In this post, the following simplification was made: $$2\pi i\sum_k\lim_{x\to ia_k}(x-ia_k) \frac{x^2e^{itx}}{\prod_j(x^2+a_j^2)}\\ =-\pi\sum_ke^{-ta_k}a_k\sum_{j\neq k}\frac1{a_j^2-a_k^2}.$$
What I have done is the following, but am unsure why it is wrong: $$ 2\pi i\sum_k\lim_{x\to ia_k}(x-ia_k) \frac{x^2e^{itx}}{\prod_k(x^2+a_k^2)}\\ =2\pi i\sum_k \frac{(ia_k)^2e^{-ta_k}}{\prod_k (ia_k+ia_k)}. $$ Even after reduction it does not matched the stated closed form. Where does the $\sum_{i\neq j}$ come from? I am unfamiliar with this notation.
Thank you!
For a given $k$, the $k$-th inner term of the sum is (before taking the limit) $$\begin{split} (x-ia_k)\frac{x^2e^{itx}}{\prod_j(x^2+a_j^2)} &= \frac {(x - ia_k)}{x^2+a_k^2} \cdot\frac{x^2e^{itx}}{\prod_{j \neq k}(x^2 + a_j^2)}\\ &=\frac {1}{x+ia_k}\cdot\frac{x^2e^{itx}}{\prod_{j\neq k}(x^2+a_j^2)}\\ \end{split}$$ where we used the fact that $x^2+a_k^2=(x-ia_k)(x+ia_k)$.
This implies $$\lim_{x\rightarrow ia_k}(x-ia_k)\frac{x^2e^{itx}}{\prod_j(x^2+a_j^2)}=-\frac 1 {2ia_k}\frac{a_k^2e^{-ta_k}}{\prod_{j\neq k}(a_j^2-a_k^2)}$$ which yields the final result.