Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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

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

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.