Characteristics Function of Cauchy Distribution

Let X ~ C(μ,λ). To find the characteristic function of X, we first find the characteristic function of the standard Cauchy variate

$$Z \ = \ \frac{X- \mu}{\lambda}$$

The probability density function (pdf) of Z is

$$f(z) \ = \ \frac{1}{\pi} \ . \ \frac{1}{1+z^2} \ \ \ \ ; \ - \ \infty \ < \ z \ < \ \infty$$

The characteristics function of Z is given by

$$\phi_z(t) \ = \ E({e^i}^{tz})$$

$$ = \ \int_{-\infty}^{\infty} \ {e^i}^{tz} \ f(z) \ dz$$

$$= \ \frac{1}{\pi} \ \int_{-\infty}^{\infty} \ {e^i}^{tz} \frac{1}{1+z^2} \ dz$$

To find Φ_z(t), we consider a standard Laplace distribution given by the probability density function

$$f_1(z) \ = \ \frac{1}{2} \ {e^-}^{|z|}, \ \ \ \ -\infty \ < \ z \ < \ \infty$$

Then, the characteristic function of the standard Laplace variable is given by

$$\phi_1(t) \ = \ \frac{1}{2} \ {e^i}^{tz} \ {e^-}^{|z|} \ dz$$

$$= \ \frac{1}{2} \ \int_{-\infty}^{\infty} \ (Cos \ tz \ + \ i \ sin \ tz) \ {e^-}^{|z|} \ dz$$

$$= \ \frac{1}{2} \ \int_{-\infty}^{\infty} \ Cos \ tz \ {e^-}^{|z|} \ dz$$

$$ = \\int_{0}^{\infty} \ Cos \ tz \ {e^-}^{|z|} \ dz$$

Integrating by parts, we get,

$$ = \ \left [ \ - \ Cos \ tz \ {e