I would like to get some feedback and hints for my answer below:
The problem is from a textbook I'm reading: problem 3.2.7 of Probability Theory Lecture Notes by Panchenko. Here's the question:
Find the characteristic function of the law on $\mathbb{R}$ with density $\max(1-|x|,0)$. Then use the Fourier transform formula to conclude that $\max(1-|t|,0)$ is a characteristic function of some distribution.
Here's my attempt:
The characteristic function of a random variable $X$ is defined by $$f(t)= \mathbb{E} e^{itX}$$ Since $$\max(1-|x|,0)= \begin{cases} 1+x & \text{if } x\in (-1,0)\\ 1-x & \text{if } x\in (0,1)\\ 0 & \text{otherwise} \end{cases}$$ so here $$f(t)= \int_{(-1,0)}e^{itx}(1+x)dx+ \int_{(0,1)}e^{itx}(1-x)dx$$
which after some calculations is $$f(t)= -(\frac{e^{it/2}-e^{-it/2}}{t})^2 = \frac{1}{t^2}(2-e^{-it}-e^{it})= \frac{4}{t^2} \sin^2(t/2)$$
Next, fourier inversion formula says that if $\int|f(t)|dt<\infty$, then $\mathbb{P}$ has density $$p(x)= \frac{1}{2\pi} \int f(t)e^{-itx}dt$$ So, we substitute the characterisitc function found above and we get that \begin{eqnarray} p(x) &=& \frac{1}{2\pi} \int_\mathbb{R} \frac{1}{t^2}(2-e^{-it}-e^{it})e^{-itx} dt \end{eqnarray}
Now, this integral seems complicated and I don't see how it gives the answer. Is there a simpler way of showing what is required at the end?