Transformation to unit sphere

Question

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.

Answer 1

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 $$\int_{\epsilon\leq|y|\leq 1}{\Omega(y')\over|y|^n}\>{\rm d}(y)=\int_\epsilon^1\int_{S^{n-1}}\Omega(y')\>{\rm d}\omega(y'){1\over r}\>dr=\log{1\over\epsilon}\int_{S^{n-1}}\Omega(y')\>{\rm d}\omega(y')\ .$$

Answer 2

Edit: Below I had the idea to use Fubuni's Theorem, but I think since the area we integrate over is not in the style of $X \times Y$, this is not possible. Instead we need to use polar coordinates. See Integration of radial functions? ... Note that $\frac{\Omega(y')}{|y|}$ is a radial function. So

$\int_{\varepsilon <|y|<1}\frac{\Omega(y')}{|y|^n}\mathrm{d}y=\\ \int_{\varepsilon}^1\frac{\Omega(y')}{r^n}r^{n-1}\mathrm{d}r\\ $

And then you are back at the calculation that gives you $log(1/\varepsilon)$

---

oldpost:

It think this makes sense if it's $|y|$ instead of $|y|^n$ on the left hand side. Let $\varepsilon > 0$ then: $\int_{\varepsilon <|y|<1}\frac{\Omega(y')}{|y|}\mathrm{d}y=\\ \int_{\varepsilon}^1\int_{|y|=r}\frac{\Omega(y')}{|y|}\mathrm{d}y\mathrm{d}r=\\ \int_{S^{n-1}}\Omega(y')\int_{\varepsilon}^1\frac{1}{r}\mathrm{d}r\mathrm{d}\sigma (y')=\\ \int_{{S^{n-1}}}\Omega(y')\mathrm{d}\sigma(y') \cdot (\log(1) - \log(\varepsilon)) = \\ \int_{{S^{n-1}}}\Omega(y')\mathrm{d}\sigma(y') \cdot \log(1/\varepsilon) $

Is that calculation somewhat clear?