- Show, by contour integration, that if \(a > 0\) and \(\xi \in \mathbb { R }\) then
\[ \frac {1}{\pi} \int_ {- \infty} ^ {\infty} \frac {a}{a ^ {2} + x ^ {2}} e ^ {- 2 \pi i x \xi} d x = e ^ {- 2 \pi a | \xi |}, \]
and check that
\[ \int_ {- \infty} ^ {\infty} e ^ {- 2 \pi a | \xi |} e ^ {2 \pi i \xi x} d \xi = \frac {1}{\pi} \frac {a}{a ^ {2} + x ^ {2}}. \]
Solution
Let \[ I(\xi)=\int_{-\infty}^{\infty}\frac{a}{a^2+x^2}e^{-2\pi i x\xi}\,dx. \] For \(\xi>0\), close the contour in the lower half-plane. Since \[ |e^{-2\pi i z\xi}|=e^{2\pi \xi\Im z}, \] the exponential decays there, and the semicircular contribution tends to \(0\). The contour is clockwise and contains only the pole \(z=-ia\). Its residue is \[ \operatorname{Res}_{z=-ia} \frac{a e^{-2\pi i z\xi}}{(z-ia)(z+ia)} =\frac{a e^{-2\pi a\xi}}{-2ia} =-\frac{e^{-2\pi a\xi}}{2i}. \] Hence \[ I(\xi)=-2\pi i\left(-\frac{e^{-2\pi a\xi}}{2i}\right) =\pi e^{-2\pi a\xi}. \] For \(\xi<0\), close in the upper half-plane; the pole at \(ia\) gives \[ I(\xi)=\pi e^{2\pi a\xi}. \] At \(\xi=0\), direct integration gives \(I(0)=\pi\). Therefore \[ \frac1\pi I(\xi)=e^{-2\pi a|\xi|}. \]
For the inverse transform, split at \(0\): \[ \begin{aligned} \int_{-\infty}^{\infty}e^{-2\pi a|\xi|}e^{2\pi i\xi x}\,d\xi &=\int_0^\infty e^{-2\pi(a-ix)\xi}\,d\xi +\int_0^\infty e^{-2\pi(a+ix)\xi}\,d\xi\\ &=\frac1{2\pi(a-ix)}+\frac1{2\pi(a+ix)}\\ &=\frac1\pi\frac{a}{a^2+x^2}. \end{aligned} \]