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Just to give a few examples, we have that

$$\eqalign{ & \int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x} - 1}}dx} = \Gamma \left( s \right)\zeta \left( s \right) \cr & \int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x} + 1}}dx} = \left( {1 - {2^{1 - s}}} \right)\Gamma \left( s \right)\zeta \left( s \right) =\eta(s)\Gamma(s) \cr & \int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x}\left( {{e^x} - 1} \right)}}dx} = \Gamma \left( s \right)\left( {\zeta \left( s \right) - 1} \right) \cr & \int\limits_0^\infty {\frac{{{x^{s-1}}{e^x}}}{{{{\left( {{e^x} - 1} \right)}^2}}}dx} = \Gamma \left( {s } \right)\zeta \left( s-1 \right)\cr & \int\limits_0^\infty  {\frac{{{e^x}{x^{s - 1}}}}{{{{\left( {1 + {e^x}} \right)}^2}}}}  = \Gamma \left( s \right)\eta \left( s-1 \right) \cr} $$

Is there any theory that enables us to state that any integral of the form

$$\int\limits_0^\infty {F\left( {\frac{1}{{{e^x} - 1}},{e^x},{x^s}} \right)dx} $$

will necessarily be evaluated in terms of $\zeta$ and $\Gamma$?

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How about the integral representation of Dirichlet L-functions? – Sangchul Lee May 1 '12 at 17:44
@sos440 I'm not really into that theory, but if you want to give an answer in terms of that, I guess I can manage. I will only need you to think that it is absolutely new theory for me. – Pedro Tamaroff May 1 '12 at 17:46
What about $\int_0^\infty \frac{\sqrt{x}\exp(-x)}{\sqrt[3]{\exp(x)-1}}\mathrm dx$? – J. M. May 1 '12 at 17:48
I'm also unfamiliar to Dirichlet $L$-functions; All my relevant knowledge on this subject just amounts a one-semester-course lecture on analytic number theory. So I cannot explain a possible deep linkage between the integral representation and the corresponding number-theoretic facts. As I remember, these integrals can be generalized to represent the Dirichlet series of (completely) multiplicative arithmetic functions. – Sangchul Lee May 1 '12 at 17:57
@J.M. What about it? – Pedro Tamaroff May 5 '12 at 2:17

I can offer the following observation. Consider the following identities \begin{align} \frac{1}{e^{x} - 1} = \sum_{k \geqslant 0} e^{-kx}, \quad \frac{1}{e^{x} + 1} = \sum_{k \geqslant 1} (-1)^{k + 1} e^{-kx} \quad \text{and} \quad \int_{0}^{\infty} x^{s} e^{kx} \ d^{\times} x = \Gamma(s) \, k^{-s}, \end{align} where $d^{\times} x = \frac{dx}{x}$ is the Haar invariant measure on $\mathbb{R}$. For $\mathsf{Re}(s) > 1$, $\mathsf{Re}(a) < 1$ and $a \not \in \mathbb{Z}$, one has \begin{align} \int_{0}^{\infty} \frac{x^{s} e^{a x}}{e^{x} - 1} \ d^{\times} x & = \int_{0}^{\infty} \sum_{k \geqslant 1} x^{s} e^{(a - k) x} \ d^{\times} x \\ & = \Gamma(s) \sum_{k \geqslant 1} (k - a)^{-s} \\ & = \Gamma(s) \zeta(s,1-a), \end{align} where $\zeta(s,1-a)$ denotes Hurwitz zeta function. Similarly, one has \begin{align} \int_{0}^{\infty} \frac{x^{s} e^{a x}}{e^{x} + 1} \ d^{\times} x & = \int_{0}^{\infty} \sum_{k \geqslant 1} (-1)^{k+1} x^{s} e^{(a - k) x} \ d^{\times} x \\ & = \Gamma(s) \sum_{k \geqslant 1} (-1)^{k+1} (k - a)^{-s} \\ & = \Gamma(s) 2^{-s} \left( \zeta(s,\tfrac{1-a}{2}) - \zeta(s,1 - \tfrac{a}{2}) \right). \end{align} The interchange of the summations and the integral sign is allowable since the summations are absolutely and uniformly convergent on their domains. These two formulas generalize the first two of your quoted formulas. The third one follows by splitting the integral into two using partial fractions and the integral formula of the Gamma function.

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Thank you. This is good. If you can, try considering powers of $(e^x \pm 1)^{-1}$. – Pedro Tamaroff May 1 '12 at 17:20

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