We have the integral :


$$\int\_{0}^{\infty}\log\left(1+\frac{s^{2}}{4\pi^{2}} \log^{2}(1+ix)\right ) e^{-2\pi nx}dx$$


Where s is a complex parameter, and n is a positive integer. The integral converges by virtue of the exponential factor. I tried to deform the path of integration such that we avoid the branch cut(s) of the logarithm. But here is where i got stuck, the internal complex log makes it confusing to do so !


A different version of this problem was posted [here](https://math.stackexchange.com/questions/1168607/help-with-the-integral-int-0-infty-log-left1-fracs24-pi2-log) with no answers. 



Help with the integral $\int_{0}^{\infty}\log\left(1+\frac{s^{2}}{4\pi^{2}} \log^{2}(1+ix)\right ) e^{-2\pi nx}dx$
$$I\_n(s)=\int\_{0}^{\infty}\log\left(1+\frac{s^{2}}{4\pi^{2}} \log^{2}(1+ix)\right ) e^{-2\pi nx}dx$$


a closed-form evaluation of this integral does not look promising, but small and large-$|s|$ asymptotics is doable:


* small $|s|$ (with $\gamma$ Euler's constant and $\_3F\_3$ the [generalized hypergeometric function)](http://en.wikipedia.org/wiki/Generalized_hypergeometric_function):


$$I\_n(s)=\frac{s^2}{48\pi^3 n}\left(24 i \pi n \; \_3F\_3(1,1,1;2,2,2;2 i n \pi )-\tfrac{3}{2} [\pi +2 i \log (2 \pi n)] [2 i \log (2 \pi n)+4 i \gamma +\pi ]+6 \gamma ^2+\pi ^2\right)+{\rm order}(|s|^4)$$


* large $|s|$:


$$I\_n(s)=\frac{1}{\pi n}\log(s)+{\rm order}(|s|^0)$$


* and for large $n$ the integral decays as $1/n^3$,


$$\lim\_{n\rightarrow\infty} n^3 I\_n(s)=-\frac{s^2}{16\pi^5}$$


