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How does one determine well-definedness in analytical continuation for $\Gamma(s)\zeta(s)$ function?

Firstly:

$$\Gamma(s)\zeta(s) = \int_0^\infty dt \frac{t^{s-1}}{e^t - 1},\quad Re(s) > 1$$

Using approximation for small $t$, I can expand the equation to:

$\displaystyle \Gamma(s)\zeta(s) = \int_0^\infty dt t^{s-1}\left(\frac{1}{e^t - 1} - \frac{1}{t} + \frac{1}{2} - \frac{t}{12}\right) + \frac{1}{s - 1} - \frac{1}{2s} + \frac{1}{12(s + 1)} + \int_1^\infty dt \frac{t^{s-1}}{e^t - 1}$

Now, I need to show that the right-hand side is well-defined for $Re(s) > -2$. I can see that there are simple poles at $-1$, $0$ and $1$, but how do I determine well-definedness for terms containing integrals?

EDIT: Let's look at $\displaystyle \int_1^\infty dt \frac{t^{s-1}}{e^t - 1}$. Since it doesn't have any poles, we only have to check the behavior at 0 and $\infty$. Therefore:

$\displaystyle \lim_{t\rightarrow 1}\frac{t^{s-1}}{e^t - 1} = \frac{1}{e - 1}$ for any $s$ and $\displaystyle \lim_{t\rightarrow \infty}\frac{t^{s-1}}{e^t - 1} = ???$

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Probably worth mentioning, that this comes from physics homework ;) –  plaes May 22 '11 at 21:56
    
Why not just use the fact that Γ and ζ have analytic continuations that are independently well-defined? –  Dan Brumleve May 22 '11 at 23:03
    
He probably wants to use this to deduce that $\zeta$ has an analytic continuation. –  N. S. May 22 '11 at 23:46
    
I actually want to show that terms with integrals are convergent even when Re(s) > -2. –  plaes May 22 '11 at 23:50
    
$\lim_{t \to \infty} \frac{t^{s-1}}{e^t-1} = 0$ since an exponential dominates a polynomial. –  Amit Kumar Gupta Jun 16 '11 at 5:54
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