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distY = TransformedDistribution[1/z^2, 
z \[Distributed] ZipfDistribution[1]]; Expectation[1/(s - i),
(*(3 (Sqrt[s] - \[Pi] Cot[\[Pi]/Sqrt[s]]))/(\[Pi]^2 Sqrt[s])*)

distY = TransformedDistribution[1/z^2, 
z \[Distributed] ZipfDistribution[1]]; Expectation[1/(s - i),i\[Distributed] distY]

(*(3 (Sqrt[s] - \[Pi] Cot[\[Pi]/Sqrt[s]]))/(\[Pi]^2 Sqrt[s])*)

Using: $$\sum _{n=1}^{\infty } \frac{6}{\pi ^2 \left(-1+n^2 s\right)}=\frac{3 \left(\sqrt{s}-\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)\right)}{\pi ^2 \sqrt{s}}$$

From comments @Yaroslav Bulatov Harder example, to invert:

Using: $$\sum _{n=1}^{\infty } \frac{6}{\pi ^2 \left(-1+n^2 s\right)}=\frac{3 \left(\sqrt{s}-\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)\right)}{\pi ^2 \sqrt{s}}$$ Then:

Using: $$\sum _{n=1}^{\infty } \frac{6}{\pi ^2 \left(-1+n^2 s\right)}=\frac{3 \left(\sqrt{s}-\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)\right)}{\pi ^2 \sqrt{s}}$$

If I use Inverse Stieltjes Transform to:$\frac{3 \left(\sqrt{s}-\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)\right)}{\pi ^2 \sqrt{s}}$ i should get:$\begin{cases} \frac{6 \left(\psi ^{(1)}\left(\left\lceil \frac{1}{\sqrt{x}}\right\rceil \right)-\psi ^{(1)}\left(\left\lfloor \frac{1}{\sqrt{x}}\right\rfloor +1\right)\right)}{\pi ^2} & 0<x\leq 1 \\ 0 & \text{True} \end{cases}$.

Let try:

 PDF[TransformedDistribution[1/z^2, 
 z \[Distributed] ZipfDistribution[1]], x]

 (*Piecewise[{{(6*(PolyGamma[1, Ceiling[1/Sqrt[x]]] - 
 PolyGamma[1, 1 + Floor[1/Sqrt[x]]]))/Pi^2, 
 Inequality[0, Less, x, LessEqual, 1]}}, 0]*)

 Plot[%, {x, -2, 2}, PlotRange -> Full]
 (*I get Zero for all range*)
 

Using: $$\sum _{n=1}^{\infty } \frac{6}{\pi ^2 \left(-1+n^2 s\right)}=\frac{3 \left(\sqrt{s}-\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)\right)}{\pi ^2 \sqrt{s}}$$