# Peculiar problem about characteristic function and density function.

I'm trying to prove the following proposition about characteristic functions:

Suppose $f(x)$ is an even probability density with characteristic function $\phi$. Define $$g(x) = \left\{ \begin{array}{c l} \displaystyle\int_{x}^{\infty}\frac{f(s)ds}{s} & x>0\\ g(-x) & x<0 \end{array} \right.$$ Then $g$ is again an even density and its characteristic function is $$\gamma(t)=\frac{1}{t}\displaystyle\int_{0}^{t}\phi(s)ds.$$

I'm stuck solving it. When I try integrate over all values of $g$ I don't get an useful expression. I have tried using integration of parts in the term $\displaystyle\int_{x}^{\infty}\frac{f(s)ds}{s},$ but the resulting term is even more complicated. I suppouse that the fact of to be even function helps to resolve it. In the second part I don't have progress.

Any kind of help is thanked in advanced.

The fact that $g$ is non-negative is quite clear. Since $f$is even, it suffices to prove that $\int_0^{ +\infty}g(x)\mathrm dx=1/2$. To this aim, write $$\int_0^{ +\infty}g(x)\mathrm dx=\int_0^{ +\infty}\int_0^{ +\infty}\frac{f(s)}s\mathbf 1\left\{ s \geqslant x\right\} \mathrm ds \mathrm dx$$ and since the integrand is non-negative, we can switch the integrals.
Do a similar method for the characteristic function and use the fact that $$\frac{\sin(tx)}x=\int_0^x\cos(tu)\mathrm du.$$
• Thanks for answer @Davide Giraudo. Could you explain a little better how to reach the equality $\displaystyle\int_{0}^{\infty}g(x)dx=1/2,$ please? – Suiz96 Oct 28 '16 at 13:38