Is this function locally integrable ? Needs spherical coordinates! Let $ d\in \mathbb{N}$ be the dimension we consider. Let $\mu$ be a probability density on $\mathbb{R}^d$. Consider vector fields of the form $f:\mathbb{R}^d\to\mathbb{R}^d$
$$ f(x):=\frac{x}{\|x\|^k},~~~~k\leq d.$$
Define the convolution of $f$ and $\mu$ as
$$ f*\mu(x)=\int_{\mathbb{R}^d} f(x-y)\mu(y)dy. $$
Now is it possible to show $f*\mu\in L^1_{\text{loc}}(\mathbb{R
}^d)$ ?  Its funny because I can show $f\in L^1_{\text{loc}}(\mathbb{R}^d)$ Integrability / integrals of functions of the form $1/|x|$. Singularities  , but the non-local behaviour of the convolution "forces me to consider $f$ on non-compact sets".

Attempt : let $\Omega\subset \mathbb{R}^d$ be compact. Let $f_i*\mu$ be the i-th component of $f*\mu$.
$$ \|f_i*\mu\|_{L^1(\Omega)}\leq \int_{\Omega}\int_{\mathbb{R}^d}\Big|\frac{(x_i-y_i)}{\|x-y\|^k}\Big|\mu(y)dydx, $$
Now I want to make the substitution $y_j=x_j+\text{spherical coordinates}$ : https://de.wikipedia.org/wiki/Kugelkoordinaten#Verallgemeinerung_auf_n-dimensionale_Kugelkoordinaten  But since the $x$ coordinate can only be taken in the compact set $\Omega$ I'm stuck...
 A: I believe that you can just prove this directly. Let $K \subset \mathbb{R}^d$ be compact. Then for each $i=1,\dots,d$\begin{align*}
\int_K \vert f_i \ast \mu \vert dx & \leqslant \int_K \int_{\mathbb{R}^d} \frac{\vert x_i -y_i \vert}{\vert x-y \vert^k} \mu(y) dydx \\
&=  \int_{\mathbb{R}^d} \int_K\frac{\vert x_i -y_i \vert}{\vert x-y \vert^k} \mu(y) dxdy \\
&= \int_{\mathbb{R}^d}\mu(y) \int_K \frac{\vert x_i-y_i \vert}{\vert x-y \vert^k} d x dy
\end{align*} where I used Tonelli's Theorem to swap the integrals. Next, given some $y \in \mathbb{R}$, choose $R>0$ such that $\vert K \vert = \vert B_R(y) \vert = \vert B_1 \vert R^d$ (by $\vert A \vert $ I mean the Lebesgue measure of $A$). Note carefully that $R$ doesn't actually depend on $y$ since one could choose $R$ such that $\vert K \vert = \vert B_R(0) \vert =  \vert B_R(y) \vert $ for all $y \in \mathbb{R}$. Then  \begin{align*} 
\int_K \frac{\vert x_i-y_i \vert}{\vert x-y \vert^k} d x & \leqslant \int_{B_R(y)} \frac{\vert x_i-y_i \vert}{\vert x-y \vert^k} d x \\
& \leqslant \int_{B_R(y)} \vert x-y \vert^{1-k} d x \\
&= \int_0^R \int_{\partial B_r(y)} \vert x-y \vert^{1-k} d S dr\\
&= \frac{d\vert B_1 \vert}{d-k+1}  R^{n-k+1} \\
&= \frac{d\vert B_1 \vert^{1-\frac 1 d}}{d-k+1}  \vert K \vert^{\frac{n-k+1}{d}} =: C_0
\end{align*} via polar coordinates. The first inequality holds since the integrand is monotone decreasing as we move radially out from $y$. Thus, \begin{align*}
\int_K \vert f_i \ast \mu \vert dx & \leqslant C_0  \int_{\mathbb{R}^d} \mu(y) d y = C_0< \infty.
\end{align*}
