Let $X(t)$ denote Levy Process. It can be proves that c.f of $X(t)$ is given:
$E(e^{i\omega X(t)}) = e^{-\Phi(\omega)}$,
where
$ \Phi(\omega) = i \omega a - \int\limits^{-\infty}_{\infty} \frac{e^{i\omega x}-1-i\omega \tau(x)}{x^2} \, M(dx),$
$a$ being a real constant, $\tau(x)$ a centering function given by:
$\tau(x) = \begin{cases} -1 & \mbox{dla } x < -1, \\ x, & \mbox{dla } |x| \le 1, \\ 1 & \mbox{dla } x > 1, \end{cases}$
$M$ a canonical measure $M\{I\}<\infty$ for each bounded interval $I$, and
$M^+(x) = \int_{x^-}^{\infty} \frac{1}{y^2} M(dy) < \infty \ \ \ \ \ \ \ \ \ M^-(-x) = \int_{-\infty}^{-x^+} \frac{1}{y^2} M(dy) < \infty$
Suppose that $M$ has no atom at the origin and
$ \alpha = \int \limits_{-\infty}^{\infty} \frac{1}{x^2} \, M(dx) < \infty $
We can then write
$M^+(x) = \alpha[1-B(x-)], \ \ \ \ \ \ \ M^-(-x) = \alpha B(-x) (x>0), $
where $B(x)$ is a distribution function.
Proof that one can simplify $\Phi(\omega)$ to
$ \Phi(\omega) = - i \omega d - \alpha \int \limits_{-\infty}^{\infty} (e^{i \omega x} - 1) \, dB(x) $
Please help me.