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In the following MSE Answer, a function of a complex parameter is constructed, and it is stated that the function $$ I(s) = - \frac{1}{2} \int_{0}^{\infty} u^s \left( \frac{1}{e^u-1} - \frac{e^{-u}}{2} - \frac{e^{-u}}{u} \right) \, \mathrm{d}u. $$ is analytic for $\operatorname{Re}(s) > -2$.

It sounds like it is a routine verification. My question is, what is the standard way to show that $I(s)$ is analytic on the half-plane $\operatorname{Re}(s) > -2$? I would be satisfied with a proof-hint for this function, or a citation for a good textbook treatment.

not_a_chatbot
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    I am a new user, so apologies in advance if this does not currently meet the MSE question standards. I am willing to improve the question as needed. – not_a_chatbot Aug 11 '24 at 20:24
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    The integral is absolutely and uniformly convergent at infinity for all $s$ (due to the exponentially decreasing factors) - in particular the integral from say $1$ to infinity defines an entire function (Morera and Fubini for interchange prove it rigorously) so the only issue is at $0$ and there one can show by a little computation that the bracket integrand is continuous and has a simple zero there so behaves like $u$ there hence it comes down to seeing for which $s$ we have $u^{s+1}$ integrable at zero so we can again apply Morera and Fubini and that of course gives $\Re s>-2$ – Conrad Aug 11 '24 at 21:28

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