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) )

 

[math771]

Your thinking is on the right track! The function $f$ that you proposed, $f(x) = x_1 + x_2 + ...$, is a good choice. This is because when we integrate over the unit sphere $S^{n-1}$, we are essentially integrating over all possible directions of the vector $x$. Since the inner product $\xi \cdot x$ is just a projection of $x$ onto the direction of $\xi$, we can think of $f(x)$ as measuring the length of the projection of $x$ in all possible directions.

To prove the given equation, we can first rewrite the left side using the definition of the inner product:

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

Then, using the formula you mentioned, we can rewrite this as:

$\int_{R^n} f(\xi \cdot x) d\xi = \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} f(r \theta ) d \sigma (\theta) ) dr$

Plugging in our function $f(x) = x_1 + x_2 + ...$, we get:

$\int_{R^n} (\xi \cdot x) d\xi = \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} (r \theta_1 + r \theta_2 + ...) d \sigma (\theta) ) dr$

Now, we can use the linearity of integration to split up the integral over $S^{n-1}$ into individual integrals over each component of $\theta$:

$\int_{R^n} (\xi \cdot x) d\xi = \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} r \theta_1 d \sigma (\theta) + \int_{S^{n-1}} r \theta_2 d \sigma (\theta) + ...) dr$

But each of these individual integrals is just the integral of a constant function over the unit sphere, which we know from basic geometry is equal to the surface area of the unit sphere, $S^{