Integral $E_n(a_1,...,a_n;t)=\int_{-\infty}^{\infty}\frac{x^2\cos(tx)}{\prod_{k=1}^{n}(x^2+a_k^2)}dx$

Using Cauchy's residue theorem the computation is almost immediate. We can assume that all $a_k$ are positive. Then we have \begin{align} \int_{\mathbb R}\frac{x^2\cos(tx)}{\prod_k(x^2+a_k^2)}\,dx &=\int_{\mathbb R}\frac{x^2e^{itx}}{\prod_k(x^2+a_k^2)}\,dx \\&=2\pi i\sum_k\lim_{x\to ia_k}(x-ia_k) \frac{x^2e^{itx}}{\prod_k(x^2+a_k^2)} \\&=-\pi\sum_ke^{-ta_k}a_k\sum_{j\neq k}\frac1{a_j^2-a_k^2}. \end{align}