Could anyone shed some light on how to convert the following integral to a sum? $$ I=\int_0^{\infty} \frac{e^{-\varepsilon s} \, (s+s^2)^{\beta}}{\log^{\gamma}((1+s)/s)} \, ds; \qquad\,\,\varepsilon,\beta,\gamma>0$$
With $t=\log((1+s)/s)$, we get $$ I=I_3=e\int_0^{\infty} \frac{e^{-\frac{\varepsilon}{e^t-1}} \, e^{\beta t}}{t^{\gamma} \left(e^t-1\right)^{2\beta+2}} \, dt=??$$
Using the generalized Bernoulli polynomials $B^\alpha_n(x)$ of order $\alpha$ defined by the generating function $$\frac{t^\alpha e^{xt}}{(e^{t}-1)^\alpha} =\sum_{n=0}^{\infty} B_n^\alpha(x) \frac{t^n}{n!}, \qquad (*)$$ Then, for $x=0$ and $\alpha=2\beta+2$, we get $$ \frac{t^{2\beta+2}}{ \left(e^t-1\right)^{2\beta+2}} =\sum_{n=0}^{\infty} B_n^{2\beta+2}(0) \frac{t^n}{n!}. $$ Invert the sum & integral, $I$ read as \begin{align*} I=e\int_0^{\infty} \frac{e^{-\frac{\varepsilon}{e^t-1}} e^{\beta t}}{t^{\gamma} \left(e^t-1\right)^{2\beta+2}} \, dt= e\sum_{n=0}^{\infty} \frac{B_n^{2\beta+2}(0)}{n!} \int_0^{\infty} \frac{e^{-\frac{\varepsilon}{e^t-1}}e^{\beta t}}{t^{\gamma+2\beta-n+2}} dt. \end{align*} It remains to calculate $\int_0^{\infty} \frac{e^{-\frac{\varepsilon}{e^t-1}} e^{\beta t}}{t^{\gamma+2\beta-n+2}} dt=?$.
But some references they impose the condition $|t|<2\pi$ to use the formula $(*)$. In my case $t\geq 0$. I don't know, is this condition ($|t|<2\pi$) necessary to write the formula $(*)$?
If not, another trick for converting integral $I$ to a sum (in terms of some special functions)??