# Determine the distribution from characteristic function

Determine the distribution of $$X$$ if

$$\varphi(t) = \frac{2}{3e^{it} - 1},$$

for $$t\in \mathbb{R}$$ is its characteristic function.

I tried to use the inverse formula, which says that the density is given by

$$f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-itx} \cdot \frac{2}{3e^{it} - 1} \mathop{dx}$$

$$= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{e^{-itx}}{3e^{it} - 1} \mathop{dx}$$

But I have no idea how to compute this integral. Can someone please help me?

$$\phi (t)=\frac 2 3 e^{-it} (1-\frac 1 3 e^{-it})^{-1}$$. Expand this as $$\sum\limits_{k=0}^{\infty} \frac 2 3 e^{-it} (\frac 1 3 e^{-it})^{n}=\frac 2 3(e^{-it}+\frac 1 3 e^{-2it}+\frac 1 {3^{2}} e^{-3it}+...$$). From this you can conclude the random variable takes the values $$-1,0,1,...$$ with probabilities $$\frac 2 3 , \frac 2 {3^{2}},...$$.