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Investigating the integral $\int_{0}^{\infty}\frac{x^{s-1}}{e^x-1-x}\,dx \small \space\colon\space Re\{s\} \gt 1 \normalsize$,
With the special case ($s=3$) introduced here as $I_2=\int_{0}^{\infty}\frac{x^2}{e^x-1-x}\,dx$ : $$ \int_{0}^{\infty} \frac{x^{s-1}}{e^x-1-x} \,dx = \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1-x} \right) \,dx = \sum_{n=1}^{\infty} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{n} \,dx $$ Analyzing the last integral, it follows that: $$ \begin{align} & \small \quad \frac{0!}{\Gamma(s+0)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{1} \,dx = \zeta(s) \\ & \small \quad \frac{1!}{\Gamma(s+1)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{2} \,dx = \zeta(s) - \zeta(s+1) \\ & \small \quad \frac{2!}{\Gamma(s+2)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{3} \,dx = \zeta(s) - 3\zeta(s+1) + 2\zeta(s+2) \\ & \small \quad \frac{3!}{\Gamma(s+3)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{4} \,dx = \zeta(s) - 6\zeta(s+1) + 11\zeta(s+2) - 6\zeta(s+3) \\ & \small \quad \normalsize \text{. . .} \quad \text{. . .} \quad \text{. . .} \\ & \small \quad \frac{N!}{\Gamma(s+N)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{N+1} \,dx = \sum_{n=0}^{N} (-1)^{n}\,a_{n}\,\zeta(s+n) \\ & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \end{align} $$ Where the coefficients $a_n$ calculated from the product expansion: $$ \prod_{n=1}^{N} \left( 1 - n s \right) = \sum_{n=0}^{N} (-1)^{n}\,a_{n}\,s^{n} = 1 - a_1\,s^1 + a_2\,s^2 - a_3\,s^3 + \text{ ... } \pm\,a_{\small N \normalsize}\,s^{\small N \normalsize}\\ \boxed{ \quad \Rightarrow \sum_{n=0}^{\infty} (-1)^{n}\,a_{n}\,\zeta(s+n) = \lim_{N\rightarrow\infty} \frac{N!}{\Gamma(s+N)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{N+1} \,dx \quad } $$

And the question is about the last limit. In other words, does the limit exist and equal zero, so: $$\sum_{n=0}^{\infty}(-1)^{n}\,a_{n}\,\zeta(s+n) = 0 \qquad\colon\space Re\{s\} \gt 1 \tag{1}$$


On the other hand, we know that (except for $s=0$) the infinit product is not convergent: $$ \prod_{n=1}^{\infty} \left( 1 - n s \right) = 1 + \sum_{n=1}^{\infty} (-1)^{n}\,a_{n}\,s^{n} = 1 - a_1\,s^1 + a_2\,s^2 - a_3\,s^3 + \text{...} $$

Is it possible to utilize the exceptional case using the limit $\left\{\small\lim_{z\rightarrow\infty}\zeta(z)=1\normalsize\right\}$ to show that: $$\sum_{n=0}^{\infty}(-1)^{n}\,a_{n}\left[\zeta(s+n)-1\right] = 0 \qquad\colon\space Re\{s\} \gt 1 \tag{2}$$

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  • $\begingroup$ As written, the infinite product above is not convergent. $\endgroup$ – Olivier Oloa Oct 30 '16 at 23:39
  • $\begingroup$ Yes, except for (s=0), clearly the product of $s^n$ is not convergent. How to conclude the sum of $\zeta(s+n)$ is not convergent? $\endgroup$ – Hazem Orabi Oct 30 '16 at 23:56
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    $\begingroup$ To make sense of the coefficients, technically we'll have to define functions $a_n(k)$ via $$\prod_{n=1}^\infty\left(1-\frac{s}{n^k}\right)=1+\sum_{n=1}^\infty (-1)^na_n(k)s^n$$ for $\mathrm{Re}(k)>1$, then analytically continue each of them to get $a_n:=a_n(-1)$. I imagine the desired conclusion is a manifestation of Newton's identities. $\endgroup$ – arctic tern Oct 30 '16 at 23:57

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