# Conjectured value of a difficult integral with Dedekind eta functions

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

• 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
• @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
• 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
• – Jack D'Aurizio Jan 12 '18 at 15:59
• @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

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.