I am working on the following:
Show that if $\varphi$ the characteristic function of an integer-valued distribution then
\begin{align*}
\mathbb P(X = k) = \frac{1}{2\pi} \int_{-\pi}^\pi e^{-itk} \varphi(t)\, dt, \quad k \in \mathbb Z.
\end{align*}
Edit:
It is
\begin{align*}
\frac{1}{2\pi} \int_{-\pi}^\pi e^{-ikt} \varphi(t) \, dt
&= \frac{1}{2\pi} \int_{-\pi}^\pi e^{-ikt} \sum_{n = -\infty}^\infty e^{int}p(n) \, dt
= \frac{1}{2\pi} \int_{-\pi}^\pi \sum_{n = -\infty}^\infty e^{it(n-k)} p(n) \, dt \\
&= \frac{1}{2\pi} \sum_{n = -\infty}^\infty \int_{-\pi}^\pi e^{it(n-k)} p(n) \, dt
= \frac{1}{2\pi} \sum_{n = -\infty}^\infty p(n) \int_{-\pi}^{\pi} e^{it(n-k)} \, dt \\
&= \frac{1}{2\pi} \left(\sum_{n = -\infty}^{k-1} p(n) \int_{-\pi}^{\pi} e^{it(n-k)} \, dt + \sum_{n = k+1}^{\infty} p(n) \int_{-\pi}^{\pi} e^{it(n-k)} \, dt + 2\pi p(k)\right).
\end{align*}
I'm not sure if the last part is right and I don't know how to go on.
Best Answer
This approach can work, we have to use an expression of cumulative distribution function.
There is an approach bases on the expression of the characteristic function: we have $\varphi(t)=\sum_{n\in\mathbb Z}\mathbb P\{X=n\}e^{itn}.$ Since the series $\sum_{n\in\mathbb Z}\mathbb P\{X=n\}$ is convergent, we can switch the series and the integral.
This has been then tried in the OP: we then use the fact that $\int_0^{2\pi}e^{it(n-k)}\mathrm dt=0$ if $n\neq k$ and $2\pi$ if $n=k$.