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Jul 27, 2018 at 8:44 history edited user70925 CC BY-SA 4.0
Restrict to a compact set to make the integral definite.
Jul 27, 2018 at 8:41 comment added user70925 Hum you are right, I'll edit that to restrict to a compact subset of $\mathbb{R}^n$ then (let's say a ball).
Jul 25, 2018 at 20:22 comment added Christian Remling I don't think this can be finite for $k\ge 2$. For example if $k=2$, then for any fixed $v\in\mathbb R^3$ (say), the set of $w$ with $\|v\times w\|\le 1$ contains a tube with axis the direction of $v$, so has infinite measure.
Jul 25, 2018 at 11:43 history asked user70925 CC BY-SA 4.0