For \(x\in\mathbb R\), let \[ f_1(x)=\int_{\mathbb R}\frac{e^{2\pi i x\xi}}{1+\xi^2}\,d\xi, \] calculate \(f_1(x)\) by the method of residues.
Solution
For every real \(x\), \[ \boxed{f_1(x)=\pi e^{-2\pi|x|}}. \]
The integral converges absolutely: its absolute integrand is \((1+\xi^2)^{-1}\), bounded on \([-1,1]\) and at most \(\xi^{-2}\) outside that interval. Therefore limits of the truncated real integrals below give the ordinary integral.
1. For \(x>0\), close the contour in the upper half-plane.
Set \[ F_x(z)=\frac{e^{2\pi i xz}}{1+z^2}. \] On the upper semicircle \(z=Re^{i\theta}\), \[ |e^{2\pi i xz}|=e^{-2\pi xR\sin\theta}\le1, \qquad |1+z^2|\ge R^2-1. \] Hence the arc integral has modulus at most \(\pi R/(R^2-1)\), which tends to zero. The only pole in the upper half-plane is \(z=i\), and \[ \operatorname{Res}_{z=i}F_x =\frac{e^{2\pi i x i}}{2i} =\frac{e^{-2\pi x}}{2i}. \] The residue theorem on the counterclockwise upper contour therefore gives [@SS03] \[ f_1(x)=2\pi i\frac{e^{-2\pi x}}{2i} =\pi e^{-2\pi x}. \]
2. For \(x<0\), close in the lower half-plane; the case \(x=0\) is direct.
When \(x<0\), the same estimate holds on the lower semicircle because \(\operatorname{Im}z\le0\) and \(x\operatorname{Im}z\ge0\). The contour is clockwise and encloses only \(z=-i\). Its residue is \[ \operatorname{Res}_{z=-i}F_x =\frac{e^{2\pi i x(-i)}}{-2i} =-\frac{e^{2\pi x}}{2i}. \] The residue theorem with clockwise orientation gives [@SS03] \[ f_1(x)=-2\pi i\left(-\frac{e^{2\pi x}}{2i}\right) =\pi e^{2\pi x}=\pi e^{-2\pi|x|}. \] For \(x=0\), \[ \int_{-\infty}^{\infty}\frac{d\xi}{1+\xi^2}=\pi, \] either from the same upper contour or from the elementary antiderivative. This matches the displayed formula.