MHB Integral Over Unit Sphere of Inner Product

[joypav]

Problem:

Prove that for any $x \in R^n$ and any $0<p<\infty$

$\int_{S^{n-1}} \rvert \xi \cdot x \rvert^p d\sigma(\xi) = \rvert x \rvert^p \int_{S^{n-1}} \rvert \xi_1 \rvert^p d\sigma(\xi)$,

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^{n-1} (\int_{S^{n-1}} f(r \theta ) d \sigma (\theta) ) dr $

If that's right... what would the function be? Maybe $f: R^n \rightarrow R, x \mapsto x_1 + x_2 + ...$?

 

[Jhoneine]

Solution:We will use the following formula:$\int_{R^n} f(x) dx = \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} f(r \theta ) d \sigma (\theta) ) dr $Let $f: R^n \rightarrow R, x \mapsto \lvert \xi \cdot x \rvert^p$, where $\xi \cdot x = \xi_1 x_1 + ... + \xi_n x_n$ is the inner product in $R^n$. Then we have:$\int_{R^n} \lvert \xi \cdot x \rvert^p dx = \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} \lvert \xi \cdot x \rvert^p d \sigma (\theta) ) dr $ $= \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} \lvert \xi_1 x_1 + ... + \xi_n x_n \rvert^p d \sigma (\theta) ) dr $ $= \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} \lvert \xi_1 \rvert^p \lvert x_1 \rvert^p + ... + \xi_n \rvert x_n \rvert^p d \sigma (\theta) ) dr $ $= \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} \lvert \xi_1 \rvert^p \lvert x_1 \rvert^p d \sigma (\theta) ) dr $ $= \lvert x \rvert^p \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} \lvert \xi_1 \rvert^p d \sigma (\theta) )