How do you invert a characteristic function, when integral does not converge? I need to find the probability density of some distribution with characteristic function given by:
$$\frac{1}{9} + \frac{4}{9} e^{iw} + \frac{4}{9} e^{2iw}$$
I know the formula for inverting a characteristic function is:
$$f_X(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \phi(\omega) e^{-i\omega x} \mathop{d\omega}$$
But obviously putting this function inside the formula, will make the integral diverge. So my question is how does one invert a characteristic function, when this integral diverges? Or is this supposed to always converge and there is something wrong with my characteristic function?
 A: The inversion formula you cite is restricted to integrable characteristic functions. The case in your question is $\varphi_X(\omega)=\sum\limits_{k=1}^np_k\mathrm e^{\mathrm i \omega a_k}$ with $p_k\gt0$ and $\sum\limits_{k=1}^np_k=1$, which is never integrable. 
Assume that $n=1$, that is, that $\varphi_X(\omega)=\mathrm e^{\mathrm i \omega a}$. Can you identify the distribution of $X$ in this case? Hint: there is no density. Then the general case might be straightforward.
A: This is an old question, but I still like to write an answer.
First note that for your c.f. it holds that $\phi_x(\omega)=\phi_x(\omega+2\pi)$, indicating that your random variable takes only value in $\mathbb{Z}$ and hence is of lattice type for which a simple inversion formula is $f_X(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-i \omega x}\phi_x(\omega)d\omega$. 
An evaluation for $\phi_x(\omega)=\frac{1}{9}+\frac{4}{9}e^{i \omega}+\frac{4}{9}e^{2 i \omega}$ shows that $f_X(x)=\dfrac{2(2+x+x^2)}{9x(2-3x+x^2)}\dfrac{\sin(\pi x)}{2\pi}$.
Now note that $f_X(x)=0$ for $x\in\mathbb{Z}-\{0,1,2\}$, and that $\lim \limits_{x\rightarrow 0}f_X(x)=\frac{1}{9}$, $\lim\limits_{x\rightarrow 1}f_X(x)=\frac{4}{9}$ and $\lim\limits_{x\rightarrow 2}f_X(x)=\frac{4}{9}$. Therefore $f_X(x)=\frac{1}{9}$ for $x=0$, $f_X(x)=\frac{4}{9}$ for $x=1$, $f_X(x)=\frac{4}{9}$ for $x=2$, and $f_X(x)=0$ everywhere else.
