Formula involving Wigner's 3j symbols and integration over irreducible representations of SU(2) In some calculations, I saw the following formula
$$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,D^{j_{1}}_{m_{1}n_{1}}(g)D^{j_{2}}_{m_{2}n_{2}}(g)D^{j_{3}}_{m_{3}n_{3}}(g)=(-1)^{j_{1}+j_{2}+j_{3}}\begin{pmatrix}j_{1} & j_{2} & j_{3}\\m_{1} & m_{2} & m_{3}\end{pmatrix}\begin{pmatrix}j_{1} & j_{2} & j_{3}\\n_{1} & n_{2} & n_{3}\end{pmatrix}$$
where $\mathrm{d}g$ denotes the (normalized) Haar measure on $SU(2)$and where $D^{j}$ denote Wigner's D-matrices, i.e. the unitary irreducible representation of $SU(2)$ of dimension $2j+1$. The matrix-looking objects on the right hand side denote Wigner's 3j symbols, which are related to the Clebsch-Gordon coefficients.
The formula appears in the mathematical derivation of the so-called "Ponzano-Regge model" of 3d Euclidean Quantum Gravity. I am wondering how this formula is derived. Furthermore, I have seen in different sources that sometimes the sign factor in front is missing. Another source is mentioning that this formula is only true if $\vert j_{2}-j_{3}\vert\leq j_{1}\leq j_{2}+j_{3}$. So all in all I become very confused and would like to know whether this formula is correct and which assumptions are needed. Does maybe someone also have some reference on it?
 A: You can find a fully worked-out derivation in these lecture notes. The formula you are looking for is equation (404), written in terms of the Wigner (small)-$d$ matrix. The relationship to the (large)-$D$ matrix goes via the Euler angle parameterization,
$$D^{j}_{mm'}(\psi,\theta,\phi)=e^{-im\psi-im'\phi}d^{j}_{mm'}(\theta).$$
The integration over SU(2) with the Haar measure is given in terms of Euler angles by
$$U(\psi,\theta,\phi)=\exp(-i(\psi/2)\sigma_1)\exp(-i(\theta/2)\sigma_2)\exp(-i(\phi/2)\sigma_3,$$
$$\int_{\rm{SU}(2)}f(U)\,dU=\frac{1}{16\pi^2}\int_0^{2\pi}d\psi\int_0^\pi\sin\theta d\theta\int_0^{4\pi}d\phi\, f[U(\psi,\theta,\phi)].$$
So the desired integral over a product of three $D$-matrices vanishes unless $m_1+m_2+m_3=0=n_1+n_2+n_3$. In that case the integrations over $\psi$ and $\phi$ give a factor $8\pi$, what remains is the integration over $\theta$. Eq. (404) in the lecture notes shows how that is related to the product of $3j$-symbols,
$$\frac{1}{2}\int_0^\pi d\theta\,d^{j_1}_{m_1n_1}d^{j_2}_{m_2n_2}d^{j_3}_{m_3n_3}=\begin{pmatrix}j_{1} & j_{2} & j_{3}\\m_{1} & m_{2} & m_{3}\end{pmatrix}\begin{pmatrix}j_{1} & j_{2} & j_{3}\\n_{1} & n_{2} & n_{3}\end{pmatrix}.$$
Both sides of the equation are equal to zero unless the $j$'s satisfy the triangle inequality, $\vert j_{2}-j_{3}\vert\leq j_{1}\leq j_{2}+j_{3}$.
This formula differs from the one in the OP by a factor $(-1)^{j_1+j_2+j_3}$. As a test, I evaluated both sides of the equation with Mathematica, for $j_1=2$, $j_3=3$, $j_4=4$, $m_1=n_1=1$, $m_2=n_2=-1$, $m_3=n_3=0$, and find $5/126$, without a minus sign.

(1/2)*Integrate[
  WignerD[{2, 1, 1}, theta]*WignerD[{3, -1, -1}, theta]*
   WignerD[{4, 0, 0}, theta]*Sin[theta], {theta, 0, Pi}]    
ThreeJSymbol[{2, 1}, {3, -1}, {4, 0}]*
ThreeJSymbol[{2, 1}, {3, -1}, {4, 0}]

