Let \(f\) be a continuous element of \(L^1(\mathbb {R})\), and recall that its Fourier transform is \[ \hat {f}(t):=\int _{-\infty }^{\infty }\exp (itx)f(x)\,dx. \]
(a) Suppose also that \(\hat {f}\in L^1(\mathbb {R})\). Prove the following [inversion formula]: \[ f(x)=\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{-ixt}\hat {f}(t)\,dt \qquad \text {for all }x\in \mathbb {R}. \] In other words, \(\widehat {\widehat {f}}(-x)=2\pi f(x)\) for all \(x\in \mathbb {R}\).
(b) Choose and fix some \(\theta >0\). Use the preceding to prove that the Fourier transform of the probability density function \[ f(x):=\frac {\theta }{\pi (\theta ^2+x^2)} \qquad [-\infty <x<\infty ] \] is \(\hat {f}(t)=\exp (-\theta |t|)\).