Let $\eta(\tau)$ be the Dedekind eta function. In his Lost Notebook, Ramanujan played around with a related function and came up with some of the nice evaluations,

$$\begin{aligned} \eta(i) &= \frac{1}{2} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\ \eta(2i) &= \frac{1}{2^{11/8}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\ \eta(3i) &= \frac{1}{2\cdot 3^{3/8}} \frac{1}{(2+\sqrt{3})^{1/12}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\ \eta(4i) &= \frac{1}{2^{29/16}} \frac{1}{(1+\sqrt{2})^{1/4}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\ \eta(5i) &= \frac{1}{2\sqrt{5}}\left(\tfrac{1+\sqrt{5}}{2}\right)^{-1/2}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\ \eta(6i) &=\; \color{red}{??}\\ \eta(7i) &= \frac{1}{2\sqrt{7}}\left(-\tfrac{7}{2}+\sqrt{7}+\tfrac{1}{2}\sqrt{-7+4\sqrt{7}} \right)^{{1/4}}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\ \eta(8i) &= \frac{1}{2^{73/32}} \frac{(-1+\sqrt[4]{2})^{1/2}}{(1+\sqrt{2})^{1/8}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\ \eta(16i) &= \frac{1}{2^{177/64}} \frac{(-1+\sqrt[4]{2})^{1/4}}{(1+\sqrt{2})^{1/16}} \left(-2^{5/8}+\sqrt{1+\sqrt{2}}\right)^{1/2}\,\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\end{aligned}$$

with the higher ones $>4$ added by this OP. (Note the powers of $2$.)


  1. Similar to the others, what is the exact value of $\eta(6i)$?
  2. Is it true that the function, $$F(\sqrt{-N}) = \frac{\pi^{3/4}}{\Gamma\big(\tfrac{1}{4}\big)}\,\eta(\sqrt{-N}) $$ is an algebraic number only if $N$ is a square?

P.S. It seems strange there is a function that yields an algebraic number for square input $N$ and a transcendental number for non-square $N$. (Are there well-known functions like that?) For an example of non-square $N$, we have,

$$\eta(\sqrt{-3}) = \frac{3^{1/8}}{2^{4/3}} \frac{\Gamma\big(\tfrac{1}{3}\big)^{3/2}}{\pi} = 0.63542\dots$$

and $F(\sqrt{-3})$ seems to be transcendental.

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    $\begingroup$ Have you seen van der Poorten and Williams, Values of the Dedekind eta function at quadratic irrationalities, Canad J Math 51 (1999) 176-224? $\endgroup$ – Gerry Myerson Jun 22 '15 at 13:37
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    $\begingroup$ @GerryMyerson: No, but I just found a copy here. It has 49 pages of math, but their main result (Theorem 9.3) for $\eta\left(\frac{b+\sqrt{d}}{2a} \right)$ involves a product of gamma functions of form $\Gamma\left(\frac{m}{d}\right)$. It doesn't seem to address the special case when $d$ is a square and the only gamma factor needed (apparently) is $\Gamma \big( \tfrac{1}{4}\big)$. $\endgroup$ – Tito Piezas III Jun 22 '15 at 13:58
  • $\begingroup$ @TitoPiezasIII. Can this title be improved upon? See the comments of math.stackexchange.com/questions/2852787/… $\endgroup$ – Mason Jul 17 '18 at 3:08

Since we know the value of $\eta(3i)$, the point is just to compute the value of the product: $$ \prod_{n\geq 0}(1+e^{-6\pi n})=\exp\sum_{n\geq 0}\log\left(1+e^{-6\pi n}\right)=\exp\sum_{n\geq 0}\int_{n}^{n+1}\frac{6\pi n}{1+e^{6\pi s}}\,ds$$ where: $$\sum_{n\geq 0}\int_{n}^{n+1}\frac{6\pi n}{1+e^{6\pi s}}\,ds = \int_{0}^{+\infty}\frac{6\pi s\,}{1+e^{6\pi s}}-\int_{0}^{+\infty}\frac{6\pi\{s\}}{1+e^{6\pi s}}$$ and the first integral in the RHS equals $\frac{\pi}{72}$ by the residue theorem, while expanding the fractional part as its Fourier series, $\{s\}=\frac{1}{2}-\sum_{n\geq 1}\frac{\sin(2\pi n s)}{\pi n}$, we get:

$$\begin{eqnarray*}\int_{0}^{+\infty}\frac{6\pi\{s\}\,ds}{1+e^{6\pi s}}&=&\frac{\log 2}{2}-\sum_{n\geq 1}\int_{0}^{+\infty}\frac{6 \sin(2\pi n s)}{n(e^{6\pi s}+1)}\,ds\\&=&\frac{\log 2}{2}-\sum_{n\geq 1}\frac{6}{n}\sum_{m\geq 0}(-1)^m\int_{0}^{+\infty}\sin(2\pi n s)\,e^{-6\pi m s}\,ds\\&=&\frac{\log 2}{2}-\frac{3}{\pi}\sum_{n\geq 1}\sum_{m\geq 0}\frac{(-1)^m}{9 m^2+n^2}\\&=&\frac{\log 2}{2}-\frac{\pi}{2}-\frac{3}{\pi}\sum_{m,n\geq 1}\frac{(-1)^m}{9m^2+n^2}\end{eqnarray*}$$ and the last series just depends on the number of ways to represent a positive integer $\not\equiv 2\pmod{3}$ through the binary quadratic form $n^2+9m^2$: it is, with minor manipulations, just a Dirichlet convolution. I have just applied the same techniques of this answer, just in reverse.

This shows a clear connection between the evaluation of the Dedekind eta function at quadratic irrationals and the class number problem: $\eta(\sqrt{-N})$ depends on $\sum_{n\geq 1}(-1)^n\frac{r(n)}{n}$, where $r(n)$ counts the number of ways to represent $n$ as $a^2+Nb^2$. If $N$ is a square or $a^2+Nb^2$ is the only reduced quadratic form of discriminant $-4N$ (class number one) we may explicitly compute such series, and it turns out that $F(\sqrt{-N})$ is an algebraic number. Otherwise, $\sum_{n\geq 1}(-1)^n\frac{r(n)}{n}$ is not even a convolution of Dirichlet series, hence your conjecture is very likely to hold.

Ultimately, the computation of $\eta(6i)$ can be carried on by recalling that:

$$j(\tau)=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^8+2^8\left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^{3} $$ and by computing the Klein $j$-invariant $j(3i)$. The Wikipedia page gives:

$$ j(3i) = \frac{1}{27}(2+\sqrt{3})^2(21+20\sqrt{3})^3.$$

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    $\begingroup$ The Wikipedia page for $j(\tau)$ is currently inconsistent; the value given is the one for $J(3\mathrm{i}) = j(3\mathrm{i})/1728$. Therefore the $1/27$ should actually be a factor $64$. $\endgroup$ – ccorn Jul 4 '15 at 5:44
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    $\begingroup$ Note that $$\sum_{m, n\in\mathbb{Z}} '\frac{(-1)^{m}}{9m^{2}+n^{2}}=4\sum_{m,n\geq 1}\frac{(-1)^{m}}{9m^{2}+n^{2}}+\frac{17\pi^{2}}{54}$$ and the sum on the left in above equation can be evaluated using this answer: math.stackexchange.com/a/2482542/72031 $\endgroup$ – Paramanand Singh Oct 29 '17 at 12:24

After persevering with a Mathematica session, I found that $F(6i)$ is the root of $96$-deg eqn (no wonder it was hard to find!) but could be prettified as,

$$\eta(6i) = \frac{1}{2\cdot 6^{3/8}} \left(\frac{5-\sqrt{3}}{2}-\frac{3^{3/4}}{\sqrt{2}}\right)^{1/6}\,\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

However, the second question is still open.

  • $\begingroup$ (+1) I think that our two answers together reply to the the first part of the question. I found a good evaluation technique but I wasn't brave enough to carry the computations till the end, but with this closed form AND my answer there is very little to add. $\endgroup$ – Jack D'Aurizio Jun 22 '15 at 14:05
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    $\begingroup$ @JackD'Aurizio: Thanks, but could you elaborate more on how your answer addresses Question 2? $\endgroup$ – Tito Piezas III Jun 22 '15 at 14:09
  • $\begingroup$ I haven't touched yet the second part. $\endgroup$ – Jack D'Aurizio Jun 22 '15 at 14:09
  • $\begingroup$ @AccidentalFourierTransform: I was using an old Mathematica version plus some intuition to get this $96$-deg algebraic factor. I believe the new RootApproximant has a default upper bound for the degree, and which may be adjusted. $\endgroup$ – Tito Piezas III Jul 16 '18 at 16:34

The answer to the second question is very simple. Assume the complete elliptic integral of the first kind $$ K(x)=\int^{\pi/2}_{0}\frac{d\theta}{\sqrt{1-x^2\sin^2(\theta)}}=\frac{\pi}{2}\cdot {}_2F_1\left(\frac{1}{2},\frac{1}{2};1;x^2\right). $$ Then if we define the elliptic singular modulus $0<k_r<1$ to be the solution of $$ \frac{K\left(\sqrt{1-k_r^2}\right)}{K(k_r)}=\sqrt{r}\textrm{, }r>0, $$ we know that $k_r$ is algebraic for $r$ positive rational. Assume the notation: $$ K[r]:=K(k_r)\textrm{, }r>0 $$ Also for $r_1,r,N$, such that $N$ positive integer, $r\in Q^{*}_{+}$, with $r_1=N^2r$, we have $$ \frac{K[N^2r]}{K[r]}=M_N(r) $$ The function $M_N(r)$ is called: "the multiplier" and for $r,N$ as above it is an algebraic number.

The Dedekind eta function is just $$ \eta(z)=\eta_1(q):=q^{1/24}\prod^{\infty}_{n=1}\left(1-q^n\right)\textrm{, }q=e^{2\pi i z}\textrm{, }Im(z)>0.\tag 1 $$ We want to find $\eta(i\sqrt{r})$, $r>0$. For this we define the Ramanujan eta function $$ f(-q):=\prod^{\infty}_{n=1}\left(1-q^n\right)\textrm{, }q=e^{-\pi\sqrt{r}}\textrm{, }r>0.\tag 2 $$ The function $f(-q)$ can evaluated at $q=e^{-\pi\sqrt{r}}$, $r>0$ as (see [W,W] Chapter 21 pg.488): $$ f(-q)=\frac{2^{1/3}}{\sqrt{\pi}}q^{-1/24}(k_r)^{1/12}(k'_r)^{1/3}K[r]^{1/2}\tag 3 $$ Assume that $q_1=e^{-\pi\sqrt{r_1}}$, with $r_1=N^2r$ and $GCD(N,r)=1$, ($r_1,N,r$ are integers, $N>1$ and $r$ is square free), then from (1),(2),(3) we have: $$ \eta(i\sqrt{r_1})=\eta(i\sqrt{N^2r})=q_1^{1/12}f(-q_1^2)=algebraic\cdot \frac{\sqrt{K[4r_1]}}{\sqrt{\pi}}=alg\cdot\frac{\sqrt{K[4N^2r]}}{\sqrt{\pi}}= $$ $$ =alg\cdot \sqrt{M_2(r)}\sqrt{M_{N}(r)}\frac{\sqrt{K[r]}}{\sqrt{\pi}} $$ Hence if $r_1$ is a square (which means $r=1)$, then $$ \eta(i\sqrt{r_1})=alg\cdot\sqrt{\frac{K[1]}{\pi}}=alg\cdot \frac{\Gamma\left(\frac{1}{4}\right)}{\pi^{3/4}} $$ Q.E.D.

Note: $k'_r=\sqrt{1-k_r^2}$. $k'_r$ is called the coplementary modulus.


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


While giving this and this answer I got all the ingredients necessary to solve the current problem.

Consider the nome $q=e^{-\pi} $ so that elliptic modulus is $k=2^{-1/2}$ and the function $f(q) $ defined by $$f(q)=q^{1/6}\prod_{n=1}^{\infty}(1-q^{4n})\tag{1}$$ which is expressed in terms of modulus $k$ and complete elliptic integral $K=K(k) $ as $$f(q) =2^{-2/3}\sqrt{\frac{2K}{\pi}}k^{1/3}k'^{1/12}\tag{2}$$ (see first answer linked above).

If $l$ is the elliptic modulus corresponding to nome $q^{3}$ then we have $$\eta(6i)=f(q^{3})=2^{-2/3}\sqrt{\frac{2L}{\pi}}(l^2\sqrt{l'})^{1/6}\tag{3}$$ The values $$\sqrt{\frac{2L}{\pi}} =\frac{(27+18\sqrt{3})^{1/4}\Gamma(1/4)} {3\sqrt{2}\pi^{3/4}} ,l=\frac{(\sqrt{3}-1)(\sqrt{2}-\sqrt[4]{3})}{2},\sqrt{l'}=\frac{\sqrt{2}+\sqrt[4]{3}(\sqrt{3}-1)}{2^{5/4}}$$ are available from the second answer linked above which help us to get the value of $\eta(6i)$ in explicit form. The calculations are somewhat lengthy but can be performed using pen and paper.


This is a question about the findings of this post as well as this one. If we like the question I can ask it as a new question for the site but it may be that this type of thing is trivially shown not to be the case with some counterexample or maybe it follows quickly: I am not sure.

Let $\alpha = (-\frac{3}{4})!\pi^{-\frac{3}{4}}$ then we have seen that for the set $S=\{\frac{1}{n} : n\in \mathbb{N} \}\cup \mathbb{N}$ we have that

$\eta (Si)\subset A[\alpha]$. So a natural question might be: is it the case that $\eta(\mathbb{Q}i)\subset A[\alpha]$?

I am already changing my completely unverified conjecture:

Maybe it's more like there is some transcendental number $t_m$ we can associate with each $m\in \mathbb{N}$ such that $\eta(i\{x:x \in \{ \frac{m}{n}\} \cup \{ \frac{n}{m} \} \})\subset A[t_m]$? What we know is that this $\eta$ function needs to respect multiplicative inverses in someway.

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    $\begingroup$ The answer to your question is settled thanks to Chowla & Selberg. If $q=e^{-\pi\sqrt{r}}, r\in\mathbb {Q}^{+} $ then the corresponding elliptic integral $K$ can be expressed in terms of values of the Gamma function at rational points. Thus the value of the corresponding eta function is expressible in terms of values of Gamma function and $\pi$ (and some algebraic factor of course). $\endgroup$ – Paramanand Singh Jul 17 '18 at 15:47

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