Rational improper integrals without real poles are upper-half-plane residue sums

By [L3], the real improper integral converges. The same degree gap implies $zR(z)\to 0$ as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$, because the numerator degree is at least two less than the denominator degree.

Let ${\mathrm{\Gamma }}_T$ be the contour formed by $[-T,T]$ and the upper semicircle [L1] ${\gamma }_T^{+}$. For all sufficiently large $T$, the contour avoids every pole of $R$ and encloses exactly the poles of $R$ with positive imaginary part. So gives ${\int }_{-T}^{T}R(x)dx+{\int }_{{\gamma }_T^{+}}R(z)dz=2\pi i{\sum }_{\mathrm{\Im }a>0}\mathrm{Res}(R,a)\mathrm{.}$

By step 1.1 and [L2], the arc integral tends to $0$ as $T\to \mathrm{\infty }$. The straight-piece integral tends to ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}R(x)dx$ by step 1.1. Passing to the limit in step 2.1 yields the residue formula.