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I would like to know if there is a closed-form for \begin{align*} \prod_{n=1}^{\infty} \frac{e^{\frac{x}{2^n}}-1}{\frac{x}{2^n}}, \end{align*} where $x\in\mathbb{R}$. The closest results I could find in the literature were \begin{align*} \prod_{n=1}^{\infty} \frac{1+e^{\frac{x}{2^n}}}{1+e^{\frac{y}{2^n}}} &= \frac{e^x-1}{x} \frac{y}{e^y-1},\\ \prod_{n=1}^{\infty} \left(1+e^{-x 2^n}\right) &= \frac{1}{2} (1+\coth(x)). \end{align*}

The motivation for this problem comes from this post. I was interested in calculating the moment-generating function for the random variable $L = \sum\limits_{n=1}^{\infty} \frac{U_n}{2^n}$, where $U_n$ are iid random variables that follow a uniform distribution in $[0,1]$. After some calculations, I ended up with this infinite product.

Update: This is related to the Fabius function, which is an infinitely differentiable function that is nowhere analytic. It probably does not have a closed-form.

P.S. Dester
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1 Answers1

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Define the function $f(x)=\prod_{n=0}^\infty g(\frac{x}{2^n})$, where $g(x)=\frac{e^x-1}{x}=e^{x/2}\frac{\sinh(x/2)}{x/2}$.

So $f(2x)=\prod_{n=0}^\infty g(\frac{x}{2^{n-1}})=g(2x)f(x)=e^x\frac{\sinh(x)}{x}f(x)$.

Make $f(x)=e^x\sum_{n=0}^\infty a_nx^n$, so we have that $$e^{2x}\sum_{n=0}^\infty 2^na_nx^n=e^{2x}\frac{\sinh(x)}{x}\sum_{n=0}^\infty a_nx^n\iff\sum_{n=0}^\infty 2^na_nx^n=\left(\sum_{n=0}^\infty a_nx^n\right)\left(\sum_{k=0}^\infty \frac{x^{2k}}{(2k+1)!}\right)\iff$$ $$\iff\sum_{n=0}^\infty \left[2^na_n-\sum_{k=0}^{\lfloor n/2\rfloor} \frac{a_{n-2k}}{(2k+1)!}\right]x^n=0\iff a_n=\frac{1}{2^n-1}\sum_{k=1}^{\lfloor n/2\rfloor} \frac{a_{n-2k}}{(2k+1)!}$$ and we have that $$f(x)=e^x\sum_{n=0}^\infty a_nx^n=\left(\sum_{n=0}^\infty a_nx^n\right)\left(\sum_{k=0}^\infty\dfrac{x^k}{k!}\right)=\sum_{n=0}^\infty\left(\sum_{k=0}^n \frac{a_k}{(n-k)!}\right)x^n$$

It isn't the prettiest result, but if $a_n$ have a closed form, then your initial product is $\frac{xf(x)}{e^x-1}$ (or $f(x/2)$), since the $k$ at $f(x)$ starts at 0, and not 1. Then to make it as a Maclaurin series then just make the same Cauchy product, since $\frac{x}{e^x-1}=\sum_{n=0}^\infty \dfrac{B_n}{n!}x^n$, where $B_n$ are the Bernoulli numbers. Or just put $x/2$ in f.

UPDATE: if you want to compute some $a_n$ values, you will need the value of $a_1$, since to evaluate it we can't use the provided formula. Since $$\ln(f(x))=\sum_{n\geq0}\ln\left(g(\frac{x}{2^n})\right) \xrightarrow{f(x)\frac{d}{dx}(\cdot)}f'(x)=f(x)\sum_{n\geq0}\frac{1}{2^n}g'(\frac{x}{2^n})$$ putting x=0 will lead us to the value of f'(0). Since $f(x)=e^x\sum_{n\geq0}a_nx^n$, so $f'(x)=f(x)+e^x\sum_{n\geq0}na_nx^{n-1}$, put x=0 and then get that $a_1=f'(0)-f(0)$. Substituting the values, you will see that $a_1=0$ (and, in that point, we can see that is trivial that $a_0=1$).

tzk
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