For what values of $\alpha$ is $x_1\lvert x \rvert^{-\alpha}$ in $L^p$?

Question

Suppose $U=B(0,1)$ is the open ball in $n$ dimensions. Then for what values of $\alpha$ is $x_1\lvert x \rvert^{-\alpha}$ in $L^p$?. Here $\alpha$ will be given in terms of $p$ and $n$.

I'm not sure how to calculate:

$$\int_{B(0,1)} \lvert x_1\rvert^p \lvert x \rvert ^{-p\alpha} \ dx.$$ I tried to use the polar coordinates integration formula to transform the above into:

$$\int_{0}^1 \int_{\partial B} \lvert x_1\rvert^p\lvert x \rvert^{-p\alpha} dS dr$$

However, I am not clear on how to proceed here. Wont the value of $\lvert x_1\rvert^p \lvert x \rvert^{-p\alpha}$ on the boundary of the unit ball be equal to $1$?

Answer 1

Yes, the function is always equal to $1$ on the boundary but the boundary is not the problem; the problem is the origin where $1/\lvert x \rvert^\alpha$ has a singularity (when $\alpha > 0$). From the bound $\lvert x_1 \rvert \le \lvert x \rvert$, we have $$\int_{B(0,1)} \lvert x_1 \rvert^p \lvert x \rvert^{-p\alpha} dx \le \int_{B(0,1)} \lvert x\rvert^{p(1-\alpha)} dx.$$ Using polar coordinates (if we are in $\mathbb R^n$), we see $$\int_{B(0,1)} \lvert x\rvert^{p(1-\alpha)} dx = \omega_n \int_0^1 r^{p(1-\alpha)} r^{n-1}dr = \omega_d \int^1_0 r^{p(1-\alpha)+n-1} dr$$ which converges iff $p(1-\alpha)+n-1 > -1$. Rearranging gives $$\alpha < \frac{n +p}{p}.$$ Thus by comparison, we certainly have convergence for these $\alpha$.

Your mistake is that $$\int_{B(0,1)} \cdots dx \neq \int_0^1 \int_{\partial B(0,1)} \cdots dS dr$$ rather $$\int_{B(0,1)} \cdots dx =\int_0^1 \int_{\partial B(0,r)} \cdots dS dr.$$ So now $\lvert x \rvert = r$ on the surface of the ball which you are integrating on.