My aim is to understand all three coefficients arising in the Chudnovsky-Formula (see also Question [300385](https://mathoverflow.net/questions/300385/why-does-this-quasi-modular-function-have-integral-values)). Two of them are easily computed, but I failed with the third:


It is known that for all $\tau$ with $Im(\tau)>1.25$ we have
\begin{align\*}
 \frac{1}{2\pi Im(\tau)}\sqrt{\frac{J(\tau)}{J(\tau)-1}} &= \sum\_{n=0}^\infty \left( \frac{1-s\_2(\tau)}{6} + n \right)\cdot \frac{(6n)!}{(3n)!(n!)^3}\cdot \frac{1}{\left(1728J(\tau)\right)^n}\\
\text{with }s\_2(\tau) &:= \frac{E\_4(\tau)}{E\_6(\tau)}\left(E\_2(\tau)-\frac{3}{\pi Im(\tau)}\right)
\end{align\*}
Then for $\tau=\frac{1+i\sqrt{163}}{2}$ it is known that $1728J(\tau)=-640320^3$.


This gives us
\begin{align\*}
 \frac{1}{\pi} &= \sum\_{n=0}^\infty \frac{(-1)^n(6n)!}{(3n)!(n!)^3}\cdot \frac{A + B\cdot n}{640320^{3n+3/2}}
\end{align\*}
with $B = \sqrt{163\cdot(1728+640320^3)} = 12\cdot 545140134$


$$\text{and } A = 12\cdot 545140134\cdot\left( \frac{1-s\_2(\tau)}{6} \right)$$


It can be easily computed that $A$ is *approximately* $12\cdot13591409$


**Question:** How can I prove that the value of $A$ is *exactly* this number?


*Edit*: Thanks to the answer of @HenriCohen, the only thing left to prove is [this](https://mathoverflow.net/questions/301476/why-are-values-of-eisenstein-e-2-algebraic-integers): a reference is needed, why $\sqrt{D}E\_2^\*(\tau)/\eta^4(\tau)$ is an algebraic integer.


**EDIT: Answer:**


Thanks to the answers of Henri Cohen (see below) and Michael Griffin (see [here](https://mathoverflow.net/questions/311879/complex-multiplication-and-algebraic-integers)), an explicit calculation of the coefficients along with a complete proof of their exactness can now be found in chapter 10 and the appendix of [this](https://arxiv.org/abs/1809.00533) arXiv-preprint.



How to compute Coefficients in Chudnovsky's Formula?
Let $\tau$ be any CM point. By basic theorems of complex multiplication, if you choose a suitable period
$\omega(\tau)$, $E\_4(\tau)/\omega(\tau)^4$, $E\_6(\tau)/\omega(\tau)^6$,
and $\sqrt{D}E\_2^\*(\tau)/\omega(\tau)^2$ (with $E\_2^\*(\tau)=E\_2(\tau)-3/(\pi\Im(\tau))$ and $D$ the discriminant of $\tau$) will be algebraic numbers of known degree, and if you choose
$\omega(\tau)=\eta(\tau)^2$, they will even be algebraic integers.
Incidentally (but this is not needed) the Chowla--Selberg formula gives
$\eta(\tau)$ explicitly as a product of gamma values.


In the case $\tau=(1+\sqrt{-163})/2$, we easily find that
$E\_4/\omega^4=-640320\rho$, $E\_6/\omega^6=-40133016\sqrt{-163}$, and
$\sqrt{-163}E\_2^\*/\omega^2=-8688\rho^2$ with $\rho=(-1+\sqrt{-3})/2$ a cube
root of unity. This gives $s\_2(\tau)=77265280/90856689$ which implies
the result.


