# Integral Identities for $S(x,y)=\sum_{k=1}^\infty\frac{y}{k^x(y+k)}$ and $\zeta(x)$ \ $\zeta(y)$

Define $$S(x,y)=\sum_{k=1}^\infty\frac{y}{k^x(y+k)}$$ This is a generalization of the harmonic function ($$n=2$$). There are many ways we could relate this to $$\zeta(x)$$. For example $$\lim_{y\rightarrow0}\frac{S(x,y)}y=\zeta(x+1)$$and$$\lim_{y\rightarrow\infty}S(x,y)=\zeta(x)$$The first limit is simple to prove (use either L'hopital's rule or simplify) while the second limit is verified by numerical computation. Here is one special value: $$S(2,1)=\zeta(2)-1$$(note the order of $$2$$ and $$1$$ on the LHS and RHS. Very cool coincidence). But I personally prefer integrals over limits as they are usually more intuitive (limits are usually evaluated using L'hopital but there are many ways to calculate integrals). So what are some integral identities that relate $$S(x,y)$$ to $$\zeta(x)$$ (or $$S(x,y)$$ and $$\zeta(y)$$)? I want something similar to the relation of $$\zeta(x)$$ with $$\Gamma(x)$$: $$\zeta(x)=\frac1{\Gamma(x)}\int_0^\infty\frac{t^{x-1}}{e^t-1}dt$$

• In terms of Lerch's transcendent, $$S(x,y) = \frac{1}{{\Gamma (x)}}\int_0^{ + \infty } {t^{x - 1} (y\,\Phi ({\rm e}^{ - t} ,1,y) - 1){\rm d}t}$$ with $\operatorname{Re}(x)>1$, $y \neq 0,-1,-2,\ldots$.
– Gary
Commented Mar 15, 2023 at 2:34

For $$\Re(s) > 0,\Re(y)>0$$, the residue theorem gives $$\int_C \frac{z^{-s}}{y+z} (\frac1{e^{2i\pi z}-1}-\frac1{2i\pi z})dz=\sum_{k\ge 1} \frac{k^{-s}}{y+k}$$ where $$C$$ is a contour $$+\infty+i\to 1/2+i\to 1/2-i \to +\infty-i$$.
With the Cauchy integral theorem for $$\Re(s)\in (0,1)$$ we can move the contour to $$\sum_{k\ge 1} \frac{k^{-s}}{y+k}=\int_{+i\infty}^{-i\infty} \frac{z^{-s}}{y+z} (\frac1{e^{2i\pi z}-1}-\frac1{2i\pi z})dz$$ $$= e^{-i\pi (s+1)/2}\int_{-\infty}^{\infty} \frac{t^{-s}}{y+it} (\frac1{e^{-2\pi t}-1}+\frac1{2\pi t})dt$$
• I just realized that this answer doesn't relate $S(x,y)$ to $\zeta(x)$ or $\zeta(y)$. Commented Mar 15, 2023 at 20:39