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Let $F(z)=p(z)/q(z)$. Its only possible finite poles are the distinct nonreal roots ${\alpha }_j$, and the residue of $F(z)e^{iz}$ there is $p({\alpha }_j)e^{i{\alpha }_j}/q^{\prime }({\alpha }_j)$; this is zero if the numerator cancels that pole. Close the real segment by a positively oriented semicircle in the upper half-plane, where $|e^{iz}|=e^{-\mathrm{Im}z}$ decays.

For sufficiently large radius $R$, all roots lie strictly inside or outside the contour as appropriate, and the degree bound gives $|F(z)|\le C/R$ uniformly on the arc. Parametrizing by $z=R{e}^{i\theta }$ bounds the arc integral byHere $\sin \theta \ge 2\theta /\pi$ on $[0,\pi /2]$, and symmetry handles the other half. The residue theorem therefore givesTo justify that this is the ordinary improper integral rather than just its symmetric principal value, note $F(x)=O(|x{|}^{-1})$ and $F^{\prime }(x)=O(|x{|}^{-2})$. On either tail, integration by parts writes the integral as a vanishing endpoint term $F(x)e^{ix}/i$ minus an absolutely convergent integral involving $F^{\prime }$. Thus both one-sided limits exist. When the degree bound is stricter the original integral is already absolutely convergent. This proves the Fourier integral of a rational function with simple nonreal poles formula with all tail and arc limits accounted for.