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

Question

We have the integral :

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

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 was stuck, the internal complex $\log$ makes it confusing to do so !

EDIT

The integral is equivalent to : $$\frac{1}{2\pi n}\int_{0}^{\infty}\frac{e^{2\pi i nx}}{(1+x)\left(\frac{4\pi^{2}}{s^{2}}+\log(1+x) \right )}dx$$ Setting $y=\log(1+x)$, it's also equivalent to : $$\frac{1}{2\pi n}\int_{0}^{\infty}\frac{\exp[{2\pi i n \left(e^{y}-1 \right )]}}{\left(\frac{4\pi^{2}}{s^{2}}+y \right )}dy$$

Answer 1

I am not able to comment due to lack of reputation, but I did manage to split up the imaginary and real components (which is something you seem interested in):

$$ \int^\infty_0 \frac{1}{2}\log\left(\left(1+\frac{s^2\log\left(x^2+1\right)}{8\pi^2}\right)^2+\frac{s^4\arctan\left(x\right)^2}{16\pi^4}\right)\text{e}^{-2\pi n x}+i\arctan\left(\frac{s^2\arctan\left(x\right)}{4\pi^2\left(1+\frac{s^2\log\left(x^2+1\right)}{8\pi^2}\right)}\right)\text{e}^{-2\pi n x}\,\text{d}x $$

Although the real term will still have some imaginary components after integration (from the log functions).

Answer 2

Starting with your integral in terms of $y$;

change $e^{2\pi in(e^y-1)}$ into $e^{2\pi ine^y}e^{-2\pi in}=e^{2\pi ine^y}$ which simplifies it a bit.

Then change variables with $u=y+\frac{4 \pi^2}{s^2}$

Then change variables with v= ln(u)

You get the integral in the form $\int_b^{\infty}exp[ce^{e^v}] dv$

Are there any theorems about integrals of triple exponentials?