Applying an inversion technique to Characteristic Functions I am struggling with this concept (self-study). Could someone show me how to explicitly apply the inversion formula for these examples? I am working through about 15 examples, but these 3 seemed sufficiently different to help me do the rest.
$\phi_1(t)=(1-|t|)_+$
$\phi_2(t)=\sum_{n=-\infty}^{\infty}\phi_1(t+2n\pi)$
$\phi_3(t)=(1-\frac{t^2}{2})e^{-t^2/2}$
Work/Thoughts
The only reference to the inversion formula that I have found is the following theorem:
Assumptions: 
1-$\phi$ is a characteristic function of a given probability distribution $F$
2- $F$ has continuity points $a, b$ with $a<b$.
$F(b)-F(a)=\lim_{n\to\infty}\frac1{2\pi}\int_{-\infty}^{\infty}\frac{e^{-ita}-e^{-itb}}{it}\phi(t)e^{-t^2/n}dt$
And I believe this simplifies to:
$\lim_{n\to\infty}\frac1{2\pi}\int_{-n}^{n}\frac{e^{-ita}-e^{-itb}}{it}\phi(t)$
From here I am not sure what to do. Thanks for any help.
more thoughts
I have read about two kinds of invertible CFs- those that are integrable, and those that are periodic. $\phi_2(t)$ is obviously of the periodic nature.
I also understand the following properties about characteristic functions:
If $F$ and $G$ are probability distributions and $G$ is absolutely continuous, then $F*G$ has density $\int_{-\infty}^{\infty}g(u-x)F(dx)$
This this helpful at all, perhaps for number 3?
 A: In Durrett's book's, we have the following inversion formula:
$$\lim_{T\to +\infty}(2\pi)^{-1}\int_{-T}^T\frac{e^{-ita}-e^{-itb}}{it}\varphi(t)dt=\mu(a,b)+\frac 12\mu(\{a,b\}).$$
Let $\mu_i$ the measure associated with $\varphi_i$.


*

*As $\varphi_1$ is integrable, we have that $\mu_1$ has density 
$$f(y)=\frac 1{2\pi}\int_{\Bbb R}e^{-ity}(1-|t|)^+dt=\frac 1{2\pi}\int_{-1}^1e^{-ity}(1-|t|)dt.$$
We will find  Polya's distribution.

*As $\varphi_2$ is not integrable, we have to use the classical inversion formula. We have 
\begin{align}
\mu_2(a,b)+\frac 12\mu_2(\{a,b\})&=\lim_{T\to +\infty}(2\pi)^{-1}\int_{-T}^T\frac{e^{-ita}-e^{-itb}}{it}\varphi_2(t)dt\\
&=\lim_{n\to +\infty}(2\pi)^{-1}\int_{-(2n+1)\pi}^{(2n+1)\pi}\frac{e^{-ita}-e^{-itb}}{it}\varphi_2(t)dt\\
&=\lim_{n\to +\infty}(2\pi)^{-1}\sum_{j=-n}^{n}\int_{(2j-1)\pi}^{(2j+1)\pi}\frac{e^{-ita}-e^{-itb}}{it}\varphi_2(t)dt\\
&=\lim_{n\to +\infty}(2\pi)^{-1}\sum_{j=-n}^{n}\int_{-\pi}^\pi\frac{e^{-i(t+2j\pi)a}-e^{-i(t+2j\pi)b}}{it}(1-|t|)^+dt\\
&=\lim_{n\to +\infty}(2\pi)^{-1}\sum_{j=-n}^{n}\int_0^1\frac{e^{-i2j\pi b}\sin(tb)-e^{-i2j\pi a}\sin(ta)}{it}(1-t)dt.
\end{align}

*As $\varphi_3$ is integrable, we have 
$$
f_3(y)=\frac 1{2\pi}\int_{\Bbb R}e^{-ity}\left(1-\frac{t^2}2\right)\exp(-t^2/2)dt,$$
where $f_3$ is a density of the measure with Fourier transform $\varphi_3$. The integral 
$$\int_{\Bbb R}e^{-ity}\exp(-t^2/2)dt$$
is classical; the other term can be found by integration by parts. 

