$-\Delta u=f$ in $L^1$ but $u_{x_ix_j}$ not in $L^1$ ($i\neq j$)

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I want to show the following function is a counterexample for Poisson equation with $L^1$ RHS but its solution is not in $W^{2,1}$:

Let $n\geq3$ and $x\in\mathbb{R}^n$. Let $f(x)=|x|^{-n}(\log|x|)^{-2}$ for $0<|x|<\frac{1}{2}$, $f(x)=0$ for $|x|\geq\frac{1}{2}$.

Let $u=c_n|x|^{2-n}\ast f$ be the convolution of the fundamental solution and $f$, and $c_n$ is a suitable constant. I want to show that $-\Delta u=f$ and $u_{x_ix_j}\notin L^1(\mathbb{R}^n)$ for $i\neq j$.

Any hint will be appreciated.

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If I am not mistaken in my computation, then one has: $$\frac{2}{c_n(n-2)n}u_{x_ix_j} = \int_{|y|<\frac 12}\dfrac{(x_i-y_i)(x_j-y_j)f(y)}{|x-y|^{n+2}}dy=\int_{|y|<\frac 12}\dfrac{x_ix_jf(y)}{|x-y|^{n+2}}dy+\int_{|y|<\frac 12}\dfrac{(y_iy_j-x_iy_j-x_jy_i)f(y)}{|x-y|^{n+2}}dy= u_1(x)+u_2(x).$$

Now, you can see that $u_1$ is not integrable because of the two $x_i,x_j$ terms, which result in $\sim\dfrac{1}{r}$ term in spherical coordinates whilst $u_2$ is integrable because the integrand becomes $\sim\dfrac{1}{r^2}$ in spherical coordinates.