Characteristic function of $\frac{1-\cos(x)}{x^2}$. I am trying to compute the characteristic function of the density $f(x)=\frac{1-\cos(x)}{x^2}$. But I do not know how to do it, I was trying to use the residue theorem  to compute
$$\phi(t)=\int_{-\infty}^\infty e^{itx}\frac{1-\cos(x)}{x^2}dx$$ but in this case it is not possible because there is no singularities, or I don't know how to use it in this case. Any help is welcome; Thank you!.

Without residues
You can compute the antiderivative first.
Using integration by parts
$$I_1=\int \frac{e^{i t x}}{x^2}\,dx=i t \text{Ei}(i t x)-\frac{e^{i t x}}{x}$$ where appears the exponential integral function.
$$I_2=\int \frac{e^{i t x}e^{ix}} {x^2}\,dx=\int\frac{e^{i (t+1) x}}{x^2}dx=\frac{i \left((t+1) x \text{Ei}(i (t+1) x)+i e^{i (t+1) x}\right)}{x}$$
$$I_3=\int \frac{e^{i t x}e^{-ix}} {x^2}\,dx=\int \frac{e^{i (t-1) x}}{x^2}dx=\frac{i (t-1) x \text{Ei}(i (t-1) x)-e^{i (t-1) x}}{x}$$ At this point, yo have all required elements. Use the bounds.
