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?