Region of convergence for $\Gamma(s)\zeta(s)$ in analytic continuation of Riemann zeta function

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?