Referring to [a previous question](https://mathoverflow.net/questions/199831/help-with-the-integral-int-0-infty-log-left1-fracs24-pi2-log/200106#200106), I am trying to do the following integral :


$$\phi(s)=i\int\_{0}^{\infty}\frac{\log \left[1+\frac{\left(s\log\sqrt{1+ix} \right )^{2}}{\pi ^{2}} \right ]-\log \left[1+\frac{\left(s\log\sqrt{1-ix} \right )^{2}}{\pi ^{2}} \right ]}{e^{2\pi x}-1}dx\;\;\;\;\;\;\;\;\;(s\in\mathbb{C})$$


I tried applying the Abel-Plana formula to :
$$\phi(s)=i\int\_{0}^{\infty}\frac{\log \left[1+\frac{\left(s\log\sqrt{1+ix} \right )^{2}}{\pi ^{2}} \right ]e^{iax}-\log \left[1+\frac{\left(s\log\sqrt{1-ix} \right )^{2}}{\pi ^{2}} \right ]e^{-iax}}{e^{2\pi x}-1}dx$$
And then take the limit $a\rightarrow 0$, but it didn't help. I tried applying the residue theorem, but i don't seem to know how !! any help is appriciated.



Help with a difficult integral
![](https://ilorentz.org/beenakker/MO/logintegral.jpg)


If asymptotics is enough, an accurate large-$s$ approximation is $\phi(s)\approx 2.8-\ln |s|$, see the plot above (the error is of order $1/s$, so almost invisible on the scale of the plot for the largest $s$); for small $s$ the function starts off quadratically, $\phi(s)\approx −0.0021\,s^2$. Not sure if there is much more to say.


For imaginary $s=i\sigma$ the asymptotics, with an overall minus sign, is pretty much the same ($-2.7+\ln|s|$ fits better, the plot shows $-2.8+\ln|s|$)


![](https://ilorentz.org/beenakker/MO/logintegral_imaginary.jpg)


