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Suppose $F(x) = P(x)/Q(x)$ is an integrable rational function on $\mathbb R$, that is, $\deg Q \geq \deg P + 2$, and $Q$ has no real roots.

Does there exist an expression for the definite integral $I_F = \int_{-\infty}^\infty F(x)dx$ in terms of only coefficients of $P$ and $Q$?

So far my idea was to apply the the residue theorem and the standard semicircle contour. This expresses $I_F$ as a symmetric function of half of all roots of $Q$, namely, those in the upper half-plane. Could we somehow upgrade it to a symmetric function of all of the roots, and thus the function of coefficients?

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  • $\begingroup$ What sort of expression do you seek? Since the integral already depends only on the coefficients, the question is just what kinds of expressions are allowed. $\endgroup$
    – LSpice
    Commented Oct 15, 2022 at 0:49
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    $\begingroup$ I think to get an "expression" for that integral, you would first need an "expression" for the complex zeros of the denominator? Once you have the zeros of the denominator, you can do partial fractions (with complex coefficients). $\endgroup$
    – Gerald Edgar
    Commented Oct 15, 2022 at 0:49
  • $\begingroup$ @GeraldEdgar Of course, one could do it, but my hope was that we could avoid it somehow, and evaluate the integral without first finding zeroes of the denominator. $\endgroup$
    – Troshkin Michael
    Commented Oct 15, 2022 at 1:00
  • $\begingroup$ So you are essentially asking if you can express the roots in terms of the coefficients? By Abel-Ruffini, if $\operatorname{deg} Q\ge 5$, then there is no such expression as an algebraic function of the coefficients. There might be more complicated expression for the roots, however. $\endgroup$
    – Dispersion
    Commented Oct 15, 2022 at 1:46
  • $\begingroup$ @Zachary, an algebraic function will almost never be possible: even in the simplest case, $\int_{-\infty}^\infty dx/(x^2+1)=\pi$ $\endgroup$
    – user44143
    Commented Oct 15, 2022 at 11:53

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