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When I asked for integrals involving the tribonacci constant $T$, user @nospoon gave the nice answer, $$\int_0^{\infty} \eta( i \, x)\,\eta(i \,11 x) \,dx = \frac{ \ln T}{\sqrt{11}} $$ However, the author of that integral and myself are having trouble proving the general case. Let $\zeta_3 =e^{2\pi\, i /3}$ and Dedekind eta function $\eta(\tau)$. In summary, how do we show that, $$-\frac{1}{16\,\zeta_3}\left(\frac{\eta(\tau) }{\eta(2\tau)}\right)^8\; \overset{\color{red}?}=\; \frac{\alpha+1}{\alpha^2-4}\;\overset{\color{red}?}=\; \frac{\beta+1}{\zeta_3\,(\beta^2-4)}\tag1$$ where, $$\alpha=u+u^{-1},\quad u=\frac{\eta\big(\tfrac{\tau+2}{6}\big)\,\eta\big(\tfrac{\tau-3}{6}\big)}{\eta\big(\tfrac{\tau}{6}\big)\,\eta\big(\tfrac{\tau-1}{6}\big)}\tag2$$ and, $$\beta=-1-\left(\frac{\eta\big(\tfrac{2\tau}{3}\big)\,\eta^4(\tau)\,\eta(6\tau)\,}{\eta^2\big(\tfrac{\tau}{3}\big)\,\eta^2(2\tau)\,\eta^2(3\tau)}\right)^2\tag3$$

P.S. The relation $(1)$ allows the eta quotients $\alpha$ and $\beta$ to be expressed in terms of each other, with $\beta$ relatively well-known and described in OEIS A227587. But $\alpha$ for $\tau=\frac{1+\sqrt{-y}}{2}$ leads to the integral considered by @nospoon, $$A(y)=\int_0^{\infty} \eta( i x)\,\eta(i x y)\,dx =\frac{\ln u}{\sqrt{y}}$$ where the tribonacci example was just the case $y=11$.

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    $\begingroup$ Rewriting $-\frac{1}{16\,\zeta_3}\left(\frac{\eta(\tau) }{\eta(2\tau)}\right)^8\; =\; \frac{\alpha+1}{\alpha^2-4}$, you want to show that $-\frac{1}{16^3} \eta^{24} ((u^2+1)^2-4u^2)^3 = \eta(2\tau)^{24} u^3 (u^2+2)^3$ an equality in $S_{12}(\Gamma_0(2))$, a finite dimensional vector space generated by ... so it reduces to computing a basis of $S_{12}(\Gamma_0(2))$ and the first few coefficients of the LHS and the RHS. The same method should work for $\frac{\alpha+1}{\alpha^2-4}=\frac{\beta+1}{\zeta_3\,(\beta^2-4)}$. $\endgroup$
    – reuns
    Commented Dec 9, 2016 at 17:57
  • $\begingroup$ (I wonder who invented such term as "Tribonacci". It seems somehow a lack of respect for the name of a person, even if he lived 800 years ago. What if e.g. one day they will call you "Trito Piezas" instead of "Piezas III" :-) ?) $\endgroup$
    – Pietro Majer
    Commented Dec 17, 2022 at 8:56
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    $\begingroup$ @PietroMajer Looking at OEIS, it is either Martin Gardner (1961), David Wells (1997), or Simon Finch (2003). Mathematicians and physicists can be quite the irreverent and whimsical bunch. Consider WIMPs (weakly interacting massive particles) and MACHOs (massive compact halo objects). There's a lot of other examples. P.S. By coincidence, I just recently made a MSE post on the tribonacci constant. :) $\endgroup$
    – Tito Piezas III
    Commented Dec 17, 2022 at 15:36
  • $\begingroup$ and who knows, maybe fibonacci would have appreciated ;) $\endgroup$
    – Pietro Majer
    Commented Dec 17, 2022 at 20:00

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