Can the integral $\int_0^{\infty} \sin\big( [x-y]p \big) dp$ be related to a Dirac Delta function? One of the most famous integral representations of the Dirac Delta is:
$$
\delta(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i x p} dp
$$
The above can be expanded using $e^{ixp} = \cos(xp) + i \sin(xp)$ to get:
$$
\delta(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \cos(xp) dp + \frac{i}{2\pi} \int_{-\infty}^{\infty} \sin(xp) dp
$$
This seems to imply that:
$$
\delta(x) \ = \frac{1}{2\pi} \int_{-\infty}^{\infty} \cos(xp) dp \\
0 \ = \ \frac{1}{2\pi} \int_{-\infty}^{\infty} \sin(xp) dp
$$
The second of which is quite surprising to me! Is the statement $0 = \frac{1}{2\pi} \int_{-\infty}^{\infty} \sin(xp) dp$ true? Is the following statement true too?
$$
0 \ = \ \frac{1}{2\pi} \int_{0}^{\infty} \sin(xp) dp \ ?
$$
I would have guessed that an integral of the form $\int_{0}^{\infty} \sin(xp) dp$ would be proportional to two Dirac deltas, so I am just surprised.
(This question is related to this one of my previous questions, where I sought (and still seek) an expression in terms of Dirac-Deltas for the integral $\int_0^{\infty} \sin(xp) \cos(y p) dp$).
 A: There is a classical meaning for your $\delta$ statement found in the Fourier transform inversion. Assuming $f$ is 'nice' at $x$ with some global integrability condition, the following inversion will hold:
\begin{align}
         f(x) & = \lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-R}^{R}e^{isx}\int_{-\infty}^{\infty}e^{-isy}f(y)dy \\
   & = \lim_{R\rightarrow\infty}\int_{-\infty}^{\infty}\left(\frac{1}{2\pi}\int_{-R}^{R}e^{is(x-y)}ds\right) f(y)dy
\end{align}
This would typically be expressed in Physics as
$$
         \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{is(x-y)}dx=\delta(x-y),
$$
and it has meaning to the Mathematician in the first equation. It's the symmetric integral over $[-R,R]$ that is considered in Math. Because $\sin$ is odd,
$$
        \int_{-R}^{R}e^{is(x-y)}ds= \int_{-R}^{R}\cos(s(x-y))ds
$$
So, in your terminology,
\begin{align}
       \delta(x-y) & = \frac{1}{2\pi}\int_{-\infty}^{\infty}\cos(s(x-y))ds \\
     0 & = \frac{1}{2\pi}\int_{-\infty}^{\infty}\sin(s(x-y))ds.
\end{align}
This is confusing without the full details of the underlying symmetric limit.
