Prove $ \int \frac{d^d k}{k^2} \propto \int k^{d-3} dk$ I have recently started wondering about the Mermin-Wagner theorem and the correction of the first-order leads to the following integral: $$ \int \frac{d^d k}{k^2}$$ And it is said in the Wikipedia article that it is proportional to: $$ \int k^{d-3} dk$$ Which proves the Mermin-Wagner theorem However, I did not understand which change of variable got applied to turn an integral of dimension $d$ into an integral of dimension $1$.

Spherical coordinates is used to express $d^d k = k^{d-1} dk\, d\sigma$, where $d\sigma$ is the surface element of the $(d-1)$-sphere. Then $$\int \frac{d^d k}{k^2} = \int d\sigma \int \frac{k^{d-1}\, dk}{k^2} = C_d \int k^{d-3}\, dk$$ where $C_d$ is the surface area of the $(d-1)$-sphere.