Consider $U$ to be a unitary $n \times n$ matrix and let $p \in \Lambda^{*} =\frac{2\pi}{L}(\mathbb{Z}+\frac{1}{2})$ be fixed. Suppose $f_{ij}$ (the subindices $i,j$ don't play any important role here) is a real-valued function which satisfies the following equality: $$\frac{1}{2}\int_{-L}^{L}dx e^{ipx}f_{ij}(x) = \sum_{k=1}^{N}U_{i,k}U^{*}_{j,k}\bigg{(}\frac{1}{-ip+\lambda_{k}}+\frac{1}{ip}\bigg{)},$$ where the $\lambda_{k}$ are real-valued constants and $U_{i,k}$ denotes the $(i,k)$-entry of $U$. I know that the right hand side of the above expression is absolutely convergent in $\Lambda^{*}$: $$\sum_{p\in \frac{2\pi}{L}(\mathbb{Z}+\frac{1}{2})} \bigg{|}\bigg{(}\frac{1}{-ip+\lambda_{k}}+\frac{1}{ip}\bigg{)}\bigg{|} \le \sum_{p\in \frac{2\pi}{L}(\mathbb{Z}+\frac{1}{2})}\bigg{(}\frac{1}{p^{2}+\lambda_{k}}+\frac{1}{p^{2}}\bigg{)} < \infty$$ so we can invert the Fourier transform for $f$ in the first expression: $$f_{ij}(x) = \sum_{p\in \frac{2\pi}{L}(\mathbb{Z}+\frac{1}{2})}\sum_{k=1}^{N}U_{i,k}U_{j,k}^{*}e^{-ipx}\bigg{(}\frac{1}{-ip+\lambda_{k}}+\frac{1}{ip}\bigg{)}$$
Here comes my question: if I want to take the limit $x \to 0$ in this last expression to calculate $f_{ij}(0)$, I would formally have: $$f_{ij}(0) = \sum_{p\in \frac{2\pi}{L}(\mathbb{Z}+\frac{1}{2})}\sum_{k=1}^{N}U_{i,k}U_{j,k}^{*}\bigg{(}\frac{1}{-ip+\lambda_{k}}+\frac{1}{ip}\bigg{)}$$ but the left hand side can have an imaginary part. Should I just choose the real part of the right hand side then? And what justify picking the real part of the series when taking the limit $x \to 0$?