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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?2012-05-01
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    @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.2012-05-01
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    What about $\int_0^\infty \frac{\sqrt{x}\exp(-x)}{\sqrt[3]{\exp(x)-1}}\mathrm dx$?2012-05-01
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    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.2012-05-01
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    @J.M. What about it?2012-05-05
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    Peter: it looks to me that the integral I gave doesn't readily admit a representation like the one described in the OP.2012-05-05
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    @J.M. I see.The integral is around 0.792012-05-05
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    @J.M. Just to be clear, I meant integral of $\exp x\pm 1$.2012-12-10

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