I have the following quantity:
$$\frac{1}{\pi\ i}\left[\frac{1}{2}\log[e^{-2ix}]-\log[e^{-ix}]\right]$$
with $x$ being a real number.
I am a bit confused as to how I can treat this. I know that $\log[e^{iy}]=iy+2in\pi,\ n\in\mathbb{Z}$
, but what I must calculate confuses me. It is obviously equal to an integer multiple of $\pi$, although I cannot figure out if it is an even or an odd multiple of $\pi$.
Thank you.
$$\frac12\ln e^{-2ix}=\frac12(-2ix+2n\pi i)=-ix+n\pi i$$ $$\ln e^{-ix}=-ix+2k\pi i$$ Thus the quantity is $$n-2k$$ which can be any integer.
Therefore, the quantity equals any integer.