I want to show that the characteristic function of Cauchy's distribution ( ) is
The second integral evaluates to zero because the integrand is odd.
The integrand of the first integral is even, so I need to evaluate
Any suggestions?
I want to show that the characteristic function of Cauchy's distribution ( ) is
The second integral evaluates to zero because the integrand is odd.
The integrand of the first integral is even, so I need to evaluate
Any suggestions?
Usual proofs are:
- Fourier inversion formula. It is easy to compute the Fourier transform of and it relates simply to Cauchy's distribution, so we can have it work backwards using Fourier inversion formula...
- contour integration in the complex plane. The integral is completed into a contour integral using a half circle in the upper half-plane; the integral along the circle is seen to converge to 0 as the radius goes to infinity (dominated convergence theorem) while the value of the contour integral is obtained by computing the residue at the pole (Residue theorem).
It's hidden in your computation. The integral along the upper half-circle converges to 0 only when . Otherwise, you have the consider the lower half-circle hence the pole is (or you can note from the beginning that the characteristic function is even by symmetry of the Cauchy distribution).