I've found an article that essentially states that an integral of the form

$$ I[f] = \int_{\mathbb{R}^N} f(x_1,\ldots,x_N) dx_1 \ldots dx_N = \int_{\mathbb{R}^N} f(r) dx_1 \ldots dx_N $$

where $r = \sqrt{x_1^2 + \ldots x_N^2}$, can always be split as product of two integrals, where the first integral is the surface of the $n$ dimensional sphere, say $A_n$ and the second integral is the radial integral

$$ R[f] = \int_0^1 f(r) r^{N-1}dr $$

The article I've found only states the result, but it doesn't actually prove it, but assuming this results holds it proves that

$$ A_N = \frac{2 \pi^{N/2}}{\Gamma\left( \frac{N}{2}\right)} $$

Is there a way to prove the statement which doesn't rely on generalized polar coordinates, which are a bit hard to remember and manipulate, if there's no other way is there an easy way to derive to generalized polar coordinates and again derive the result?

I would assume the proof is a calculus fact.


Since $f$ is radially symmetric, $f(\vec{x})d^N\vec{x}=r^{N-1}f(r)drd\Omega$, with $d\Omega$ an $(N-1)$-dimensional infinitesimal element over the angle coordinate(s). We don't need to work out how $d\Omega$ looks at all, so don't get out your $\sin\theta$s here. The claimed factorisation is immediate.

If you prefer a different proof that an $(N-1)$-sphere of radius $R$ has measure $\frac{2\pi^{N/2}}{\Gamma(N/2)}R^{N-1}$, let's prove instead that an $N$-ball of radius $R$ has volume $\int_0^R\frac{2\pi^{N/2}}{\Gamma(N/2)}r^{N-1}dr=\frac{\pi^{N/2}}{\Gamma(N/2+1)}R^N$. (To clarify, an $n$-ball is $n$-dimensional, but an $n$-sphere is the $n$-dimensional surface of an $(n+1)$-ball.) Equivalently, the unit $N$-ball has measure $V_N:=\frac{\pi^{N/2}}{\Gamma(N/2+1)}$. We'll proceed by induction.

The result is correct for $N=0$; a "$0$-sphere" is the single point in $0$-dimensional space, and the definition of measure is counting that point. (If that argument seems to strange, take $N=1$ as your base step, for which measure is a line segment's width, or failing that $N=2$, for which you just want a circle's area.) By slicing an $N$-ball into hypercylinders of infinitesimal thickness with $(N-1)$-ball cross sections of radius $\sqrt{1-x^2}$,$$V_N=\int_{-1}^1V_{N-1}(1-x^2)^{(N-1)/2}dx=2V_{N-1}\int_0^1(1-x^2)^{(N-1)/2}dx\\=V_{N-1}\int_0^1y^{-1/2}(1-y)^{(N-1)/2}dx=V_{N-1}\operatorname{B}\left(\frac12,\,\frac{N+1}{2}\right).$$(The intuition of this is easiest with slicing a sphere into cylinders, since you can visualise that.) So the inductive step is$$V_{N-1}=\frac{\pi^{(N-1)/2}}{\Gamma((N+1)/2)}\implies V_N=\frac{\pi^{(N-1)/2}}{\Gamma((N+1)/2)}\frac{\sqrt{\pi}\Gamma((N+1)/2)}{\Gamma(N/2)+1},$$which is what we needed.

  • $\begingroup$ How did you get your very first equality? $\endgroup$ – user8469759 Oct 19 '19 at 18:22
  • $\begingroup$ @user8469759 Well, that's just the polar form of $d^N\vec{x}$. It needs to be proportional to $d(r^N)$, because an $N$-ball of radius $R$ wouldn't otherwise have measure $\propto R^N$. $\endgroup$ – J.G. Oct 19 '19 at 18:24
  • $\begingroup$ So you implicity applied a coordinate transformation am I right? $\endgroup$ – user8469759 Oct 19 '19 at 18:28
  • $\begingroup$ @user8469759 Yes, like when you write $f(x)dx=f(x)g^\prime(u)du$ if $x=g(u)$. $\endgroup$ – J.G. Oct 19 '19 at 18:30
  • $\begingroup$ Ok, but to prove that equality you want to know the coordinate transformation. Or proving the existance maybe. I wanted to avoid an explicit computation if possible. $\endgroup$ – user8469759 Oct 19 '19 at 21:38

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