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In looking at the sum $g(s) = \sum_{n=2}^{\infty} \frac{\zeta(n)}{s^n}$ for $|s|>1$, I found that $g(s)$ may be expressed as, $$\int_1^{\infty} \frac{x-1}{x^s -1} \frac{\,dx}{x}$$ and this is the integral I am trying to solve. Wolfram alpha appears to give closed form solutions for values of $s$ which makes me think there is some known closed form of $g$, but I can't seem to solve it. I thought that maybe I could do some "term cancelling" between $x-1$ and $x^s -1$ and then do PFD, but I just don't see how to do that for an arbitrary natural $s$. When I tried the term cancelling, I got something like $\int_1^{\infty} \frac{\,dx}{\sum_{m=0}^{s-1} x^{m+1}}$, but got stuck there,

Some values reported by Wolfram after inputting the integral are:

$g(2)=\ln(2)$, $g(3)=\frac{1}{18}(9\ln(3) - \sqrt{3}\pi$), $g(4)=\frac{1}{8}(3\ln(4) - \pi)$

(As an aside, my motivation behind this problem was simply exploring the sum $\sum_{n=2}^{\infty} \zeta(n)\{\zeta(n)\}$, which results from $\sum_{s=2}^{\infty} g(s)$), where $\{x\}$ is the fractional part of $x$).

jimjim
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Ryan Goulden
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    The original expression you gave for $g(s)$ has the closed form $$g(s)=-\frac{H_{-1/s}}s$$where $H_z$ is the complex harmonic number defined by $$H_z=\psi(1+z)+\gamma$$This follows directly from the series expansion of $H_z$ (or equivalently $\psi(z)$). – Peter Foreman Jun 08 '20 at 22:21
  • Ah, I did not know that. I guess that serves as a closed form of the integral as well, then! Though I would still be curious about some method for solving the integral itself. In any case, do you know where this formula comes from? – Ryan Goulden Jun 08 '20 at 22:27
  • Oh I see that now. – Ryan Goulden Jun 08 '20 at 22:31
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    I think you want this series. – Peter Foreman Jun 08 '20 at 22:32

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