Bound this integral: $\int_{B_x\cap(\mathbb R^d\backslash B_0)}\frac{1}{|x-y|^{d+a}}dy\leq \frac{C}{\operatorname{dist}(x,\partial B_0)}$ For $x\in\mathbb R^d$, let $B_x$ be the open ball with radius 1 with center $x$. For all $x\in B_0$ and all $y\in \mathbb R^d\backslash B_0$, we have $$\operatorname{dist}(x,\partial B_0)\leq|x-y|.$$  where $\operatorname{dist}(x,\partial B_0)$ denotes the distance from from $x$ to $\partial B_0$.

Show that for all $a<1$ there exists $C\geq 0$ with $$I:=\int\limits_{B_x\cap(\mathbb R^d\backslash B_0)}\frac{1}{|x-y|^{d+a}}dy\leq \frac{C}{\operatorname{dist}(x,\partial B_0)}$$ for all $x\in B_0$.

I do not know how to show this - I started with
$$I\leq \int\limits_{B_x\cap(\mathbb R^d\backslash B_0)}\frac{1}{\operatorname{dist}(x,\partial B_0)^{d+a}}dy$$
but I don't know how to continue.
 A: Assume $|x| < 1$. Clearly ${\mathrm{dist}}(x,\partial B_0) = 1-|x|$.
Note that if $y \in B_x \cap ({\mathbf R^d} \setminus B_0)$ then $|x-y| < 1$ and $|y| \ge 1$.  In particular $|x-y| \ge |y| - |x| \ge 1 - |x|$, so that
$$B_x \cap ({\mathbf R^d} \setminus B_0) \subset \{y : 1-|x| \le |x-y| < 1\}$$
and thus
$$\int_{B_x \cap ({\mathbf R^d} \setminus B_0)} \frac 1{|x-y|^{d+a}} \, dy \le \int_{1-|x| \le |x-y| < 1} \frac 1{|x-y|^{d+a}} \, dy$$
The latter integral is treated easily with spherical coordinates and (since $a < 1$) there is a constant $C_a$ with
$$\int_{1-|x| \le |x-y| < 1} \frac 1{|x-y|^{d+a}} \, dy = C_a [1 - (1-|x|)^{1-a}].$$
Thus your question is whether or not there is a constant $C$ independent of $x$ satisfying $$C_a [1 - (1-|x|)^{1-a}] \le \frac{C}{1-|x|}$$ or equivalently, whether $$1-|x| - (1-|x|)^{2-a}$$ is bounded on $B_0$. An obvious bound is $1$.
A: Let $j_0\ge 0$ be such that $2^{-j_0}\le d(x,\partial B_0) < 2^{-j_0+1}$, and for each $0\le j \le j_0$, consider the set $E_j = \{y:2^{-j}\le|x-y|<2^{-j+1}\}$. As the $\{E_j\}_{0\le j\le j_0}$ are disjoint and cover $B_x\setminus B_0$, we can bound the integral by
\begin{align*}
I&\le \sum_{j=0}^{j_0}\int_{E_j}|x-y|^{-(d+a)}\,dy\\
&\le\sum_{j=0}^{j_0}\int_{E_j}2^{j(d+a)}\,dy\\
&=\sum_{j=0}^{j_0}2^{j(d+a)}|E_j|.
\end{align*}
For each $j$, $|E_j|\le c_d2^{-jd}$, so
$$
I\le c_d\sum_{j=0}^{j_0}2^{ja} \lesssim 2^{j_0a}\sim \frac{1}{d(x,\partial B_0)^a},
$$
which is slightly better than the advertised inequality, since $d(x,\partial B_0)^a \ge d(x,\partial B_0)$ for $a \le 1$.

For two nonnegative quantities $X$ and $Y$, $X\lesssim Y$ means $X\le CY$ for some absolute constant $C$, and $X\sim Y$ means both $X\lesssim Y$, and $Y\lesssim X$. For a set $E$, $|E|$ denotes the Lebesgue measure of $E$.
A: Let's introduce a little notation. Let $n\in\mathbb N$ be fixed and let
$$U( x):= \mathbb B(x,1)\cap \big(\mathbb R^n\setminus \mathbb B(0,1)\big)$$
We would like to show that, $\forall p\in(0,1)$, $\exists C(p)>0$ such that
$$\Phi(x,p):=\int\limits_{U(x)}\frac{1}{|x-x'|^{n+p}}\mathrm d^nx'\leq \frac{C(p)}{\operatorname{dist}\big(x, \partial \mathbb B(0,1)\big)}\tag{*}$$
Where
$$\operatorname{dist}(x,S):=\inf_{s\in S}|x-s|$$
NB: It is important to require $x\in\mathbb B(0,1)$ to avoid singularities in the left hand side of $(*)$.
Also note: The inequality is trivial when $x=0$ since $\Phi(0,p)=0$.

First point: Since $x\in\mathbb B(0,1)$ then
$$\operatorname{dist}\big(x, \partial \mathbb B(0,1)\big)=1-|x|$$
We can also see readily that
$$U(x)\subset \{x'\in\mathbb R^n: 1<|x'|\leq 1+|x|\}\equiv \Omega(x)$$
See the below diagram:

And so
$$\Phi(x,p)\leq \int\limits_{\Omega(x)} \frac{1}{|x-x'|^{n+p}}\mathrm d^n x'$$
Converting to spherical coordinates, this is roughly
$$\int_{\text{angle ranges}}\int_{1}^{1+|x|}\frac{1}{r^{1+p}}\mathrm dr~\{\text{angular integrals}\}=K_n~\frac{1-(1+|x|)^{-p}}{p}$$
The precise value of $K_n$ is not important.
So, the problem boils down to, when $|x|<1$, can we find a $C(p)$ such that
$$K_n~\frac{1-(1+|x|)^{-p}}{p}\leq \frac{C(p)}{1-|x|}$$
Or
$$\frac{1}{p}\big(1-|x|\big)\big(1-(1+|x|)^{-p}\big)\leq\frac{C(p)}{K_n}$$
Or simpler, if the quantity
$$\big(1-|x|\big)\big(1-(1+|x|)^{-p}\big)$$
Is bounded for $|x|<1$. And it is - clearly $1$ is an upper bound.
