I found the following statement: If a $\varphi$ characteristic function is integrable $\left(\varphi\in L^{1}\right)$, then there is an $f$ continuous bounded density function that is $$f\left(x\right)=\frac{1}{2\pi}\int_{\mathbb{R}}\exp\left(-itx\right)\varphi\left(t\right)dt\;\;\;\forall{x}\in\mathbb{R}.$$ So this is a kind of inversion formula. I'm surprised, that $f$ is continuous, but I saw the proof and it is indeed continuous.
Is that statement also mean if $\mathbf{Q}$ is a discrete probability distribution, then $\varphi$ is not integrable?
I wanted to find a good example for a symmetric distribution, because I wanted $\varphi$ to be real valued. I don't really know well known symmetric discrete distribution but the Rademacher distribution. $\xi$ is Rademacher distributed if $\xi=\left\{ 1,-1\right\}$ with $\mathbf{P}\left(\xi=1\right)=\mathbf{P}\left(\xi=-1\right)=\frac{1}{2}$. Here the characteristic function is $\cos t$, which is indeed not integrable on $\mathbb{R}$, so I think it is a good example.
A real-valued random variable whose characteristic function is integrable indeed admits a probability density function whose expression is given by the formula you wrote.
This implies indeed that if $\mathbb Q$ is a discrete probability distribution, then the characteristic function of a random variable whose distribution is $\mathbb Q$ is not integrable.