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I am trying to solve the following question:

Verify that the integral $\int_{0}^{\infty} \, \frac{t^{z}}{e^{\,t\,}+1}dt$ represents an analytic function in the half plane $Re(z)>-1$. Show also that this function can be continued analytically to the strip $Re(z)>-2$ except for the point $-1$ and the point $-1$ is a first order pole of this function, find the residue at this point.

We have

$|\frac{t^{z}}{e^t+1}|\leq\frac{t^{Re(z)}}{2}$ for $t>0$ but the integral $\int_{0}^{\infty} \frac{t^{Re(z)}}{2}dt$

is not convergent. I don't know how to approach otherwise. Any help would be great.

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  • $\begingroup$ In general, $~\displaystyle\int_0^\infty\frac{t^n}{e^t+1}~dt~=~n!~\eta(n+1),~$ and $~\displaystyle\int_0^\infty\frac{t^n}{e^t-1}~dt~=~n!~\zeta(n+1),~$ see the Dirichlet $\eta$ and Riemann $\zeta$ functions for more information, both of which possess well-known analytic continuations. $\endgroup$
    – Lucian
    Apr 20, 2015 at 23:36

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