# Compact Formula for Wigner D-matrix

I have done an elementary calculation that appears to be giving me a simple (intuitive) formula for an arbitrary Wigner D-matrix. I can't seem to find this formula anywhere.

In the following, $\mathscr{R}$ represents an active rotation, $\Omega$ is solid angle, and $|\Omega\rangle$ are an orthonormal basis.

\begin{align} D^{l}_{m,m'}(\mathscr{R})&=\langle l,m'|\mathscr{R}|l,m\rangle\\ &=\int d\Omega\,\langle l,m'|\mathscr{R}|\Omega\rangle\langle\Omega|l,m\rangle\\ &=\int d\Omega\,\langle l,m'|\mathscr{R}\Omega\rangle\langle\Omega|l,m\rangle\\ &=\int d\Omega\, \,Y_l^{m'*}(\mathscr{R}\Omega)\,Y_l^m(\Omega) \end{align}

So it's a convolution of spherical harmonics. Is this right? I know my calculation is a bit cavalier but they seem more-or-less sound.

The bigger question is why this is useful. Usually, an expression like your last integral would arise when you're trying to evaluate the spherical-harmonic modes of a function in a rotated coordinate system. But we already know how to rotate modes. If you're suggesting that this might be an easier way to actually calculate the $D$ matrices, or just a simpler way to define them, I'm skeptical. But it's certainly another valid way to think of them, if you like.
(*) My one quibble is that I'm not sure that the order of your $m, m'$ indices is standard. Then again, I've never seen much consistency at all about any conventions involved in rotations. I've tried to stick with your ordering as best I can. But if you see swapper indices, that's probably why.
Your expression relies on the relation $$D^\ell_{m0}(\alpha,\beta,0)= \sqrt{\frac{4\pi}{2\ell+1}} Y_\ell^{m*}(\beta,\alpha)$$ so that you get \begin{align} &\int \sin\beta d\beta\,d\alpha \frac{2\ell+1}{4\pi} D^\ell_{m'0}(\omega\cdot(\alpha,\beta,0))D^{\ell*}_{m0}(\alpha,\beta,0)\, ,\\ &=\sum_{k} \int \sin\beta d\beta\,d\alpha \frac{2\ell+1}{4\pi} D^\ell_{m'k}(\omega)D^\ell_{k0}(\alpha,\beta,0))D^{\ell*}_{m0}(\alpha,\beta,0)\, ,\\ &=\sum_{k}D^\ell_{m'k}(\omega)\frac{2\ell+1}{4\pi} \int \sin\beta d\beta\,d\alpha D^\ell_{k0}(\alpha,\beta,0))D^{\ell*}_{m0}(\alpha,\beta,0)\, \end{align} The integral is proportional to $\delta_{km}$ so the result is indeed $D^{\ell}_{m'k}(\omega)$.
Given there are closed form expressions for $D^{\ell}_{m'k}(\omega)$, and given it is defined quite intuitively by $$D^{\ell}_{m'k}(\omega) = \langle \ell m'\vert R(\omega)\vert \ell k\rangle$$ I'm not sure what is gained by expressing this as a convolution.