TITLE: Simplifying $i \left[ \ln(x+i)-\ln(x-i)-\ln(1+ix)+\ln(1-ix) \right]$
QUESTION [5 upvotes]: Can the following expression further be simplified and expressed in terms of usual functions such as inverse hyperbolic or inverse trigonometric functions?
$$
f(x) = i \left[ \ln(x+i)-\ln(x-i)-\ln(1+ix)+\ln(1-ix) \right] \, , 
$$
where $x\ge 0$ is a real number. 
Inputs and ideas welcome.
Thanks

REPLY [2 votes]: Another approach to simplifying $f(x)$ is to simply use the expansion in power series of $x$.
Specifically,
\begin{align}
i \left( \ln(x+i)-\ln(x-i) \right) &=-\pi+ 2\left( x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots \right) , \\
i \left(-\ln(1+ix)+\ln(1-ix) \right) &= 2\left( x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots \right).
\end{align}
Thus
$$
f(x) = -\pi + 4 \left( x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots \right) = -\pi+4\arctan x.
$$