# Can $\prod\limits_{k=1}^\infty \left(1- \frac{1}{e^{ \sqrt{2} \pi k}}\right)$ be put into closed form?

Let $$\alpha = \prod_{k=1}^{\infty} \left(1- \frac{1}{e^{ \sqrt{2} \pi k}}\right)$$ and $$\beta = \prod_{k=1}^{\infty} \left(1 + \frac{1}{e^{ \sqrt{2} \pi k}}\right) = \frac{\exp \left(\frac{\sqrt{2}\pi }{24}\right)}{\sqrt[4]{2}}$$

Then $$\sum_{n=1}^{\infty} \text{arctanh} \left(\frac{1}{e^{ \sqrt{2} \pi n}} \right) = -\text{arctanh} \left( \frac{\alpha - \beta}{\alpha + \beta}\right) \approx 0.0119025137323901862$$

Is there a way to solve for the product $$\alpha$$? I figure that it's likely given that $$\beta$$ can be.

• That's $f(z)=\prod(1-z^i)$ for $z=e^{-\sqrt2}.$ I have a vague recollection that that is related to partition functions, and has a name, or at least is related to a function with a name. Specifically, $1/f(z)=\sum_{n=0}^{\infty} p_nz^n$ where $p_n$ is the number of partitions of $n.$ Commented Nov 5, 2022 at 3:27
• How did you get the closed formula for $\beta?$ Commented Nov 5, 2022 at 3:30
• It belongs to the theory of modular forms. $j(i\sqrt 2)$ and $\Delta(2i \sqrt2)/\Delta(i \sqrt2)$ have a closed-form but $\Delta(i\sqrt 2)$ probably doesn't @ThomasAndrews Commented Nov 5, 2022 at 8:30
• With Mathematica: $\prod _{k=1}^{\infty } \left(1-\frac{1}{\exp \left(\sqrt{2} \pi k\right)}\right)=\exp \left(\sum _{k=1}^{\infty } \ln \left(1-\frac{1}{\exp \left(\sqrt{2} \pi k\right)}\right)\right)=e^{\frac{\pi }{12 \sqrt{2}}} \eta \left(\frac{i}{\sqrt{2}}\right)$ where: $\eta \left(\frac{i}{\sqrt{2}}\right)$ is Dedekind eta modular elliptic function. Commented Nov 5, 2022 at 12:26
• Then we have: $\sum _{n=1}^{\infty } \tanh ^{-1}\left(\frac{1}{\exp \left(\sqrt{2} \pi n\right)}\right)=-\frac{1}{8} \ln \left(2 \eta \left(\frac{i}{\sqrt{2}}\right)^4\right)$ Commented Nov 5, 2022 at 12:33

Let $$q=e^{-\pi\sqrt{2}}$$ and then the product $$\alpha$$ in question is $$\alpha=\prod_{n=1}^{\infty}(1-q^n)\tag{1}$$ This is related to the famous Dedekind eta function given by $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty}(1-q^n)\tag{2}$$ which in turn has a closed form in terms of elliptic modulus $$k$$ associated with nome $$q$$ and complete elliptic integral of first kind $$K(k)$$: $$\eta(q) =2^{-1/6}\sqrt{\frac{2K(k)}{\pi}}k^{1/12}k'^{1/3}\tag{3}$$ where \begin{align} k&=\frac{\vartheta_{2}^2(q)}{\vartheta_3^2(q)}\tag{4a}\\ k'&=\sqrt{1-k^2}\tag{4b}\\ K(k)&=\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}\tag{4c}\\ \vartheta_2(q)&=\sum_{n\in\mathbb{Z}}q^{(n+(1/2))^2}\tag{4d}\\ \vartheta_3(q)&=\sum_{n\in\mathbb{Z}}q^{n^2}\tag{4e} \end{align} Both Jacobi and Ramanujan established that if $$p$$ is a positive rational number and $$q=e^{-\pi\sqrt{p}}$$ then the modulus $$k$$ in $$(4a)$$ is an algebraic number. Chowla and Selberg further proved (in this paper) that for such $$q$$ and corresponding $$k$$ the elliptic integral $$K(k)$$ can be expressed in closed form containing $$\pi$$ and values of Gamma function at rational points.
The evaluation of $$k$$ for $$p=2,q=e^{-\pi\sqrt{2}}$$ is done via modular equation of degree $$2$$ and one can show that $$k=\sqrt{2}-1,k' =\sqrt{2(\sqrt{2}-1)}$$ The value of $$K(k)$$ is given by $$K(k) = \frac{\sqrt{\sqrt{2} +1} \Gamma (1/8)\Gamma (3/8)}{2^{13/4}\sqrt{\pi}}$$ and is evaluated in this answer. Using these values the value of $$\eta(q)$$ in $$(2)$$ is obtained and then $$\alpha =q^{-1/24}\eta(q)$$ gets evaluated in closed form.
The product $$\beta$$ is simpler to handle because we have $$\beta=q^{-1/24}\cdot\frac{\eta(q^2)}{\eta(q)}\tag{5}$$ and we have $$\eta(q^2)=2^{-1/3}\sqrt {\frac{2K(k)}{\pi}}(kk')^{1/6}\tag {6}$$ and thus $$\beta$$ is $$e^{\pi\sqrt {2}/24}$$ times an algebraic number (which turns out to be $$2^{-1/4}$$ using given values of $$k, k'$$).
• I don't understand how you relate $\Gamma(1/8)\Gamma(3/8)$ with $\int_0^{\pi/2} \frac{dx}{\sqrt{1-(\sqrt2-1)^2 \sin^2 x}}$ please avoid linking to very long answers linking to more long answers not giving much insights Commented Nov 28, 2022 at 11:09
• This doesn't answer my question. The evaluation of $\Delta(i)$ is based on the beta function identity. How does it generalize, and why do we get 2 or 3 (or more) $\Gamma$ terms. Commented Nov 28, 2022 at 11:20