I have the following problem: I have two d-dimensional integrals, where the integrand has the following dependence:
$$\int\mathrm{d}^{d}x\mathrm{d}^{d}y\,f(\vert\vec{x}\vert,\vert\vec{y}\vert,\vert\vec{x}+\vec{y}\vert)$$
The term $\vert\vec{x}+\vec{y}\vert$ can be written as
$$\vert\vec{x}+\vec{y}\vert=\sqrt{\vert\vec{x}\vert^{2}+\vert\vec{y}\vert^{2}+2\vert\vec{x}\vert\vert\vec{y}\vert\cos{(\Theta_{12})}},$$
where $\Theta_{12}$ denotes the relative angle between $\vec{x}$ and $\vec{y}$.
If we are working in $d=3$ dimensions, it is quite clear what to do. We can choose w.l.o.g. the $\vec{x}$ direction as our integration axis and then:
$$\mathrm{d}^{3}x=\vert\vec{x}\vert^{2}\mathrm{d}\vert\vec{x}\vert\mathrm{d}\Omega_2$$ $$\mathrm{d}^{3}y=\vert\vec{y}\vert^{2}\mathrm{d}\vert\vec{y}\vert\mathrm{d}\cos(\Theta_{12})\mathrm{d}\Omega_1$$
where $\mathrm{d}\Omega_{d}$ denotes the $d$ dimensional sphere-surface element. Integrating over $\mathrm{d}\Omega_{1,2}$ we get
$$\int\mathrm{d}^{3}x\mathrm{d}^{3}y\,f(\vert\vec{x}\vert,\vert\vec{y}\vert,\vert\vec{x}+\vec{y}\vert)=8\pi^{2}\int\vert\vec{x}\vert^{2}\vert\vec{y}\vert^{2}\mathrm{d}\vert\vec{x}\vert\mathrm{d}\vert\vec{y}\vert\int_{-1}^{1}\mathrm{d}\cos(\Theta_{12})f(\vert\vec{x}\vert,\vert\vec{y}\vert,\cos(\Theta_{12}))$$
But how can I generalize this to $d$-dimensions? Using the same idea we can write for the $x$-integration
$$d^{d}x=\vert\vec{x}\vert^{d-1}\mathrm{d}\vert\vec{x}\vert\mathrm{d}\Omega_{d-1}\rightarrow\vert\vec{x}\vert^{d-1}\mathrm{d}\vert\vec{x}\vert\frac{2\pi^{d/2}}{\Gamma(d/2)}$$
But how can I write the second integration? How can I "factorize" the relative angle $\Theta_{12}$ out of $\mathrm{d}\Omega_{d-1}$?
Thanks in advance!
EDIT: I think I have an idea, but I am not sure if this is right. The $n$-dimensional volume element of a sphere is given by
$${\displaystyle d_{S^{n-1}}V=R^{n-1}\sin ^{n-2}(\varphi _{1})\sin ^{n-3}(\varphi _{2})\cdots \sin(\varphi _{n-2})\,d\varphi _{1}\,d\varphi _{2}\cdots d\varphi _{n-1}.}$$
When I compare this with my formula above for $3$ dimemsion, I see that I can identify $\varphi_{1}$ with the relative angle $\Theta_{12}$. Therefore I can write:
$$d^{d}x=\vert\vec{x}\vert^{d-1}\mathrm{d}\vert\vec{x}\vert\sin^{d-2}(\Theta_{12})\mathrm{d}\Theta_{12}\Omega_{d-2}=\vert\vec{x}\vert^{d-1}\mathrm{d}\vert\vec{x}\vert(1-\cos^{2}(\Theta_{12}))^{(d-2)/2)}\mathrm{d}\Theta_{12}\Omega_{d-2}$$
Does this look right?