Which of these reasonings is correct? If $K(x)=-\ln\|x\|$, we know that $\Delta K=\delta_0$
in $\mathbb{R^2}.$ Furthermore, we consider $f$ as the characteristic function of the ball $B(0,1)$.  I have obtained two different results when I use divergence theorem, but I am not able to find which way is correct. Any help would be welcome because I am really stuck with this question.
On the one hand
\begin{align}
\int_{\mathbb{R}^2}(K*f)(x)f(x)dx&=\int_{\mathbb{R}^2}(K*f)(x)(\delta_0*f)(x)dx=\int_{\mathbb{R}^2}(K*f)(x)\Delta K*f(x)dx\\&=\int_{\mathbb{R}^2}(K*f)(x)div (\nabla K*f(x))dx
\\&=\lim_{r\to\infty}\Big[ \int_{C(0,r)}(K*f)(x)(\nabla K*f)(x)\cdot n(x)dx   \Big]-\int_{\mathbb{R}^2}(\nabla K*f)(x)\cdot(\nabla K*f)(x)dx.
\end{align}
On the other hand
\begin{align}
\int_{\mathbb{R}^2}(K*f)(x)f(x)dx&=\int_{B(0,1)}(K*f)(x)f(x)dx=\int_{B(0,1)}(K*f)(x)(\delta_0*f(x))dx\\&=\int_{B(0,1)}(K*f)(x)(\Delta K*f(x))dx=\int_{B(0,1)}(K*f)(x)div (\nabla K*f(x))dx
\\&= \int_{C(0,1)}(K*f)(x)(\nabla K*f(x))\cdot n(x)dx-\int_{B(0,1)}(\nabla K*f(x))\cdot(\nabla K*f(x))dx.
\end{align}
 A: There is an error in your first calculation.  What is actually true is that
\begin{align*}
\int_{\mathbb{R}^2} (K\ast f)(x) \textrm{ div}(\nabla K\ast f)(x)\,dx
&= \lim_{R\to\infty} \int_{B(0,R)} (K\ast f)(x) \textrm{ div}(\nabla K\ast f)(x)\,dx \\
&= \lim_{R\to\infty} \big[\int_{C(0,R)} (K\ast f)(x) (\nabla K\ast f)(x) \cdot n(x)\,dx \\
&\qquad \qquad - \int_{B(0,R)}(\nabla K\ast f)(x)(\nabla K\ast f)(x)\,dx.\big]
\end{align*}
The reason this is important is that $\nabla K(x) =\frac{x}{|x|^2}$ decays like $|x|^{-1}$, so also $(\nabla K\ast f)(x)$ decays like $|x|^{-1}$ and therefore the function $|(\nabla K \ast f)(x)|^2$ is not integrable.  Indeed, the integral
$$
\int_{B(0,R)} |(\nabla K\ast f)(x)|^2 \,dx 
$$
grows like $\log(R)$.  On the other hand, the integral over the circle $C(0,R)$ also grows like $\log(R)$ because the $K\ast f(x)$ term has magnitude roughly $\log(R)$, while $\nabla K\ast f$ contributes a factor $R^{-1}$ and the integration is over a circle of perimeter on the order $R$.
The point is that although the limit as $R\to\infty$ exists of the expression above, you cannot separate the terms because there is significant cancellation.
The second derivation you gave is correct.
Asymptotics of $\nabla K\ast f$
I'll add here a computation showing that $|(\nabla K\ast f)(x)|\geq (20|x|)^{-1}$ for $|x|\geq 20$.
We first write out explicitly the expression for $(\nabla K\ast f)(x)$,
$$
(\nabla K\ast f)(x) = 
\int f(y) \frac{x-y}{|x-y|^2} \,dy.
$$
The idea is that when $x$ is large, $x-y$ has an angle that is very close to $x$.  So to get a lower bound on the magnitude it makes sense to take a dot product against $x$ itself.
\begin{align*}
|\langle x, \nabla K\ast f(x)\rangle|
&= 
\int f(y) \frac{|x|^2-x\cdot y}{|x-y|^2} \,dy.
\end{align*}
To get a lower bound on this quantity we estimate
$||x|^2 - x\cdot y|\geq |x|^2 - |x\cdot y| \geq |x|^2 - |x| |y|$.
Since $|y|\leq 1$ in the integral (due to the presence of $f(y)$),
we have
$$
|\langle x, \nabla K\ast f(x)\rangle|
\geq \int f(y) \frac{|x|^2}{|x-y|^2} \,dy
- \int f(y) \frac{|x|}{|x-y|^2}\,dy.
$$
The other observation is that $\frac{1}{4}|x|^2 \leq |x-y|^2 \leq 4|x|^2$ when $|x|>2$ and $|y|\leq 1$.  Therefore, for $|x|>2$ we have
$$
|\langle x, \nabla K\ast f(x)\rangle|
\geq \frac{1}{4}\int f(y) \,dy
- 4|x|^{-1} \int f(y)\,dy.
$$
Thus, for $|x|>20$ we can conclude that
$|\langle x, \nabla K\ast f(x)\rangle| \geq \frac{1}{20}$.  Now using Cauchy-Schwarz we conclude
$$
\frac{1}{20} \leq |\langle x, \nabla K\ast f(x)\rangle| \leq |x|
|(\nabla K\ast f)(x)|,
$$
so dividing by $|x|$ gives the result we wanted.
