Asymptotics of $\int_{B(0,1)}\frac{dy}{|x-y|^{n+\alpha}}$ as $|x|\to 1^+$ In a study of Fractional Laplacian, I encounter the integral
$$I(x):=\int_{B(0,1)}\frac{dy}{|x-y|^{n+\alpha}}$$ where

*

*$B(0,1)\subset\mathbb R^n$ is the $n$-dimensional unit ball centered at the origin

*$x\in\mathbb R^n$ does not depend on $y$, $|x|>1$

*$\alpha\in(0,1)$ is a constant.

I conjecture that $$I(x)\le C(|x|-1)^{-\alpha}\qquad (*)$$ as $|x|\to 1^+$, i.e. for $|x|>1$ sufficiently close to $1$, there exists a constant $C>0$ indepednent of $x$ such that the inequality $(*)$ holds.
Question: Is my conjecture true? If not, what is the asymptotics of $I(x)$ as $|x|\to 1^+$?
(NB: This conjecture arises in an attempt to prove Proposition 3.1 (whose proof is omitted) in the paper On the superharmonicity of the first eigenfunction of the
fractional Laplacian for certain exponents.)

Failed Attempt
Using the reverse triangle inequality, one sees that
$$\begin{align}
I(x)&=\int_{B(0,1)}\frac{dy}{|x-y|^{n+\alpha}} \\
&\le\int_{B(0,1)}\frac{dy}{||x|-|y||^{n+\alpha}} \\
&=n\cdot \text{vol}(B(0,1))\int_{0}^1\frac{r^{n-1}}{||x|-r|^{n+\alpha}} dr \qquad (1)\\
&=n\cdot \text{vol}(B(0,1))\int_{0}^1\frac{r^{n-1}}{(|x|-r)^{n+\alpha}} dr \\
&=n\cdot \text{vol}(B(0,1))\left(\frac{1}{(|x|-1)^{n+\alpha-1}}-\frac{n-1}{n+\alpha-1}\int_{0}^1\frac{r^{n-2}}{(|x|-r)^{n+\alpha-1}} dr\right) \qquad (2)\\
\end{align}$$
(1): Switching to spherical coordinates.
(2): Integrating by parts. The boundary term is $O((|x|-1)^{-n+1-\alpha})$; through integration by parts, the integral term can be shown to be $O((|x|-1)^{-n+2-\alpha})$. Thus, this bound only shows that $I(x)=O((|x|-1)^{-n+1-\alpha})$ which is weaker than the conjecture, unless $n=1$.
 A: Rotate to a suitable coordinate system such that $x$ only has a $y_n$ component and instead consider the equivalent integral with spherical coordinates centered at $x$
$$I(x)=\int_{B_1(x)}\frac{r^{n-1}\sin\theta}{r^{n+\alpha}}\:d\theta\:dr\:d\Omega$$
where $\theta$ is the "last" angular coordinate defined by $y_n = r\cos\theta$ and $d\Omega$ is the measure on $S^{n-2}$. The equation for the sphere in this coordinate system is given by
$$s^2+(y_n-x)^2 = 1$$
where $s^2+y_n^2 = r^2$. Then in our new translated spherical coordinates we have that
$$r^2-2xr\cos\theta +x^2 = 1 \implies \cos\theta = \frac{r^2+x^2-1}{2rx}$$
From here we can set up our integral
$$I(x) = \operatorname{vol}(S^{n-2})\cdot\int_{x-1}^{x+1}\int_0^{\cos^{-1}\left(\frac{r^2+x^2-1}{2rx}\right)}\frac{\sin\theta}{r^{1+\alpha}}\:d\theta\:dr = \operatorname{vol}(S^{n-2})\cdot\int_{x-1}^{x+1}\frac{1-(r-x)^2}{2xr^{2+\alpha}}dr$$
which, after some Mathematica (the integral is completely doable by hand though, but I don't trust myself with erroneous constants and minus signs) and reinserting understood norms, exactly evaluates to
$$I(x) =\frac{\operatorname{vol}(S^{n-2})}{2\alpha(1-\alpha^2)}\cdot\frac{1}{|x|}\cdot\left(\frac{|x|-\alpha}{(|x|-1)^\alpha}-\frac{|x|+\alpha}{(|x|+1)^\alpha}\right)$$
So your conjecture is indeed correct as when $|x|\to1^+$ the resulting expression tends to
$$\sim \frac{C_1}{(|x|-1)^\alpha}-C_2 \lesssim \frac{1}{(|x|-1)^{\alpha}}$$
for $C_1,C_2>0$.
A: Layer cake representation gives
\begin{align*}
I(x)=\int_{B(0,1)}\frac{1}{|x-y|^{n+\alpha}}\mathrm{d}y = (n+\alpha)\int_{0}^\infty s^{n+\alpha-1}A(s)\mathrm{d}s,
\end{align*}
where
$$
A(s)=m\left\{y\in B(0,1);|x-y|^{-1}>s \right\} = m\left(B(0,1) \cap B(x,1/s) \right).
$$
If $s>(|x|-1)^{-1}$, we have $A(s)=0$ so
$$
I(x)=(n+\alpha)\int_{0}^{(|x|-1)^{-1}}s^{n+\alpha-1}A(s)\mathrm{d}s.
$$
Bounding trivially $A(s) \leq m(B(x,s^{-1}))=c s^{-n}$, we find
\begin{align*}
I(x)&\leq c(n+\alpha)\int_{0}^{(|x|-1)^{-1}}s^{\alpha-1}\mathrm{d}s \\
&=\frac{c(n+\alpha)}{\alpha}(|x|-1)^{-\alpha},
\end{align*}
where $c=m(B(0,1))$, which is what you conjectured.
