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In my research on quantum groups I have the following conjecture: \begin{equation} \int_0^\infty\frac{\eta (2 i x)^8}{\eta (i x)^2 \eta (4 i x)^2}\,dx\,{\stackrel?=}\,\frac{K(\tfrac{1}{\sqrt{2}})}{\pi}\tag{1} \end{equation} where \begin{equation} \eta(ix)=e^{-\frac{\pi x}{12}}\prod_{n=1}^\infty (1-e^{-2n\pi x}) \end{equation} is the Dedekind eta function and \begin{equation} K(\tfrac{1}{\sqrt{2}})=\frac{\Gamma^2(\tfrac14)}{4\sqrt{\pi}} \end{equation} is the Elliptic integral singular value.

This value has been guessed using Inverse symbolic calculator and then checked numerically to a high precision, but I do not have a proof.

Question: Is (1) true?

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    In the first post, you have been looking quite aggressive to several people who wanted to help you. Now, I delete my answer. – Claude Leibovici Jan 12 '18 at 10:45
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    @Hans: as already said, it is enough to invoke the residue theorem, special values for the Dedekind eta function and the known relation between $K\left(\frac{1}{\sqrt{2}}\right)$ and $\Gamma\left(\frac{1}{4}\right)^2$. In my humble opinion, what is really missing here is some context, namely why the LHS of $(1)$ is somewhat relevant and some attempts made to prove such identity. – Jack D'Aurizio Jan 12 '18 at 14:27
  • @JackD'Aurizio unfortunately the context is very hard to explain, because it involves very technical stuff like conformal field theory, Lie algebras and so on, and also some guess work. Can you please eloborate on this remark it is enough to invoke the residue theorem? Can you at least write down the residue of which function and at what points? –  Jan 12 '18 at 14:34
  • @Hans: consider the third and the fourth row of the table after equation $(95)$ here - http://mathworld.wolfram.com/JacobiThetaFunctions.html. – Jack D'Aurizio Jan 12 '18 at 14:56
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    Let me ask you a simpler question: can you compute $$ \int_{0}^{+\infty}\left[\sum_{n\geq 1}(-1)^{n+1} e^{-n^2 x}\right]^2,dx $$ ? Your question is not really different from this one. – Jack D'Aurizio Jan 12 '18 at 15:25
  • @JackD'Aurizio so you convert the LHS of (1) to a quadruple sum, and then what? Where does the residue theorem come into play? –  Jan 12 '18 at 15:41
  • @Hans: no, I do not convert the integrand function as a quadruple sum, I convert it into a double product. And where the residue theorem comes into play is pretty clear from my last counter-question. I am going to open a thread about it, let us cover the basics before attacking more sophisticated problems. – Jack D'Aurizio Jan 12 '18 at 15:45
  • @JackD'Aurizio can you please stop being criptic and just give an outline of the answer detailing main steps, provided you have an answer? Or maybe you don't have an answer and just speculating? –  Jan 12 '18 at 15:55
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    @Hans: https://math.stackexchange.com/questions/2602500/an-interesting-identity-involving-jacobi-theta-4-and-zeta2 – Jack D'Aurizio Jan 12 '18 at 15:59
  • @JackD'Aurizio in your case you convert the integral into a double sum. In the present case there will be a quadruple sum or a double sum (probably both of them divergent), as I mentioned earlier. I still don't see any residues nor in your question nor in this present case either. Also how do you plan to get an elliptic integral out of an dilogarithm is beyond my understanding. –  Jan 12 '18 at 16:10
  • @Hans: you have to compute the $L^2$ norm over $\mathbb{R}^+$ of $\theta_2(\sqrt{q})\theta_4(q^2)q^{-1/8}$, which is a product. In your case the elliptic integral comes from the square of $\eta(i)$. – Jack D'Aurizio Jan 12 '18 at 16:15
  • @JackD'Aurizio why wouldn't you just give an answer? I bet that's because you don't have it. –  Jan 12 '18 at 16:24
  • @Hans: No, that is not the correct reason. I mentioned that your question lacks some crucial context, and guided you through a solution nevertheless. Even reposting the exactly same question (with the same issues) is kind of debatable. And honestly speaking, if you do not care about providing crucial details, why should I? What is the purpose of having an answer without grasping the techniques behind it? – Jack D'Aurizio Jan 12 '18 at 16:56
  • @JackD'Aurizio your method in the linked question do not carry over to the present integral. The case you considered is too trivial. And what context do you need? I said it is related to very technical stuff from CFT. –  Jan 12 '18 at 17:19
  • Share your CFT knowledge with us :) – Mariusz Iwaniuk Jan 12 '18 at 17:27
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    @Hans: First step: simplify the integrand function. For any $x>0$ we have $$ \eta(ix) = q^{\frac{1}{24}}\prod_{n\geq 1}(1-q^{n}),\qquad q=e^{-2\pi x} $$

    $$ \frac{\eta(2ix)^8}{\eta(ix)^2\eta(4ix)^2} = \frac{q^{\frac{2}{3}}\prod_{n\geq 1}(1-q^{2n})^8}{q^{\frac{5}{12}}\prod_{n\geq 1}(1-q^n)^2(1-q^{4n})^2}=\frac{1}{4}\left[\frac{\vartheta_4(q^2)\vartheta_2(\sqrt{q})}{q^{\frac{1}{8}}}\right]^2$$ where the last identity has been deduced through the third and fourth line of the table after eq.(95) [here][1]. Now

     – Jack D'Aurizio Jan 12 '18 at 18:08
  • $$ \vartheta_2(q)=\sum_{n\in\mathbb{Z}}q^{(n+1/2)^2},\qquad \vartheta_4(q)=\sum_{n\in\mathbb{Z}}(-1)^n q^{n^2} $$ give $$\frac{\vartheta_4(q^2)\vartheta_2(\sqrt{q})}{q^{\frac{1}{8}}} = \sum_{m,n\in\mathbb{Z}}(-1)^n q^{\frac{m^2+m}{2}+2n^2}$$ $$\left[\frac{\vartheta_4(q^2)\vartheta_2(\sqrt{q})}{q^{\frac{1}{8}}}\right]^2 = \sum_{m,M,n,N\in\mathbb{Z}}(-1)^{n+N} q^{\frac{m^2+m+M^2+m}{2}+2(n^2+N^2)}$$ such that the wanted integral equals $$\frac{1}{8\pi}\sum_{n,N,m,M\in\mathbb{Z}}\frac{(-1)^{n+N}}{\frac{m+m^2+M+M^2}{2}+2(n^2+N^2)}. $$ – Jack D'Aurizio Jan 12 '18 at 18:08
  • The problem boils down to recalling the series representation of $\vartheta_3^2$ and the value of $\eta(i)^2$. – Jack D'Aurizio Jan 12 '18 at 18:08
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    How to compute such series through the residue theorem is explained in the other thread. Now, respectfully, I disengage. – Jack D'Aurizio Jan 12 '18 at 18:39
  • @skbmoore amazing solution! Do you also happen to know a proof for this one too? –  Jan 13 '18 at 07:34
  • I don't think there is any need to challenge anyone to produce an answer. @JackD'Aurizio is only trying to help here by providing an approach which he thinks is suitable for the current problem. If you see my answer below then your problem appears not that complicated, but the same technique is not helpful for the question Jack D'Aurizio is asking in the linked thread and thus I don't consider his integral as trivial. – Paramanand Singh Jan 13 '18 at 10:00

1 Answers1

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The following solution uses the link between elliptic integrals and theta functions.

Let's use $q=e^{-\pi x} $ and then we define function $f$ via $$f(q) =q^{1/12}\prod_{n=1}^{\infty}(1-q^{2n})=\eta(ix)\tag{1}$$ Let $x=K(k') /K(k) $ where $k, k'$ are elliptic moduli complementary to each other and $K(k)=K $ is the complete elliptic integral of the first kind. From the theory of elliptic integrals and theta functions we have \begin{align} f(q)=\eta(ix)&=2^{-1/3}\sqrt{\frac{2K} {\pi}} (kk') ^{1/6}\tag{2a}\\ f(q^2)=\eta(2ix)&=2^{-2/3}\sqrt {\frac {2K}{\pi}}k^{1/3}k'^{1/12}\tag{2b}\\ f(q^4)=\eta(4ix)&=2^{-13/12}\sqrt{\frac{2K}{\pi}}\frac{k^{2/3}k'^{1/24}}{(1+k')^{1/4}}\tag{2c} \end{align} The integrand in question is $$\frac{f^8(q^2)} {f^2(q)f^2(q^4)}=2^{-5/2}\left(\frac{2K}{\pi}\right)^2kk'^{1/4}(1+k')^{1/2} $$ and we have $$\frac{dx} {dk} =\frac{dx}{dq}\cdot\frac {dq} {dk} =-\frac{1} {\pi q} \cdot\frac{\pi^2 q}{2kk'^2K^2} =-\frac{\pi} {2kk'^2K^2}\tag{3}$$ so that the desired integral is equal to $$\frac{1}{2\pi\sqrt{2}}\int_{0}^{1}k'^{-7/4}(1+k')^{1/2}\,dk=\frac{1}{2\pi\sqrt{2}}\int_{0}^{1}t^{-3/4}(1-t)^{-1/2}\,dt\tag{4}$$ (via substitution $k'=\sqrt{1-k^2}=t$) which is evaluated easily via Beta/Gamma functions to obtain desired result.

We were lucky that after switching from $x$ to $k$ the elliptic integral $K$ got cancelled and the resulting integral was having an algebraic function as an integrand. Thus the original integral appears to be designed to have such a nice evaluation. See this answer for another instance of this technique.

Paramanand Singh
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