zeta sum identity and its general form

Question

I noticed that $$\frac{\pi}{2}=\exp\left(\sum_{n=1}^\infty{\frac{\zeta(2n)}{n}\left(\frac{1}{4}\right)^n}\right)$$ I came across this while trying to find the general form $$\exp\left(\sum_{n=1}^\infty{\frac{\zeta(sn)}{n}x^n}\right)$$ Can we get a general formula for this function?

Edit I thought it might be useful to note that for $s=1$ we have $$\exp\left(\sum_{n=2}^\infty{\frac{\zeta(n)}{n}x^n}\right)=e^{-\gamma x}\Gamma(1-x)$$ but I'm not quite sure how we would use it.

EDIT:

I figured it out for $s=4$

Start with

$$-\sum_{n=1}^\infty\ln\left(1+\frac{x}{n^2}\right)=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{\left(-\frac{x}{m^2}\right)^n}{n}=\sum_{n=1}^\infty\frac{(-1)^n\zeta(2n)}{n}x^{n}$$

Using WolframAlpha I figured out that this equals

$$\ln\left(\frac{\pi\sqrt{x}}{\sinh(\pi\sqrt{x})}\right)$$ Amazing resemblance to our $s=2$ formula! (Shown in the comments) Of course, this part isn't rigorous. A rigorous proof of this is still needed.

Now since we already know the answer for $s=2$ we can combine the two...

$$\sum_{n=1}^\infty\frac{\zeta(2n)}{n}x^{n}+\sum_{n=1}^\infty\frac{(-1)^n\zeta(2n)}{n}x^{n}=\sum_{n=1}^\infty\frac{\zeta(4n)}{n}x^{2n}=\ln\left(\frac{\pi\sqrt{x}}{\sin(\pi\sqrt{x})}\right)+\ln\left(\frac{\pi\sqrt{x}}{\sinh(\pi\sqrt{x})}\right)$$

So then

$$\sum_{n=1}^\infty\frac{\zeta(4n)}{n}x^{n}=\ln\left(\frac{\pi x^{1/4}}{\sin(\pi x^{1/4})}\right)+\ln\left(\frac{\pi x^{1/4}}{\sinh(\pi x^{1/4})}\right)$$

and exponentiating we have

$$\exp\left(\sum_{n=1}^\infty\frac{\zeta(4n)}{n}x^{n}\right)=\left(\frac{\pi x^{1/4}}{\sin(\pi x^{1/4})}\right)\cdot\left(\frac{\pi x^{1/4}}{\sinh(\pi x^{1/4})}\right)$$

Answer 1

By the Taylor Series for the PolyGamma Function, we have the representation: $$ \psi^{(0)}(1+x) = -\gamma + \sum_{k = 1}^\infty (-1)^{k+1}\zeta(k+1)x^k \\ -\gamma - \psi^{(0)}(1-x) = \sum_{k = 1}^\infty \zeta(k+1) x^k $$ Upon integration: $$ -\gamma x + \ln(\Gamma(1-x)) = \sum_{k = 2}^\infty \frac{\zeta(k)}{k} x^k $$ Sending $x \to x \ \zeta_s$, where $\zeta_s$ is an $s$th root of unity, for $s>1$, and summing over all roots, we obtain: $$ \ln \left ( \prod_{k=0}^{s-1} \Gamma \left(1-xe^{\frac{2 \pi i k}{s}}\right)\right) = \sum_{k = 1}^\infty \frac{\zeta(sk)}{k} x^{sk} $$ And thus by exponentiating, we obtain the desired quantity as a finite product of Gamma Functions, which will only simplify in degenerate cases.

Answer 2

I don't think there is a simple formula for $s \ne 2$.

We have the closed form $$\zeta(2n) = (-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$$ Where $B_n$ are the Bernouilli numbers appearing from $$\frac{x}{e^x-1}=\sum_{n=0}^\infty \frac{B_n}{n!}x^n, \qquad |x| < 2\pi$$

Since $\frac{x}{e^x-1}-\frac{x}{2}$ is even $B_{2n+1} = 0, n \ge 1$.

Thus, for $|x| < 1$ $$F(x)=-2\sum_{n=1}^\infty \zeta(2n) x^{2n}= \sum_{n=1}^\infty (2i\pi x)^{2n}\frac{B_{2n}}{(2n)!}=-B_0 - B_1 2 i \pi x +\sum_{n=0}^\infty (2i\pi x)^{n}\frac{B_{n}}{n!}\\= \frac{2i\pi x}{e^{2i\pi x}-1}-1+ i\pi x$$ Finally $$G(x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n} x^{2n}, \qquad G'(x)=2\sum_{n=1}^\infty \zeta(2n) x^{2n-1}= -\frac{F(x)}{x}$$

$$G(x) = -\int \frac{F(x)}{x}dx = \log(x)-i \pi x-2i\pi\int \frac{dx}{e^{2i\pi x}-1}\\ = \log(x)-i \pi x- \log(e^{-2i\pi x}-1)+C$$ $G(0) =0$ means $C=-\log(-2i\pi)$