# Eisenstein $E_2$ at imaginary quadratic arguments

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, $$E_2\left(\frac{1+5i}{2}\right)$$?

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.
• Suppose that I'm able to numerically calculate $\sqrt{D}((E_2^*)^2/E_4)(\tau)$ to a "sufficiently high precision" and the algdep command suggests an algebraic integer $a$. How can I then prove that it is really equal to $a$ and not some other algebraic integer? And thank you for your answer. Commented Apr 17 at 15:33
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 $$(1)$$.