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Evaluate $$E_n(a_1,...,a_n;t)=\int_{-\infty}^{\infty}\frac{x^2\cos(tx)}{\prod_{k=1}^{n}(x^2+a_k^2)}dx$$ for $i\ne j\iff a_i^2\ne a_j^2$, and $a_i\in\Bbb R\setminus \{0\}$.

I've been able to find the similar integral $$J_n(a_1,...,a_n;t)=\int_{-\infty}^{\infty}\frac{\cos(tx)}{\prod_{k=1}^{n}(x^2+a_k^2)}dx$$ by seeing that we can write $$\prod_{k=1}^{n}\frac1{z+\alpha_k}=\sum_{k=1}^{n}\frac{1}{z+\alpha_k}\prod_{k\ne j=1}^{n}\frac{1}{\alpha_j-\alpha_k}\ ,$$ And then using $z\equiv x^2$, $\alpha_k\equiv a_k^2$ to get $$J_n(a_1,...,a_n;t)=\sum_{k=1}^{n}J_1(a_k;t)\prod_{k\ne j=1}^{n}\frac{1}{a_j^2-a_k^2}\ .$$ Then we take the Laplace transform of the remaining integral: $$\begin{align} \mathcal{L}\{J_1(q;t)\}(s)&=2\int_0^\infty e^{-st}\int_0^\infty \frac{\cos(tx)}{x^2+q^2}dxdt\\ &=2\int_0^\infty\frac{1}{x^2+q^2}\int_0^\infty \cos(tx)e^{-st}dtdx\\ &=2\int_0^\infty\frac{1}{x^2+q^2}\text{Re }\int_0^\infty e^{-(s-ix)t}dtdx\\ &=2\int_0^\infty\frac{1}{x^2+q^2}\text{Re }\left[\frac1{s-ix}\right]dx\\ &=2s\int_0^\infty\frac{1}{(x^2+q^2)(x^2+s^2)}dx\\ &=\frac{2s}{s^2-q^2}\left[\int_0^\infty\frac{dx}{x^2+q^2}-\int_0^\infty\frac{dx}{x^2+s^2} \right]\\ &=\frac{\pi s}{s^2-q^2}\left[\frac{1}q-\frac{1}s \right]\\ &=\frac{\pi}{q}\left[\mathcal{L}\{\cosh(qt)\}(s)-\mathcal{L}\{\sinh(qt)\}(s)\right]\\ &=\frac{\pi}{q}\mathcal{L}\{e^{-qt}\}(s)\ . \end{align}$$ Which gives $$J_1(q;t)=\frac{\pi}{q}e^{-qt}$$ and $$J_n(a_1,...,a_n;t)=\pi\sum_{k=1}^{n}\frac{e^{-a_k t}}{a_k}\prod_{k\ne j=1}^{n}\frac{1}{a_j^2-a_k^2}\ ,$$ from which many remarkable results may be derived.

I have the suspicion however, that we may be able to find $E_n$ in a similar way, but I don't know how. One idea: $$\int_{-\infty}^{\infty}\frac{(x^2+1)\cos(tx)}{(x^2+1)(x^2+\mathbf{a}_n^2)}dx=J_n(\mathbf{a}_n;t)=E_{n+1}(1,\mathbf{a}_n;t)+J_{n+1}(1,\mathbf{a}_n;t)$$ Where we used the shorthand notation $$x^2+\mathbf{a}_n^2\equiv\prod_{k=1}^{n}(x^2+a_k^2)$$ $$\mathbf{a}_n\equiv(a_k)_{k=1,...,n}\ .$$ It should be noted that the above holds for $1\not\in \mathbf{a}_n$.

Any Ideas? Thanks.

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    $\begingroup$ I see you've been busy generalizing that other integral! $\endgroup$ – dxdydz Jun 3 '19 at 16:53
  • $\begingroup$ @dxdydz I just can't help myself :) $\endgroup$ – clathratus Jun 3 '19 at 17:08
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    $\begingroup$ $x^2=(x^2+a_1^2)-a_1^2$ in the numerator?.. $\endgroup$ – metamorphy Jun 3 '19 at 17:46
  • $\begingroup$ (Yet another idea is to look at $\partial^2 J_n/\partial t^2$.) $\endgroup$ – metamorphy Jun 3 '19 at 17:50
  • $\begingroup$ Take the second derevative with respect to t $\endgroup$ – Gustave Jun 3 '19 at 20:28
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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}

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  • $\begingroup$ This works! thanks :) $\endgroup$ – clathratus Jun 3 '19 at 21:08
  • $\begingroup$ Quick clarification: your contour is the real line? $\endgroup$ – clathratus Jun 4 '19 at 14:11
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    $\begingroup$ @clathratus You should first take the semicircle contour (see the first picture here: mathworld.wolfram.com/ContourIntegration.html), and then take the limit $R\to\infty$. The integral over the arc vanishes in the limit, and thus the integral over the semicircle tends to the integral over the real line as $R\to\infty$. $\endgroup$ – user246336 Jun 4 '19 at 18:35
  • $\begingroup$ Aah. I see. Thanks :) $\endgroup$ – clathratus Jun 4 '19 at 18:38
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You have just to take the second derevative with respect to t to find $E$ in fonction of $J$.

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