# Find the values of $p\ge 1$ such that $|x|^{-\frac{4}{5}p}(1+|x|^3)$ is integrable on unit ball and complement of unit ball, respectively

Consider the function $$w:\mathbb{R}^4\setminus\{0\}\to \mathbb{R}, w(x)=|x|^{-\frac{4}{5}p}(1+|x|^3)$$. Firstly, I am asked to find the values of $$p\ge 1$$ such that $$w\in L^1(B_1(0))$$ and then I am asked to find the values of $$p\ge 1$$ such that $$w\in L^1(\mathbb{R}^4\setminus\overline{B_1(0)})$$, where in both cases $$B_1(0)$$ stand for the open unit ball in $$\mathbb{R}^4$$.

According to what I learned in the course where I found this problem, I have to use the coarea formula. To be clear, I have to use the fact that $$\int_{B_{R}(0)}f dx=\int_0^R\left(\int_{\partial B_r(0)}fd\sigma \right)dr$$ for an integrable function $$f$$ (I think that it also works if $$f$$ is not integrable, but only positive).

So, off we go. According to the coarea formula, $$\int_{B_1(0)}|w(x)|dx=\int_0^1\left(\int_{\partial B_r(0)}|x|^{-\frac{4}{5}p}(1+|x|^3)d\sigma(x) \right)dr=\int_0^1r^{-\frac{4}{5}p}(1+r^3)|\partial B_r(0)|dr,$$ where I used that $$w$$ is a radial function. So, if we denote by $$\omega_4$$ the area of the unit sphere in $$\mathbb{R}^4$$, we have that $$|\partial B_r(0)|=\omega_4\cdot r^3$$, so we want $$\int_0^1 r^{-\frac{4}{5}p+3}(1+r^3)dr<\infty\iff \int_0^1 (r^{4-\frac{4}{5}p}+r^{6-\frac{4}{5}p})dr<\infty \iff p<\frac{25}{4}.$$ Thus, $$1\le p <\frac{25}{4}$$ is the answer to the first part of the problem.

Now, I am a bit unsure about the second part. I don't know if by the coarea formula I am allowed to write $$\int_{\mathbb{R}^4\setminus B_1(0)}|w(x)|dx=\int_1^\infty\left(\int_{\partial B_r(0)}|w|d\sigma \right)dr$$ and then of course proceed as above. I sure know that I can write $$\int_{\mathbb{R}^4}|w(x)|dx=\int_0^\infty \left(\int_{\partial B_r(0)}|w|d\sigma \right)dr,$$ but I don't know about the other one.

For the first question, your answer is wrong because you added a superfluous factor of $$r$$ (the exponent went from 3 to 4). The correct result is that $$1\le p< 5$$. For the second question, just use the formula you know applied to $$w(x)\,\cdot1_{|x|\ge 1}$$
• thank you! Where did the exponent go from $3$ to $4$? Not that it matters, I was kind of sure that I messed up the computations. And for the second question: that's a really nice idea! Just to be sure, applying my formula to $w(x)\cdot 1_{|x|\ge 1}$ gives me precisely what I wanted, i.e. that $\int_{\mathbb{R}^4\setminus B_1(0)}|w(x)|dx=\int_1^\infty\left(\int_{\partial B_r(0)}|w|d\sigma \right)dr$, right? Jan 27, 2023 at 22:14