On the integral $\int_0^\infty \eta^2(i x) \,dx = \ln(1+\sqrt{3}+\sqrt{3+2 \sqrt{3}})$ and its cousins

Question

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?

Edit.

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.

Answer 1

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,

$$\frac{2^{12}}{w}=\frac{(u^2-1)^6}{(-u^3-u^2-u)^3}\tag7$$

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.

$\color{green}{Update:}$

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 $(5)$.

Answer 2

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 $v$,

$$v^3-3h^2v-2(h^3-128)=0\tag2$$

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,

$$z^2-(h+v)(z-1)=4\tag3$$

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.