So I'm reading a book about PDE and I have following question: For Poisson's equation in $\mathbb{R}^2: -\bigtriangleup u=f$
The solution is $u(x)=-\frac{1}{2\pi}\int_{\mathbb{R}^2} log(|x-y|)f(y)dy$
At the beginning of the proof it sais: Following calculation would be wrong:
$\bigtriangleup_xu(x)=-\frac{1}{2\pi}\int_{\mathbb{R}^2}\bigtriangleup_x log(|x-y|)f(y)dy=0$
There is no further explanation why this is wrong, so I tried to find out:
$\frac{d^2}{dx_i^2}log(|x-y|)=\frac{1}{(x_1-y_1)^2+(x_2-y_2)^2}-\frac{2(x_i-y_i)^2}{((x_1-y_1)^2+(x_2-y_2)^2)^2}$
So $\bigtriangleup_x log(|x-y|)=0$. This means the first equation must be wrong and it has to be false to simply pull the laplacian into the integral.
I tried to find the reason on the Internet, but the only comment on this I found was "The second derivative cannot be integrated." But no explanation why.
Can anyone explain to me why this cannot be integrated, please? For me this looks like a "normal" function.