So the point is that you obtain density function through the characteristic function with the inversion formulas. That's the part of the task. Then it supposed to be uniformly continuous.
That is, if $\phi(t)$ is absolutely inegrable, show that $\frac{1}{2\pi}\int \phi(t) e^{-itx} dt$ is unformly continuous.
I have got the exercise in my book, I’ve tried and it led to nothing as follows: $|p_\xi(x)-p_\xi(y)| \leq \frac{1}{ 2\pi}|x -y|\int |t \phi_\xi(t)| dt$. If the last integral was without $t$ then it would be it. But it contains $t$ and may not converge since that. I’ve been stuck on this for a few days, please, help.
As you noticed, it would be great if $\int |t \phi_\xi(t)| dt$ was finite. Under the assumptions of the question, it is not the case. But $t\mapsto t\phi(t)$ is locally integrable.
Take a positive $\varepsilon$. Since $\phi$ is integrable, we can find $R$ such that $\int_{\mathbb R\setminus[-R,R]}\lvert\phi(t)\rvert dt<2\pi\varepsilon$. Denoting $$f(x):=\frac{1}{2\pi}\int \phi(t) e^{-itx} dt,$$ one has for each $x,y\in\mathbb R$: $$ \lvert f(x)-f(y)\rvert\leqslant 2\varepsilon+\frac 1{2\pi}\left\lvert\int_{-R}^R\phi(t)\left(e^{-itx}-e^{-ity} \right)dt \right\rvert\leqslant 2\varepsilon+\frac{R^2}{\pi}\lvert x-y\rvert. $$