Clarification about proof involving product/sum manipulation

Question

In this post, the following simplification was made:
$$2pi isum_klim_{xto ia_k}(x-ia_k)
frac{x^2e^{itx}}{prod_j(x^2+a_j^2)}\
=-pisum_ke^{-ta_k}a_ksum_{jneq k}frac1{a_j^2-a_k^2}.$$
What I have done is the following, but am unsure why it is wrong:
$$
2pi isum_klim_{xto ia_k}(x-ia_k)
frac{x^2e^{itx}}{prod_k(x^2+a_k^2)}\
=2pi isum_k frac{(ia_k)^2e^{-ta_k}}{prod_k (ia_k+ia_k)}.
$$
Even after reduction it does not matched the stated closed form. Where does the $sum_{ineq j}$ come from? I am unfamiliar with this notation.
Thank you!

Stefan Lafon · Accepted Answer

For a given $k$, the $k$-th inner term of the sum is (before taking the limit)
$$begin{split}
(x-ia_k)frac{x^2e^{itx}}{prod_j(x^2+a_j^2)} &= frac {(x - ia_k)}{x^2+a_k^2}
cdotfrac{x^2e^{itx}}{prod_{j neq k}(x^2 + a_j^2)}\
&=frac {1}{x+ia_k}cdotfrac{x^2e^{itx}}{prod_{jneq k}(x^2+a_j^2)}\
end{split}$$
where we used the fact that $x^2+a_k^2=(x-ia_k)(x+ia_k)$.
This implies
$$lim_{xrightarrow ia_k}(x-ia_k)frac{x^2e^{itx}}{prod_j(x^2+a_j^2)}=-frac 1 {2ia_k}frac{a_k^2e^{-ta_k}}{prod_{jneq k}(a_j^2-a_k^2)}$$
which yields the final result.

Clarification about proof involving product/sum manipulation

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