Define $f(x,y)=(x^2+y^2)\ln(\sqrt{x^2+y^2})$ when $(x,y)\neq(0,0)$ and $f(0,0)=0$.
Show that $\nabla^{(4)}f=\delta_0$ in the sense of distribution where $\nabla^{(4)}$ denotes the bilaplacian.
My work
Denoting $\phi(r)=r^2\ln(r)$ with $\phi(0)=0$, I have verified that its bilaplacian vanishes for $r\neq 0$ using the formula for polar coordinates, that is $$\nabla^{(4)}\phi=\partial_r^4\phi+\frac2r\partial_r^3\phi-\frac1{r^2}\partial_r^2\phi+\frac1{r^3}\partial_r\phi$$ However, I can't seem to show that $\int_\mathbb{R^+}\nabla^{(4)}\phi(r)\varphi(r)\mathrm{d}r$ is equal to $\varphi(0)$. I have tried integrations by parts but it just gets out of hands and nothing comes off of it.