we have $P(q^2) =E_2(\tau)$. Further we have $$\eta(\tau) =q^{1/12}\prod_{j=1}^{\infty}(1-q^{2j}),q=e^{\pi i\tau}\tag{4}$$ It is well known that the eta function can be expressed in terms of $k, K$ as $$\eta(\tau)=2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{5}$$ and therefore the equation $(2)$ can be written as $$A_n=\dfrac{E_{2}(\tau_n)-\dfrac{3}{\pi\Im{\tau_n}}}{\eta^4(\tau_n)}$$ which is an algebraic number.

Note that if $\tau_n=\dfrac{n+\sqrt{-n} }{2}$ then we have $E_2(\tau_n)=P(q^2)$ if $n$ is even and $q=e^{\pi i\tau_n} $ and if $n$ is odd then $E_2(\tau_n)=P(-q^2)$. Using Ramanujan's technique one can prove that $$P(e^{-\pi\sqrt{n}})=\left(\frac{2K}{\pi}\right) ^2B_n+\frac{6}{\pi\sqrt{n}}\tag{6}$$ (just replace $n$ in $(2)$ by $n/4$ and $B_n=A_{n/4}$) and $$P(-e^{-\pi\sqrt{n}}) =\left(\frac{2K}{\pi}\right)^2C_n+\frac{6}{\pi\sqrt{n}}\tag{7}$$ where $B_n, C_n$ are algebraic numbers and $n$ is a positive rational number and thus the expression mentioned in the question is an algebraic number. Proving that it is an algebraic integer is unfortunately not possible via Ramanujan methods.

  • $\begingroup$ This is very nice. I will search in your posts. $\endgroup$ Nov 17, 2018 at 8:54
  • $\begingroup$ "Proving that it is an algebraic integer is unfortunately not possible via Ramanujan methods." - Thank you for this warning! $\endgroup$
    – L. Milla
    Nov 24, 2018 at 13:12
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    $\begingroup$ @L.Miller: Ramanujan mostly used processes of calculus (differentiation and integration) combined with a huge amount of algebraic manipulation (in most economic fashion) to arrive at his results and didn't have any idea of the theory of quadratic number fields (algebraic number theory) and thus there are some limitations to what can be achieved using his methods. But it is still astonishing that he used these ideas and simple tools to find beautiful series for $1/\pi$ way back during British rule in India with modest resources. $\endgroup$ Nov 27, 2018 at 8:34
  • $\begingroup$ @ParamanandSingh: You may be interested in this new MO post. L. Miller brought my attention to his work which enabled me to find a missing piece of the puzzle. $\endgroup$ Aug 21, 2019 at 5:15

I would like to outline a conceptual idea which does not answer this question directly.

The contents of Section 14, Chapter VIII of Weil's book Elliptic Functions according to Eisenstein and Kronecker implies that $$E_2(\tau)-\frac{3}{\pi\, Im(\tau)}=\frac{3}{\pi^2}K_2(0,0,2),$$

where $K_2(0,0,s)$ is defined by $$K(0,0,s):=\sum_{w\in W}{^*}\frac{\bar{w}^2}{|w|^{2s}}$$ with $Re(s)>2$, $W$ is the lattice spanned by $1,\tau$ and $K(0,0,2)$ is defined by analytic continuation of $K(0,0,s)$.

Let $1,\tau$ be the basis of algebraic integers of some imaginary quadratic field of class number $1$. Then $K(0,0,s)$ is an integral multiple of Hecke $L$-function whose character is of $(2,0)$-type. We would like to remark that to evaluate $K(0,0,2)$ is to evaluate a critical value of a Hecke L-function. The observation attributed to Deligne et al. asserts that the critical values of these L-functions are some algebraic periods times an algebraic number, and the claim in the OP might be solved by refiner studies of these Hecke L-functions. See p. 12 of M. Watkins' note for an example.

P.S. This idea can be extended to other Ramanujan series arising from quadratic fields of class number 2.

  • $\begingroup$ This sounds promising, thank you! Could you please state the mentioned Hecke L-Function more explicitly? $\endgroup$
    – L. Milla
    Jun 18, 2018 at 16:42
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    $\begingroup$ @L.Miller If $O_K \subset \mathbb{C}$ is an imaginary quadratic field which is a PID and with $O_K^\times = \pm 1$ then $\sum_{w\in O_K^*}\frac{\bar{w}^2}{|w|^{2s}} = 2\sum_{I\subset O_K} \chi(I) N(I)^{1-s}$ where $\chi(a O_K) = \overline{a}^2$ is a Hecke character (with trivial finite part). If $O_K$ doesn't have class number one then the series is a linear combination $\sum_\psi L(s-1,\psi\chi)$ with $\psi$ the characters of the class group allowing to isolate the class of principal ideals $\endgroup$
    – reuns
    Oct 4, 2018 at 2:11
  • 1
    $\begingroup$ @Y.Zhao Similar to Dirichlet L-functions the period should depend on the symmetries of the character $\endgroup$
    – reuns
    Oct 4, 2018 at 2:24

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