When asking for validation on your answers, it is often good to give an explanation and a thought process so that others can more easily see how you found the answers that you did.

You can actually explicitly evaluate these integrals by changing to polar coordinates

@peek-a-boo Which integrals are you referring to?

@peek-a-boo is referring to the $L^p$ norm, wich is defined through an integral. And the answer to your question is no, this is not correct. The correct answer depends on $n$.

For the first case, I set up the integral: $\|f\|_{p}:=\left(\int_{B(0,1)}|x|^{-p} d x\right)^{\frac{1}{p}}$. How do I find p from this?

Suppose $0\leq a<b\leq \infty$ and we consider the annulus $E_{a,b}:=\{x\in\Bbb{R}^n\,:\, a<|x|<b\}$. Then, for any $\lambda\in\Bbb{R}$, we have $\int_{E_{a,b}}\frac{1}{|x|^{\lambda}}\,dx=\int_a^b\frac{1}{r^{\lambda}}A_{n-1}r^{n-1}\,dr=A_{n-1}\int_a^b\frac{1}{r^{\lambda+1-n}}\,dr$, where $A_{n-1}$ is the surface area of the unit sphere $S^{n-1}\subset\Bbb{R}^n$ (just FYI: the exact formula is $A_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$, but this is irrelevant for determining finiteness of the integrals). Just look at the different cases and ask yourself when this is finite.

@peek-a-boo I don't know when it is finite. How can I find p from this?

This is a basic definite integral to evaluate... so I think you need to try harder to attempt the problem yourself.

@peek-a-boo Taking the integral, I get $A_{n-1} \frac{1}{r^{\lambda+2-n}-(\lambda+1-n)}\bigr\vert_{a}^{b}$. How do I find p from here?

Well, that's the answer assuming $\lambda+1-n\neq 1$ (since $\int\frac{1}{r}=\log r$). Now, look at your question. WHat values of $a,b$ should you be choosing? For what values of $\lambda$ (i.e $p$) is the integral finite?

$\lambda$ should be positive for the integral to be finite?

@peek-a-boo We would choose b>a? So it is the same answer for all three cases?

@peek-a-boo I am still not sure where to go with this. How do I find what a and b I should be choosing? and p>0?

FOr the unit ball $X=B(0;1)=\{x\in\Bbb{R}^n\,:\, \|x\|<1\}$, there is a very clear choice of $a,b$ (Hint: $a=0$, $b=$____). Likewise for the complement $\Bbb{R}^n\setminus B(0;1)$, there is once again a very clear choice (different from the previous one). For $\Bbb{R}^n$, there is again a clear choice (different from the previous two). Compare with the definition of $E_{a,b}$ I gave previously (note that Lebesgue measure of single points and spheres is 0, so in the definition of $E_{a,b}$, weak/strict inequalities aren't really important).

@peek-a-boo For B(0,1), a = 0, b = 1. For $\mathbb{R}^{n} \backslash B(0 ; 1)$, a = 1, b = $\infty$, and for $\mathbb{R}^{n}$, a = 0, b = $\infty$. Is this correct?

yes that's right

Thank you. So the final answer is p=$\lambda$>0, $\left.A_{n-1} \frac{1}{r^{\lambda+2-n}-(\lambda+1-n)}\right|_{a} ^{b}$ (filling in a,b for the different cases)?

@peek-a-boo Is my condition p =$\lambda$> 0 correct?

no the answer depends on $n$, and the domain of integration.

So p can take any value given my integral above?

@peek-a-boo Please let me know! Want to have this problem finalized