
#1
April 20th, 2019,
11:31
Problem:
Prove that for any $x \in R^n$ and any $0<p<\infty$
$\int_{S^{n1}} \rvert \xi \cdot x \rvert^p d\sigma(\xi) = \rvert x \rvert^p \int_{S^{n1}} \rvert \xi_1 \rvert^p d\sigma(\xi)$,
where $\xi \cdot x = \xi_1 x_1 + ... + \xi_n x_n$ is the inner product in $R^n$.
Some thinking...
I believe I'd like to define a function $f$ on $R^n$ so that I can utilize the formula...
$\int_{R^n} f(x) dx = \int_0^{\infty} r^{n1} (\int_{S^{n1}} f(r \theta ) d \sigma (\theta) ) dr $
If that's right... what would the function be? Maybe $f: R^n \rightarrow R, x \mapsto x_1 + x_2 + ...$?
Last edited by joypav; April 20th, 2019 at 17:51.

April 20th, 2019 11:31
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