Eisenstein $E_2$ at imaginary quadratic arguments

Question

In the paper On Epstein's zeta-function, Chowla and Selberg give a formula for evaluating the Dedekind eta function $$\eta (\tau)=e^{\pi i\tau/12}\prod_{n=1}^\infty (1-e^{2\pi i n\tau}),\quad \Im\tau\gt 0$$ when its argument is imaginary quadratic; it has a closed form in terms of the gamma function at rational arguments.

In turn, this gives the normalized Eisenstein series $$E_{2k}(\tau)=1-\frac{4k}{B_{2k}}\sum_{n=1}^\infty \sigma_{2k-1}(n)e^{2\pi i\tau n}$$ ($\sigma$ is the divisor function, $B$'s are Bernoulli numbers) at imaginary quadratic arguments in terms of the gamma function at rational arguments for $k\ge 2$ only – meaning $E_2$ is left out.

This left me wondering: How to evaluate $E_2$ at imaginary quadratic arguments in terms of the gamma function at rational arguments?

It is known for $\tau=i\sqrt{r}$ where $r$ is positive rational, as suggested by the alpha function in Pi and the AGM (p. 152, p. 164 Ex. 15) by Borwein & Borwein.

But how should I evaluate, for example, $E_2\left(\frac{1+5i}{2}\right)$?

Answer 1

Exactly the same Chowla--Selberg formula is valid, but you must apply it to the modified (non-holomorphic) $$E_2^*(\tau)=E_2(\tau)-3/(\pi\Im(\tau))$$ In other words, $E_2^*(\tau)/\eta^4(\tau)$ is an algebraic number belonging to a number field of known properties. For $\tau=(1+5i)/2$ it has degree 8 (you can easily find it using for instance Pari/GP's algdep command, working at 115D). In fact, one has the additional property that if $D$ is the discriminant of $\tau$, then $\sqrt{D}((E_2^*)^2/E_4)(\tau)$ is an algebraic integer.

Answer 2

Let $r$ be a positive rational number and let $\tau=(1+i\sqrt{r}) /2$ so that $\exp(2\pi i\tau) =-e^{-\pi\sqrt{r}} =-q$ and then your $E_2(\tau)$ is nothing but $P(-q) $ with $$P(q) =1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n}$$ It can be proved that if $k$ is elliptic modulus and $K, E$ elliptic integrals corresponding to nome $q$ then $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)\tag{1}$$ Ramanujan proved in his paper Modular equations and approximations to $\pi$ that $$P(q^2)=\frac{3}{\pi\sqrt{r}}+A_r\left(\frac{2K}{\pi}\right)^2\tag{2}$$ for some algebraic number $A_r$ and he described a direct method to get these algebraic numbers.

For $r=25$ I have done the calculations in this answer and we have $$P(q^{2})=\frac{3}{5\pi}+\left(\frac{2K}{\pi}\right) ^26(5-3\phi)\sqrt{3\phi-4}$$ where $\phi=(1+\sqrt{5})/2$ is golden ratio. The value of $K$ is also available from this answer and then one can get the value of $P(-q) $ using $(1)$.