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?

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- April 28th 2010, 10:20 AMRandom Variablecharacteristic function (evaluation of an definite integral)
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? - April 28th 2010, 12:32 PMLaurent
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). - April 28th 2010, 08:45 PMRandom Variable
I hate contour integration.

But this one is not that bad because there's a nice formula.

Let

which has the simple pole in the upper half plane

so

so

and

From where is the absolute value coming? (Wondering) - April 29th 2010, 04:23 AMLaurent
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).