For $s<0, k\geq 1$ and $S(k):=\bar{B}(0;k+1)\backslash B(0;k)$ I need to show that
$$ \int_{S(k)} (1+||x||)^s dx \leq nV_n \int_k^{k+1}t^st^{n-1}dt, $$
where $V_n = \text{vol}_n(\bar{B}(0;1))$. I also know that $\text{vol}_n(rK)=r^n\text{vol}_n(K)$ for any compact Jordan measurable set $K$ in $\mathbb{R}^n$.
Estimates I got so far:
For any $k\geq 0$ we have (but remember I want to find the estimate above for $k\geq 1$) $$ \int_{S(k)} (1+||X||)^s dx \leq V_n(1+k)^s[(1+k)^n-k^n]. $$
For $k\geq 1$ I found
$$ \int_{S(k)} (1+||X||)^s dx < \int_{S(k)}||x||^sdx, $$
and also
$$ \int_{S(k)} (1+||X||)^s dx = \int_{\bar{B}(0;k+1)}(1+||x||)^sdx - \int_{B(0;k)}(1+||x||)^sdx. $$
I also found that for all $x\in S(k)$, we have $k\leq ||x|| \leq k+1$, which sounds good, since these are precisely the integration limits.
SPECULATION BELOW:
Maybe I could use the relation between the surface area and volume of an n-ball? Then $A_{n−1}(k)=\frac{d}{dk}V_n(k)=nt^{n−1}V_n$, which is what I want to get. But how do I make this mathematically rigorous? I need to formally create a change of variables?
Next I think I have to use the change of variables theorem (from wikipedia):
$$ \int_{\varphi(U)} f(\mathbf{v})\, d\mathbf{v} = \int_U f(\varphi(\mathbf{u})) \left|\det(D\varphi)(\mathbf{u})\right| \,d\mathbf{u}. $$
I think we can write both integrals in the difference above as something like this (in which we obviously miss some terms in the desired estimate), using a change of variables. I cannot come up with a good function tho, so this is just hand wavering and I know I haven't applied the change of variables theorem correctly just yet:
$\int_0^{k+1} t^s dt - \int_0^k t^s dt = \int_k^{k+1} t^s dt$.
If this is the correct approach, hopefully the determinant of the total derivative of the function used in the change of variables theorem has determinant with $nV_nt^{n-1}$, which also seems likely, since $nV_nt^{n-1}$ is the derivative of $V_nt^n$ with respect to $t$.