TITLE: Help with the integral $\int_{0}^{\infty}\log\left(1+\frac{s^{2}}{4\pi^{2}} \log^{2}(1+ix)\right ) e^{-2\pi nx}dx$
QUESTION [5 upvotes]: 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 with no answers.

REPLY [8 votes]: $$I_n(s)=\int_{0}^{\infty}\log\left(1+\frac{s^{2}}{4\pi^{2}} \log^{2}(1+ix)\right ) e^{-2\pi nx}dx$$
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):

$$I_n(s)=\frac{s^2}{48\pi^3 n}\left(24 i \pi  n \; _3F_3(1,1,1;2,2,2;2 i n \pi )-\tfrac{3}{2} [\pi +2 i \log (2 \pi  n)] [2 i \log (2 \pi  n)+4 i \gamma +\pi ]+6 \gamma ^2+\pi ^2\right)+{\rm order}(|s|^4)$$

large $|s|$:

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

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

$$\lim_{n\rightarrow\infty} n^3 I_n(s)=-\frac{s^2}{16\pi^5}$$