Bessel function, characteristic function, semicircle distribution I am trying to prove that $\dfrac{1}{x} J_1(2x)$, where $J_n$ is the Bessel function of order n, is the characteristic function of the semicircle distribution, i.e. $σ(x)=\dfrac{1}{2π}\sqrt{4-x^2} 
 \textbf{1}_{|x|<2} $. Basically, I would like to calculate the integral
$$\dfrac{1}{2π} \displaystyle\int_{\mathbb{R}}{\dfrac{1}{t}J(2t)}e^{-itx} dt $$
So, I do not want to calculate the characteristic equation of $σ$, I want to find $σ$ when I am given the characteristic function. I have tried some integral forms of $J_1$ but can't seem to find what I am looking for. Any hints?

$$J_1(z) = \dfrac{1}{iπ} \int_{0}^π e^{iz\cos θ}\cos θ  dθ $$
Now, we want to calculate the integral shown above to find the density function corresponding to the characteristic function $ \dfrac{1}{x} J_1(2x) $.
\begin{align*} 
f(x) &= \dfrac{1}{2π} \int_{\mathbb{R}} φ(t) e^{-itx} dt = \dfrac{1}{2π} \int_{\mathbb{R}} \dfrac{1}{t} J_1\left(2t\right) e^{-itx} dt \\[7pt]
&= \dfrac{1}{2iπ^2} \int_{\mathbb{R}}\int_{0}^π \dfrac{1}{t} e^{2it\cos θ }\cos θ e^{-itx} dθ dt  \\[7pt] 
&=  \dfrac{1}{2iπ^2} \int_{0}^π  \cos θ  \underbrace{ \int_{\mathbb{R}}  \dfrac{1}{t}  e^{it(2\cos θ-x)} dt }_{-iπ \text{sgn}(2cosθ-x) } dθ \\[7pt]
&= - \dfrac{1}{2π} \int_{0}^π \cos θ \text{sgn}(2\cos θ-x)  dθ
\end{align*}
If $ x > 2 $ then $ \text{sgn} (2 \cos θ-x)  = -1$, and if  $ x < -2 $ then $ \text{sgn} (2 \cos θ-x)  = 1$ and so, since $ \int_{0}^π \cos θ dθ  = 0$, $ f(x) = 0$ as well. Now, let $ -2\le x\le 2 $. If we set $ u=2\cos θ-x  $, then
\begin{align*}
&-\dfrac{1}{2π}\int_{-2-x}^{2-x} \dfrac{u+x}{2} \dfrac{1}{\sqrt{1-\left(\frac{u+x}{2}\right)^2}}  \text{sgn}(u)  du \\[7pt]
&= \dfrac{1}{2π}\int_{-2-x}^{0} \dfrac{u+x}{2} \dfrac{1}{\sqrt{1-\left(\frac{u+x}{2}\right)^2}}  du-\dfrac{1}{2π}\int_{0}^{2-x} \dfrac{u+x}{2} \dfrac{1}{\sqrt{1-\left(\frac{u+x}{2}\right)^2}}   du\\[7pt]
   &= \dfrac{1}{2π}  \left(    2 \int_{-1}^{\frac{x}{2}}  \dfrac{y}{\sqrt{1-y^2}}  dy - 2\int_{\frac{x}{2}}^{1}  \dfrac{y}{\sqrt{1-y^2}}   dy   \right)\\[7pt]
&=\dfrac{1}{2π} \left(     2 \sqrt{1-y^2} \bigg\rvert_{-1}^{x/2}  -  2 \sqrt{1-y^2} \bigg\rvert_{x/2}^{1}    \right) = \dfrac{1}{2π} \sqrt{4-x^2}
\end{align*}
