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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.

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 $$U(\psi,\theta,\phi)=\exp(-i(\psi/2)\sigma_1)\exp(-i(\theta/2)\sigma_2)\exp(-i(\phi/2)\sigma_3,$$ $$\int_{\operatorname{SU}(2)}f(U)\,dU=\frac{1}{16\pi^2}\int_0^{2\pi}d\psi\int_0^\pi\sin\theta d\theta\int_0^{4\pi}d\phi\, f[U(\psi,\theta,\phi)].$$

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, $\lvert j_{2}-j_{3}\rvert\leq j_{1}\leq j_{2}+j_{3}$.

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.

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.

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.

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 minus sign. 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, so this answer seems to be correct.

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.