First off, i appologise for the long question, but it seems this is the only way i can convey my thoughts. Referring to this unanswered question on MO, i have thought for some time about it, and came up with the following : we first denote by $\beta_{n+1}(x)$ the function : $$\beta_{n+1}(x)=(-1)^{n+1}\frac{2}{\pi}\sin(\pi x)\frac{n+1}{x^{2}-(n+1)^{2}}\;\;\;\;n=0,1,2...$$ and by $g_{n+1}(s)$ its Mellin transform : $$g_{n+1}(s)=\int_{0}^{\infty}\beta_{n+1}(x) x^{s-1}dx\;\;\;\;\;a<\Re(s)<b$$ by Mellin inversion theorem, we have for every $k=0,1,2 ... $, and for $a<\sigma<b$ : $$\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}g_{n+1}(s)(1+k)^{-s}ds=\left\{\begin{matrix} 1 &,&k=n \\ 0 &,& \text{otherwise} \end{matrix}\right. $$ Now, for any sequence $\left \{ f(n+1) \right \}_{n=0}^{\infty}$ : we define the related function : $$A_{F}(x)=\sum_{n=0}^{\infty}\frac{f(n+1)}{n+1}\beta_{n+1}(x)=- \frac{2}{\pi}\sin(\pi x)\sum_{n=0}^{\infty}(-1)^{n}\frac{f(n+1)}{x^{2}-(n+1)^{2}}$$ The Mellin transform of $A_{F}(x)$ is given by : $$\hat{A}_{F}(s)=\int_{0}^{\infty}A_{F}(x)x^{s-1}dx=\sum_{n=0}^{\infty}\frac{f(n+1)}{n+1} g_{n+1}(s)\;\;\;\;\;\;\;\;\;(A)$$ when the integral converges. By (A), we have that : $$\frac{f(k+1)}{(k+1)^{z+1}}=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\hat{A}_{F}(s)(1+k)^{-s-z}ds$$ Thus : $$\sum_{m=0}^{\infty}a_{m}\sum_{k=0}^{m}(-1)^{k}\binom{m}{k}\frac{f(k+1)}{(k+1)^{z+1}}=\frac{1}{2\pi i }\int_{\sigma-i\infty}^{\sigma+i\infty}\hat{A}_{F}(s)G(s+z)ds$$ Where : $$G(\mu)=\sum_{m=0}^{\infty}a_{m}\sum_{k=0}^{m}(-1)^{k}\binom{m}{k}(k+1)^{-\mu}=\frac{1}{\Gamma(\mu)}\int_{0}^{\infty}x^{\mu-1}e^{-x}g(1-e^{-x})dx$$ $$g(y)=\sum_{m=0}^{\infty}a_{m}y^{m}$$ A simple example of such a construction is the everywhere-convergent series for the Riemann zeta function: $$\zeta(s)-\frac{1}{s-1}=\sum_{m=0}^{\infty}\left | G_{m+1} \right |\sum_{k=0}^{m}(-1)^{k}\binom{m}{k}\frac{1}{(k+1)^{s}}$$ $\left | G_{n+1} \right |$ being the absolute Gregory coefficients. In this case, the function $\hat{A}_{F}(s)$ is given by (needs some work) : $$\hat{A}_{F}(s)=\pi^{1-s}\Gamma(s)\sin\left(\frac{\pi s}{2}\right)\frac{1}{(s-2)(s-1)}\;\;\;(-1<\Re(s)<1)$$ Thus : $$\sum_{m=0}^{\infty}\left | G_{m+1} \right |\sum_{k=0}^{m}(-1)^{k}\binom{m}{k}\frac{1}{(k+1)^{z+1}}$$$$\zeta(z+1)-\frac{1}{z}=\frac{1}{2\pi i }\int_{\sigma-i\infty}^{\sigma+i\infty}\pi^{1-s}\Gamma(s)\sin\left(\frac{\pi s}{2} \right )\frac{1}{(s-2)(s-1)}\left[\zeta(z+s)-\frac{1}{z+s-1} \right ]ds$$ The function : $$\pi^{-r}\Gamma(r)\sin\left(\frac{\pi r}{2}\right)$$ decreases rapidly as $\Im(r)\rightarrow \pm \infty$ in the half plane $\Re(r)< \frac{1}{2}$, and the zeta function is of finite order. Thus, we are allowed to shift the line of integration to the left, picking up the residues at the points $s=-1,-3,-5...$. However, the answer we get is not correct ! But, if we shift the line of integration to the right of 1, we get the correct answer.
My Questions : Is there something wrong in my derivation ? And why aren't we allowed to shift the line of integration o the left ? By the same token, are we allowed to shift the line of integration to right ?