Considering a generic p.d.f. $p(x)$, we know that the characteristic of such distribution is given by $\varphi(t)=E[e^{i t x}]$. Here $t,x\in (-\infty,\infty)$.
Now I would like to ask what is the characteristic function of a p.d.f. where its support is not the whole real numbers, for instance, the support is $\mathbb{R}^+$.
I already understand that as follows: Assume a two dimensional distribution $f(x,y)$ in $\mathbb{R}^2$ with characteristic function $E[e^{i(k_x x+k_y y)}]$. By using polar coordinates $x=r \cos \theta$ and $x=r \sin \theta$ and averaging over $\theta$ we find $$ p(r)=r\int_{0}^{2\pi} d\theta\, f(r\cos\theta,r\sin\theta). $$ Now the support of the distribution $p(r)$ is $\mathbb{R}^+$. In addition, we can average out the characteristic function too, $$ \frac{1}{2\pi} E\left[ \int_{0}^{2\pi}d\theta\, e^{i r k\cos(\theta-\theta_k)}\right]=E[J_0(k\, r)],$$ where in the above $k_x=k \cos \theta_k$ and $k_y=k \sin \theta_k$. In the above, $J_0(k\, r)$ is the Bessel function of the first kind. In other words, I think the characteristic function for a general distribution $p(r)$ with $\mathbb{R}^+$ is $E[J_0(k\, r)]$.
I am not sure that the above conclusion is true. And if it is true I am not sure it is unique. Also what is the systematic way to find the above result?
Just focusing on your primary problem without going into your example (I am afraid I don't quite follow there - maybe you can add some context there), the easiest example I can imagine is the uniform distribution $X\sim\mathcal U[a,b]$. If we choose $0<a<b\in \mathbb R$, we get a distribution which has mass only in $[a,b]$ with $\rm pdf$ (probability density function) $$ p(x)={\begin{cases}{\frac {1}{b-a}}&\mathrm {for} \ a\leq x\leq b,\\[8pt]0&\mathrm {for} \ x<a\ \mathrm {or} \ x>b\end{cases}} $$ This implies for the characteristic function $\varphi(x)$ (which always exists for all probability distributions), that we get for $t\in \mathbb R$ $$ \varphi_X(t)=E[e^{i t X}]=\int_{\mathbb R}e^{i t x}p(x)dx=\frac 1{b-a}\int_{a}^be^{i t x}dx={\displaystyle {\frac {\mathrm {e} ^{itb}-\mathrm {e} ^{ita}}{it(b-a)}}} $$ Again, this is the easiest example I can think of, but in general, there is no problem whatsoever with your $\rm pdf$ having support not on the whole real line.