Let $X$ be a r.v, with characteristic function $\phi$ and distribution function $F.$ Prove that $\forall \alpha>0,x \in \mathbb{R}$: $$\int_0^\alpha (F(x+u)-F(x-u))du=\frac{1}{\pi}\int_{\mathbb{R}}\frac{1-\cos(\alpha u)}{u^2}e^{-ixu}\phi(u)du.$$
Computing the right side gave $$\int_{\mathbb{R}}\left(\frac{1}{2}|\alpha+x-y|+ \frac{1}{2}|\alpha-x+y|-|x-y|\right)dP_X(y)$$ How to prove that this quantity is equal to the left side?
According to inversion formula $\dfrac{F(x+h) - F(x-h)}{2h} = \dfrac{1}{2\pi} \displaystyle \int_{-\infty}^{\infty}\sin(ht) e^{-itx} \phi_x(t) dt$
Hence we have: $$\int_0^{\alpha} F(x+h)-F(x-h) dh = \dfrac{1}{\pi}\int_0^{\alpha}\int_{-\infty}^{\infty} h\sin(ht)e^{-ixt}\phi_x(t)dtdh = \dfrac{1}{\pi} \int_{-\infty}^{\infty} e^{-ixt}\phi_x(t) \int_0^{\alpha} h\sin(ht)dh dt$$
Now calculate $\displaystyle \int_0^\alpha h \sin(ht) dh$