My professor during the lesson said that the following expression is the 3D delta function written in spherical coordinates: $$\sum_{l=0}^{\infty}\sum_{m=-l}^l \frac{\delta(p'-p)}{p^2}Y_{lm}(\alpha',\beta')Y_{lm}^{\star}(\alpha,\beta) = \delta{}^{(3)}(\vec{p'}-\vec{p})$$
where $(\alpha,\beta),(\alpha',\beta')$ are the polar and azimuthal angle of $\vec{p}$ and $\vec{p}'$ and $Y_{lm}$ are spherical harmonics. Now, he didn't give any explanation (and I'm unable to prove it by myself), but if it is easy to prove, then it would be interesting to see the proof.
Thank you
Let $d_L(p,\alpha,\beta|p',\alpha',\beta')=\sum_{\ell=0}^L \sum_{m=-\ell}^\ell Y_{\ell m}^*(\alpha',\beta')Y_{\ell m}(\alpha,\beta)\frac{\delta(p-p')}{p'^2}$ and let $\phi(p,\alpha,\beta)$ be a suitable test function. Then, we have
$$\begin{align} \lim_{L\to\infty}\langle d_L,\phi)&=\lim_{L\to\infty}\int_0^\infty \int_0^{2\pi}\int_0^\pi d_L(p,\alpha,\beta|p',\alpha',\beta')\phi(p',\alpha',\beta')\,p'^2\sin(\beta')\,dp'\,d\beta'\,d\alpha'\\\\ &=\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell Y_{\ell m}(\alpha,\beta)\underbrace{\int_0^{2\pi}\int_0^\pi Y_{\ell m}^*(\alpha',\beta') \phi(p,\alpha',\beta')\,\sin(\beta')\,d\beta'\,d\alpha'}_{\text{Spherical Harmonic Expansion Coefficients of }\phi}\tag1\\\\ &=\phi(p,\alpha,\beta) \end{align}$$
Hence as a distribution, $\lim_{L\to\infty}d_L(p,\alpha,\beta|p',\alpha',\beta')=\frac{\delta(p-p')}{p'^2}\frac{\delta(\beta-\beta')}{\sin(\beta')}\delta(\alpha-\alpha')$ as was to be shown!
NOTE: HERE is a reference to recognizing $(1)$ as the expansion of $\phi$ in terms of sperical harmonic basis functions.