Disclaimer: this is my first question here, english is not my first language and i'm a physicist, so sorry in advance for any mistakes!
I'm dealing with the orthogonality relations of 3-dimensional basis $e^{i\vec k \cdot\vec x}$ and Bessel functions $j_0(kr)=\frac{\sin(kr)}{kr}$, where $k=|\vec k|$ and $r=|\vec x |$.
I know that:
- \begin{equation} 4\pi \int_0^\infty r^2 dr \frac{\sin (kr)}{kr}\frac{\sin(pr)}{pr}=\frac{ 2\pi^2}{k^2}\delta(k-p),\end{equation}
- \begin{equation}\int_{\rm I\!R^3}d\vec{x}e^{-i\vec{k} \cdot\vec{x}}e^{i\vec{p}\cdot\vec{x}}=(2\pi)^3\delta(\vec{k}-\vec{p}) \end{equation} and, with test function such as $f(\vec k)=f(k)$,
- \begin{equation} \delta(\vec{k}-\vec{p})=\frac{1}{4\pi k^2}\delta(k-p) ,\end{equation} so I've concluded that (1) is equal to (2). I want to demonstrate that the integral in (1) is the integral in (2) passing in spherical coordinates, but when I change the coordinates in (2) i get a wrong (?) result. Here's my attempt: \begin{equation} \int_{\rm I\!R^3}d\vec{x}e^{-i\vec{k}\cdot\vec{x}}e^{i\vec{p}\cdot\vec{x}}=\int_{\rm I\!R^3}d\vec{x}e^{-i(\vec{k}-\vec{p})\cdot\vec{x}}=2\pi\int_0^\infty r^2 dr \int_0^\pi \sin\theta d\theta e^{-i|\vec{k}-\vec{p}|r\cos\theta}=\\ =\frac{4\pi}{|\vec{k}-\vec{p}|}\int_0^\infty dr\, r \sin(|\vec{k}-\vec{p}|r). \end{equation} I don't know how to continue, but still there seems to be something wrong. Can anyone help me to demonstrate this?