If $f$ characteristic function, then $\,\,\,\int^{\infty}_0 f(ut)e^{-u}du\,\,$ is a characteristic function.
My progress in the solution is: $$\int^{\infty}_0 f(ut)e^{-u}du= \int_{0}^{\infty}(\int_{-\infty}^{\infty}e^{itux}d\mu_{x})e^{-u}du=(fubini)=\int_{-\infty}^{\infty}(\int_{0}^{\infty}e^{itux-u}du)d\mu_{x}$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int_{-\infty}^{\infty}(\int_{0}^{\infty}e^{itux-u}du)d\mu_{x}=\int_{-\infty}^{\infty}(\lim_{U\to\infty}\int_{0}^{U}e^{u(itx-1)}du)d\mu_{x}$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int_{-\infty}^{\infty}(\lim_{U\to\infty}\frac{e^{Uitx}e^{-U}-1}{itx-1})d\mu_{x}, \,\,\,but \,\,(|e{^{Uitx}|\le 1;e^{-U}\to0}\,\, whit\,\,U\to\infty)$$ $$=\int_{-\infty}^{\infty}\frac{1}{1-itx}d\mu_{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$
my problem is to express the integral last, as a characteristic function.
thanks for any suggestions.