Show that $J_0(x)=\frac{2}{\pi}\int_0^\infty\frac{\sin\left(x+y\right)}{x+y}J_0(y)dy$ I've been trying to show, without success, that:
\begin{align*}
J_0(x)=\frac{2}{\pi}\int_0^\infty\frac{\sin\left(x+y\right)}{x+y}J_0(y)dy\quad\text{for }0\leq x<\infty,
\end{align*}
where $J_0$ is the Bessel function of the first kind of order $0$. The result has been verified numerically.
One of the things I tried is to use the integral representation $\frac{1}{x+y}=\int_0^\infty e^{-t(x+y)}dt$ an apply Fubini's. After simplying the first integral I arrive at:
\begin{align*}
\frac{1}{i\pi}\int_0^\infty e^{-tx}\left(\frac{\cos x+i\sin x}{\sqrt{(t-i)^2+1}}+\frac{-\cos x+i\sin x}{\sqrt{(t+i)^2+1}}\right)dt,
\end{align*}
which I doubt has a simple closed-form.
 A: Let $I(x)$ be represented by the integral
$$\begin{align}
I(x)&=\frac2\pi\int_0^\infty \frac{\sin(x+y)}{x+y}J_0(y)\,dy\tag1
\end{align}$$
Letting $t=-x$ and $I(-t)=J(t)$ in $(1)$ we see that
$$\begin{align}
J(t)&=\frac2\pi\int_{-\infty}^\infty \frac{\sin(t-\tau)}{t-\tau}J_0(\tau)H(\tau)\,d\tau\\\\
&=\frac2\pi (\text{sinc}*J_0H)(t)\tag2
\end{align}$$
Using the convolution theorem, the Fourier transform of $J(t)$ in $(2)$, as given by $\mathscr{F}\{J\}(\omega)=\int_{-\infty}^\infty J(t)e^{i\omega t}\,dt$, can be expressed as
$$\begin{align}
\mathscr{F}\{J\}(\omega)&=\frac2\pi\underbrace{\mathscr{F}\{\text{sinc}\}(\omega)}_{=\pi \text{Rect}(\omega/2)}\mathscr{F}\{J_0H\}(\omega)\\\\
&=2\text{rect}(\omega/2)\int_0^\infty J_0(t) e^{i\omega t}\,dt\\\\
&=2\text{rect}(\omega/2) \left(\int_0^\infty J_0(t) \cos(\omega t)\,dt+i\int_0^\infty J_0(t) \sin(\omega t)\,dt\right)\\\\
&=2\text{rect}(\omega/2) \left(\frac12\int_{-\infty}^\infty J_0(t) e^{i\omega t}\,dt+i\int_0^\infty J_0(t) \sin(\omega t)\,dt\right)\tag3
\end{align}$$
In THIS ANSWER, I showed that the Fourier Transform of $J_0(t)$ is equal to $\mathscr{F}\{J_0\}(\omega)=\frac{2\text{rect}(\omega/2)}{\sqrt{1-\omega^2}}$ while using the result found HERE, we see that $\int_0^\infty J_0(t)\sin(\omega t)\,dt=0$ when $|\omega|<1$.
Using these results in $(3)$ reveals
$$\begin{align}
\mathscr{F}\{J\}(\omega)&= \frac{2\text{rect}(\omega/2)}{\sqrt{1-\omega^2}}\tag4
\end{align}$$
Since the inverse Fourier Transform of the right-hand side of $(4)$ is $J_0(t)$ and $J_0(t)=J_0(-t)=J_0(x)$, we are done!.
