I am trying to evaluate the following expression for $\lambda \in \mathbb{R}$ :
$$f(\lambda)=\sum_{n=1}^{+\infty}e^{-i\lambda n}$$
My idea is to introduce an epsilon prescription, so I choose $\epsilon>0$, I then define a new function :
$$f_\epsilon(\lambda)=\sum_{n=1}^{+\infty}e^{-i\lambda n-\epsilon n}$$
Doing the geometric summation, I find
$$f_\epsilon(\lambda)=\frac{e^{-i\lambda -\epsilon}}{1-e^{-i\lambda-\epsilon}}$$
I now want to take the limit $\epsilon \rightarrow 0$, for this I neglect the $\epsilon$ term on the numerator and expand it on the denominator.
$$f_\epsilon(\lambda)\sim \frac{1}{e^{i\lambda}-1+\epsilon}= \frac{i}{-2e^{i\lambda/2}\sin(\lambda/2)+i\epsilon}$$
This would lead to
$$f(\lambda)= \pi \sum_{l\in \mathbb{Z}}\delta(\lambda-2\pi l)+\mathcal{P}(\frac{1}{e^{i\lambda}-1})$$
where $\mathcal{P}$ stands for Cauchy principal value. Yet, this seems in contradiction with Poisson summation formula,
$$\sum_{n\in \mathbb{Z}}e^{-i\lambda n}=2\pi \sum_{l\in \mathbb{Z}}\delta(\lambda -2\pi l)$$
as $$\sum_{n\in \mathbb{Z}}e^{-i\lambda n}=f(\lambda)+f(-\lambda)+1$$ I don't see how the principal values would cancel one with another...