Referring to a previous question, i am having a hard time trying to do the integral:
$$f(s)=-i\int_{0}^{\infty}\frac{\log \left[1+\frac{\left(s\log(1+ix) \right )^{2}}{4\pi ^{2}} \right ]-\log \left[1+\frac{\left(s\log(1-ix) \right )^{2}}{4\pi ^{2}} \right ]}{e^{2\pi x}-1}dx$$
$s$ being a complex parameter. I have tried to apply the Abel-Plana formula. Namely, we have: $$\int\log\left(1+\frac{(s\log(1+x))^{2}}{4\pi^{2}} \right ) dx=(1+x)\log\left(1+\frac{(s\log(1+x))^{2}}{4\pi^{2}} \right ) $$ $$-e^{2\pi i /s}\text{Ei}\left(\log(1+x)-\frac{2\pi i }{s} \right )-e^{-2\pi i /s}\text{Ei}\left(\log(1+x)+\frac{2\pi i }{s} \right )$$ Where $\text{Ei}\left(\cdot\right)$ is the exponential integral function. Thus, by direct application of the Able-Plana formula : $$f(s)=e^{2\pi i /s}\text{Ei}\left(-\frac{2\pi i }{s} \right )+e^{-2\pi i /s}\text{Ei}\left(\frac{2\pi i }{s} \right ) $$
$$+\lim_{N \rightarrow \infty}\left(1+N\right)\log\left(1+\frac{\left(s\log(1+N)\right)^{2}}{4\pi^{2}}\right)$$ $$-e^{2\pi i /s}\text{Ei}\left(\log(1+N)-\frac{2\pi i }{s} \right )-e^{-2\pi i /s}\text{Ei}\left(\log(1+N)+\frac{2\pi i }{s} \right )$$ $$-\sum_{n=0}^{N}\log\left(1+\frac{(s\log(1+n))^{2}}{4\pi^{2}} \right ) $$ However, i haven't been able to obtain a meaningful result of the limit yet.
EDIT: Using the asymptotic estimate of the exponential integral function, the limit reduces to: $$\lim _{N\rightarrow \infty}\sum_{n=1}^{N}\left[\log\left(1+\frac{(s\log N)^{2}}{4\pi^{2}} \right )-\log\left(1+\frac{(s\log n)^{2}}{4\pi^{2}} \right )\right]-N\left(\frac{2\log N}{(\log N)^{2}+\frac{4\pi^{2}}{s^{2}}} \right )$$ Or :
$$2\lim_{N\rightarrow \infty}\sum_{n=1}^{N} \left[\int_{\log n}^{\log N}\frac{x}{x^{2}+\frac{4\pi^{2}}{s^{2}}}dx-\left(\frac{\log N}{(\log N)^{2}+\frac{4\pi^{2}}{s^{2}}} \right ) \right ]$$