Please explain how this equality came
$$ \int_{|z| \le R} \frac{dz}{|z|^\alpha} = \int_{S^1} dS \int_0^R \frac{r^{m-1}}{r^{\alpha}} dr$$
I don't understand how degree $m-1$ came about. Sorry for my English
2026-05-14 03:51:45.1778730705
change of variables on integral
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Note that in the 2D case, you get $dz=rdrdt$ so the integral ends up $$ \int_0^{2\pi} dt \int_0^R \frac{rdr}{r^\alpha} $$ where $r = r^{2-1}$ in the numerator comes from taking the Jacobian from orthogonal to polar coordinates.
In 3D you will convert to spherical, etc. The $r^{m-1}$ is coming from the Jacobian.