While experimenting with integrals involving the Dedekind Eta function, I came across a family of integrals which seem to follow a very simple pattern. With $y \in \mathbb{N}$, define: $$A(y) = \int_0^{\infty} \eta( i x)\,\eta(i x y)\,dx.$$ The integral can be rewritten in the following infinite series forms:

\begin{align} A(y) & = \frac{12}{\pi} \sum_{(n,m) \in \mathbb{Z}^2} \frac{(-1)^{n+m}}{(6n+1)^2+y \, (6m+1)^2} \\[8pt] & =\frac{2 \sqrt{3}}{\sqrt{y}} \sum_{n \in \mathbb{Z}} \frac{(-1)^n}{6n+1} \, \dfrac{ \sinh \frac{\pi \sqrt{y}}{3} (6n+1)}{\cosh \frac{\pi \sqrt{y}}{2} (6n+1)} \\[8pt] & = \frac{2}{\sqrt{y}} \sum_{n \in \mathbb{Z}} (-1)^n \tanh^{-1} \left( \frac{\sqrt{3}}{2} \operatorname{sech}(\pi \sqrt{y} (n+1/6))\right). \end{align}

Numerical computations seem to confirm that

\begin{align} A(1) & = \ln\left(1+ \sqrt{3} +\sqrt{3+2 \sqrt{3}} \right) \tag{1} \\[8pt] A(2) & = \frac1{\sqrt{2}} \ln \left(1+ \sqrt{2} + \sqrt{2+ 2 \sqrt{2}} \right) \tag{2} \\[8pt] A(3) & = \frac1{\sqrt{3}} \ln \left( 1+ 2^{1/3} + 2^{2/3} \right) \tag{3} \end{align}

And generally, it looks like $$A(y) = \frac1{\sqrt{y}} \,\ln u \tag{4}$$ where $u$ is the root closest to $1$ from above, of a polynomial $P_y$. I've checked dozens of different $y$'s and made a list of those polynomials - check this pastebin link. Some are missing, e.g. I could not find $P_6$. Others seem to follow patterns of their own, for example the Heegner numbers. Here's the polynomial for $y=163$:

$$\small P_{163}(u) = u^{12} + 640314 u^{10} + 1280624 u^9 + 640287 u^8 - 1280736 u^7 - 2561412 u^6 - 1280736 u^5 + 640287 u^4 + 1280624 u^3 + 640314 u^2 + 1 = 0$$

Other interesting things to look at are the behaviour of $P_y(1)$ and $P_y(-1)$, with regard to $y \pmod{24}$, and approximations to $\pi$ which follow from terminating the infinite series at its first term.

However, I have got no clue how to prove it. What would be a way to prove $(4)$? What can be said about the polynomials $P_y$? Also, can you help me find $P_6$, or other missing polynomials from my list?


Finally, I was able to produce a closed form for this integral thanks to @DaveHuff's hints. The idea is to rewrite the infinite series as $$A(y) = \frac2{\sqrt{y}} \sum_{n=0}^{\infty} \tanh^{-1}\left( \dfrac{\cos \frac{\pi}{6} (2n+1)}{\cosh \frac{\pi \sqrt{y}}{6} (2n+1)}\right),$$ and then, using $\displaystyle \,\,\,\tanh^{-1}x = \frac12 \ln \left( \frac{1+x}{1-x} \right),$ proceed to factorize the summand and obtain $$\sqrt{y} \,A(y) = \sum_{n=1}^{\infty} \ln \left( \dfrac{(1-e^{5 \pi i n/6-\pi n\sqrt{y}/6})(1-e^{-5 \pi i n/6-\pi n\sqrt{y}/6})}{(1-e^{ \pi i n/6-\pi n\sqrt{y}/6})(1-e^{-\pi i n/6-\pi n\sqrt{y}/6})} \right),$$ which means: $$A(y) = \frac1{\sqrt{y}} \,\ln \left( \dfrac{\eta\left(\frac{i \sqrt{y}+5}{12}\right)\eta\left(\frac{i \sqrt{y}-5}{12}\right)}{\eta\left(\frac{i \sqrt{y}+1}{12}\right)\eta\left(\frac{i \sqrt{y}-1}{12}\right)}\right).$$

I still don't know enough eta quotient theory, so I don't know how to show that this eta quotient is in fact algebraic for every natural $y$ (let alone bring it to the implicit form in @TitoPiezasIII's answer), but this is still good progress.

  • $\begingroup$ I may have figured out a nice identity. I'll give more details later. :) $\endgroup$ Nov 26, 2016 at 20:22
  • $\begingroup$ iirc $1$-forms on modular curves correspond to $f(z)dz$ where $f$ is an automorphic function of weight $2$, and here you have $f$ of weight $1$ instead so it looks strange. $\endgroup$
    – mercio
    Nov 26, 2016 at 21:16
  • $\begingroup$ @mercio: My answer connects it to the $24$th power of a Weber modular function $\endgroup$ Nov 27, 2016 at 14:01
  • $\begingroup$ I must say, Considering my response to your last post that you are going in the right direction. $\endgroup$
    – Dave huff
    Dec 5, 2016 at 9:24
  • $\begingroup$ @TitoPiezasIII are you aware of this paper Two-dimensional series evaluations via the elliptic functions of Ramanujan and Jacobi ? Similar double sums have been considered in the literature and evaluated in the form given by eq. $(4)$. $\endgroup$ Nov 23, 2017 at 12:35

2 Answers 2


I. We assume it is true your integral $A(y)$ is, $$A(y) = \frac1{\sqrt{y}} \,\ln u \tag{4}$$ The problem is to find $u$. After some laborious manipulation, it turns out that if $\color{blue}{\tau=\frac{1+\sqrt{-y}}{2}}$, then we have the rather simple relation, $$\big(\mathfrak{f}_2(\tau)\big)^{24} =\left(\frac{\sqrt2\,\eta(2\tau)}{\eta(\tau)}\right)^{24}=\frac{(u^2-1)^6}{(-u^3-u^2-u)^3}\tag5$$ where $\mathfrak{f}_2(\tau)$ is a Weber modular function. Since for integer $y>0$, the eta quotient $\frac{\eta(\tau)}{\eta(2\tau)}$ is an algebraic number, then $u$ is also an algebraic number.

II. The advantage of using $\frac{\eta(\tau)}{\eta(2\tau)}$ is that it is well-studied and the algebraic numbers it forms are simpler than $u$. For example, let $\tau=\frac{1+\sqrt{-6}}{2}$, then $w = \Big(\frac{\eta(\tau)}{\eta(2\tau)}\Big)^{24}$ is just a root of a quartic,

$$2^{12} - 4831232 w + 108672 w^2 + 2272 w^3 + w^4 = 0\tag6$$

To find $P_6$, we use $(5)$ as,


Eliminating $w$ between $(6),(7)$ (I assume you have CAS?) and we get a high $24$th deg polynomial in $u$ and which was one reason you had trouble finding it.


As requested, here is the method to find $(5)$. It is not that "laborious" in retrospect, but it does need some effort to spot the usual patterns.

From previous experience, it has been frequently observed that the minpoly of a modular function with argument $\frac{1+\sqrt{-d}}{2}$ and $d$ a Heegner number has near multiples of a power of the j-function $j(\tau)$ amongst the coefficients. For example, one can see integers close to $640320$ in, $$\small P_{163}(u) = u^{12} + 640314 u^{10} + 1280624 u^9 + 640287 u^8 - 1280736 u^7 - 2561412 u^6 - 1280736 u^5 + 640287 u^4 + 1280624 u^3 + 640314 u^2 + 1 = 0$$ In fact, if we let, $$r = -\sqrt[3]{j(\tau)}\tag8$$ then the above has the palindromic form, $$\small 1 + (r - 6) u^2 + 2(r - 8) u^3 + (r - 33) u^4 + 2(-48 - r) u^5 + 4(-33 - r) u^6 + \\ \small2(-48 - r) u^7 + (r - 33) u^8 + 2(r - 8) u^9 + (r - 6) u^{10} + u^{12}=0\tag9$$ Checking its discriminant $D$ (one should always check this) shows it is the neat, $$D=-2^{24}\cdot3^{15}(n+18)^4(n^2+108)^6$$ where $n=r-6$. Plus, testing with non-Heegner $d$ and the same equation holds which suggests it is valid generally. Since the j-function can be expressed by eta quotients as, $$j(\tau) = \frac{(x+16)^3}{x},\quad\text{where}\quad\small x = \big(\mathfrak{f}_2(\tau)\big)^{24} =\left(\frac{\sqrt{2}\,\eta(2\tau)}{\eta(\tau)}\right)^{24}\tag{10}$$ Eliminating $r$ and $j(\tau)$ between $(8),(9),(10)$ and choosing the appropriate factor then yields $(5)$.

  • $\begingroup$ P.S. While each side of $(5)$ is a cube, there is a reason for that as I took away the root of unity $\zeta_{48}$ for simplicity. Trying to get rid of the cube power results in the eta quotient becoming complex-valued. $\endgroup$ Nov 27, 2016 at 14:04
  • $\begingroup$ Are you in the function field $\mathbb{C}(j(\tau),j(2\tau))$ with $j(2\tau)$ algebraic over $\mathbb{C}(j(\tau))$ or something like that ? (if you don't explain then it is not really useful for us) $\endgroup$
    – reuns
    Nov 27, 2016 at 14:07
  • $\begingroup$ @user1952009: For $\tau=\frac{1+\sqrt{-d}}{2}$ and integer $d>0$, the j-function $j(\tau)$ and eta quotient $w=\Big(\frac{\eta(\tau)}{\eta(2\tau)}\Big)^k$ are algebraic numbers, but $w$ gives far simpler minimal polynomials called Weber polynomials. $\endgroup$ Nov 27, 2016 at 14:13
  • $\begingroup$ It doesn't make sense. What function field are you considering ? See for example This book p.311 $\endgroup$
    – reuns
    Nov 27, 2016 at 14:23
  • 1
    $\begingroup$ @user1952009: Of course. :/ $\endgroup$ Nov 27, 2016 at 14:30

Let $\color{blue}{\tau =\frac{1+\sqrt{-y}}{2}}$ and $y$ a positive integer. The well-known the j-function $j(\tau)$ would then be an algebraic number. Consider the OP's relations, $$A(y) = \frac{2}{\sqrt{y}}\,\tanh^{-1}\sqrt{z-1} = \frac{1}{\sqrt{y}}\,\ln\frac{1+\sqrt{z-1}}{1-\sqrt{z-1}}$$ where, $$z=\frac{2}{k}\left(1-\sqrt{1-k+k^2}\right)$$ $$k =\frac{1}{4}e^{2\pi\, i /3}\left(\frac{\sqrt{2}\,\eta(2\tau)}{\eta(\tau)}\right)^8$$ It is known that, $$j(\tau) = \frac{(x+16)^3}{x}$$ where $x = \left(\frac{\sqrt{2}\,\eta(2\tau)}{\eta(\tau)}\right)^{24}$. So if $j(\tau)$ is an algebraic number, then so is $x$ and $z$. What remains (based on an update by the OP) is to show that, $$\frac{1+\sqrt{z-1}}{1-\sqrt{z-1}}=\frac{\eta\big(\tfrac{\tau+2}{6}\big)\,\eta\big(\tfrac{\tau-3}{6}\big)}{\eta\big(\tfrac{\tau}{6}\big)\,\eta\big(\tfrac{\tau-1}{6}\big)}\tag0$$ though this step seems difficult.

An alternative way to show that $z$ also is an algebraic number is by directly expressing it in terms of $j(\tau)$ itself. Define,

$$h = \big(\tfrac{1}{27}\,j(\tau)\big)^{1/3}\tag1$$

and the cubic in $v$,


The discriminant $D$ of this is $D=64-h^3$. Since $\tau=\frac{1+\sqrt{-y}}{2}$ and $y>3$ has negative $h$, this implies the cubic has only one real root. Using the real root $v$, then $z$ satisfies the simple relation,


Since $h$ is an algebraic number, then so is $z$.

P.S. Of course, this is also another way to solve for $z$. However, the appropriate root of $(3)$ has to be used.

  • $\begingroup$ @nospoon: Sorry for the delay. I was preoccupied with finding a pattern for the interesting integrals in this post. :) $\endgroup$ Dec 6, 2016 at 16:14
  • $\begingroup$ Hey Tito, I really appreciate your answers and insights! But the "OP's relations" is just an explicit form of your first answer. Do you have any idea how to prove that the eta quotient in the edit on the OP is equivalent to the the implicit form $(5)$ on your first answer? $\endgroup$ Dec 7, 2016 at 17:06
  • $\begingroup$ @nospoon: I figured it was just my answer, but it was a nice simplification you did. However, to prove its equivalence to your update (labeled as $(0)$) above) seems difficult as the LHS is a sum and has a radical sign while the RHS is a product. Someone might know how to do so, though. $\endgroup$ Dec 8, 2016 at 3:04
  • $\begingroup$ @nospoon: I've asked a summarized version about $(0)$ in this MO post. $\endgroup$ Dec 8, 2016 at 4:38
  • $\begingroup$ Neat. I am still wondering what 'laborious manipulations' you did in your original answer. How did you initially found $(5)$? $\endgroup$ Dec 8, 2016 at 8:25

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