# Thread: characteristic function (evaluation of an definite integral)

1. ## characteristic function (evaluation of an definite integral)

I want to show that the characteristic function of Cauchy's distribution ($\displaystyle f(x) = \frac{1}{\pi} \frac{b}{b^{2}+x^{2}}, \ b>0$) is $\displaystyle \phi_{X}(t) = e^{-b|t|}$

$\displaystyle \phi_{X}(t) = E[e^{itX}] = E[\cos(tX)] + i E[\sin (tX)]$

$\displaystyle = \frac{b}{\pi} \int^{\infty}_{-\infty} \frac{\cos(tx)}{b^{2}+x^{2}} \ dx + i \ \frac{b}{\pi} \int^{\infty}_{-\infty} \frac{\sin(tx)}{b^{2}+x^{2}} \ dx$

The second integral evaluates to zero because the integrand is odd.

The integrand of the first integral is even, so I need to evaluate $\displaystyle \frac{2 b}{\pi} \int^{\infty}_{0} \frac{\cos(tx)}{b^{2}+x^{2}} \ dx$

Any suggestions?

2. Originally Posted by Random Variable
I want to show that the characteristic function of Cauchy's distribution ($\displaystyle f(x) = \frac{1}{\pi} \frac{b}{b^{2}+x^{2}}, \ b>0$) is $\displaystyle \phi_{X}(t) = e^{-b|t|}$

Any suggestions?
Usual proofs are:

- Fourier inversion formula. It is easy to compute the Fourier transform of $\displaystyle e^{-b|t|}$ 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 $\displaystyle \int_{-R}^R \frac{e^{itx}}{1+t^2}dt$ 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 $\displaystyle i$ (Residue theorem).

3. I hate contour integration.

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

Let $\displaystyle f(z) = \frac{e^{itz}}{b^{2}+z^{2}}$

which has the simple pole $\displaystyle z = bi$ in the upper half plane

so $\displaystyle \int^{\infty}_{-\infty} \frac{\cos (tx)}{b^{2}+x^{2}} = -2 \pi \ Im (Res\{f,bi\})$

$\displaystyle Res\{f,bi\} = \lim_{z \to bi} (z-bi) \frac{e^{itz}}{b^{2}+z^{2}}= \lim_{z \to bi} \frac{e^{itz}}{z+bi} = \frac{e^{-bt}}{2bi} = -i \ \frac{e^{-bt}}{2b}$

so $\displaystyle \int^{\infty}_{-\infty} \frac{\cos (tx)}{b^{2}+x^{2}} = \pi \ \frac{e^{-bt}}{b}$

and $\displaystyle \phi_{X}(t) = e^{-bt}$

From where is the absolute value coming?

4. Originally Posted by Random Variable

From where is the absolute value coming?
It's hidden in your computation. The integral along the upper half-circle converges to 0 only when $\displaystyle t>0$. Otherwise, you have the consider the lower half-circle hence the pole is $\displaystyle -ib$ (or you can note from the beginning that the characteristic function is even by symmetry of the Cauchy distribution).