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One way to determine values of the (analytically continued) Riemann zeta function is to $\zeta(s)$ is to use the product $\Gamma(s)\zeta(s)$ and use our knowledge of the poles of $\Gamma(s)$.

We can show that \begin{eqnarray} \Gamma(s) \zeta(s) &=& \sum_{n=1}^{\infty} \frac{1}{n^2} \int_{0}^{\infty} e^{-t} t^{s-1} \, dt = \int_{0}^{\infty} \dfrac{t^{s-1}}{e^t - 1} \, dt \end{eqnarray} which converges for $\mathrm{Re}(s) > 1$ (because the denominator $e^{t} -1$ is $\mathcal{O}(t)$, so the integrand is of the form $t^{s-2}$).

By splitting the integral into two and massaging the terms, we can arrive at an expression that is valid in a larger region:

\begin{eqnarray} \Gamma(s) \zeta(s) &=& \int_0^{1} \dfrac{t^{s-1}}{e^t - 1} \, dt + \int_1^{\infty} \dfrac{t^{s-1}}{e^t -1} \,dt \\ &=& \int_0^1 t^{s-1} \left( \dfrac{1}{e^t - 1} - \frac{1}{t} + \frac{1}{2} -\frac{1}{12}t \right) dt + \dfrac{1}{s-1} - \dfrac{1}{2s} + \dfrac{1}{12(s+1)} + \int_1^{\infty} \dfrac{t^{s-1}}{e^t -1} \,dt \end{eqnarray}

If we examine $\left( \dfrac{1}{e^t - 1} - \dfrac{1}{t} + \dfrac{1}{2} -\dfrac{1}{12}t \right)$, we find it is $\mathcal{O}(t^3)$

Question:

I believe that this means that the final expression is valid for $\mathrm{Re}(s)>-3$, with $s\neq 1, 0 -1$. I have found a claim that this integral is valid for $\mathrm{Re}(s) > -2$, but I can't see how that's the case. Which is correct?

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