TITLE: Transformation to unit sphere
QUESTION [2 upvotes]: I read a book on harmonic analysis where there is the following transformation, which I don't understand:
$$\lim\limits_{\varepsilon\to 0}\int_{\varepsilon <|y|<1}\frac{\Omega(y')}{|y|^n}\mathrm{d}y=\lim\limits_{\varepsilon\to 0}\int_{{S^{n-1}}}\Omega(y')\mathrm{d}\sigma(y')
\cdot \log(1/\varepsilon)$$
Where $\sigma$ is the Lebesgue measure on the unit sphere and $y':=y/|y|$ and $\Omega$ defined on the unit sphere in $\mathbb{R}^n$ is integrable.
I tried to work with the transformation theorem and the transformation function  $\phi:x\mapsto x^n/|y|^n$. What confuses me is what is happening with $y'$ and why there is a $\log(1/\epsilon)$. If someone could explain me this step, it would be greatly appreciated.

REPLY [1 votes]: Use the parametric representation
$$(r,y')\mapsto y:=r\>y'\qquad(\epsilon\leq r\leq1, \quad y'\in S^{n-1})$$
of the given domain of integration. Then $|y|=r$ and $${\rm d}(y)=r^{n-1}{\rm d}\omega(y')\>dr\>,$$ whereby ${\rm d}\omega$ denotes the surface element on $S^{n-1}$. It follows that
$$\int_{\epsilon\leq|y|\leq 1}{\Omega(y')\over|y|^n}\>{\rm d}(y)=\int_\epsilon^1\int_{S^{n-1}}\Omega(y')\>{\rm d}\omega(y'){1\over r}\>dr=\log{1\over\epsilon}\int_{S^{n-1}}\Omega(y')\>{\rm d}\omega(y')\ .$$
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