Weak derivative of $\log\left(\log\left(1+\frac{1}{|x|}\right) \right)$ I want to prove that if $n\geq 2$, the function of $u(x)=\log\left(\log\left(1+\frac{1}{|x|}\right)    \right),\;|x|<\frac{1}{e}$ admits weak derivative respect of each component.
I expose my attempt and the doubts that have arisen:
I choose as candidate for $\partial_{x_i}u=\dfrac{-x_i}{\log\left(1+\frac{1}{|x|}\right)(|x|^2+|x|^3)}$.
Let $B=B(0,1/e)$ and $\phi \in C_0^{\infty}(B)$. We want to prove that $$\displaystyle\int_{B} u\partial_{x_i}\phi=-\displaystyle\int_{B} \phi\partial_{x_i}u.$$
My idea consists in taking $r>0$, so  $$\displaystyle\int_{B-B(0,r)} u\partial_{x_i}\phi=-\displaystyle\int_{B-B(0,r)} \phi\partial_{x_i}u + \displaystyle\int_{\partial B(0,r)} \phi\partial_{x_i}u.$$ (How can I justify this step?)
I have proved that  $u,\partial_{x_i}u\in L_n(B)\implies u,\partial_{x_i}u\in L_1(B)$. Thus, it would suffice to prove that $$\displaystyle\lim_{r\to0} \displaystyle\int_{\partial B(0,r)} \partial_{x_i}u=0..$$
(I don't know how to prove this last assertion).
Does anyone know an easier way to prove this? Any help will be welcome?
 A: *

*Integration by parts usually goes $$\int_{\Omega} u \partial_i \phi\, dx = -\int_{\Omega} (\partial_i u) \phi\,  dx + \int_{\partial \Omega } u\phi n^i \,d\Sigma$$ 
Here, $n^i$ is the $i$th component of the outward normal. This follows from divergence theorem  $\int_{\partial \Omega} U\cdot n=\int_{\Omega}\nabla\cdot U$, where the vector valued function $U$'s $i$th component is the only nonzero component, with $U_i(x) = u(x)\phi(x)$.

*(Revised as response to comment; the reader might have realised we were doing the wrong integration earlier.) What we need to bound is not $\int_{\partial B_r} \partial_iu\phi n^i d\Sigma$, but (see point 1)
$$\left|\int_{\partial B_r} u\phi n^i d\Sigma\right| \le \|\phi\|_{L^\infty} \left|\log\left(\log\left(1+\frac1 r\right)\right)\right| C_nr^{n-1} $$
For $r\ll 1$, $$\log(1+1/r) \le \log(2/r)=\log2+\log \frac1r \le 2 \log\frac1r \\
\log\left(\log\left(1+\frac1 r\right)\right) \le \log\left( 2\log \frac1r\right)= \log2 + \log\log\frac1r \le 2 \log\log\frac1r $$
Since for any $\epsilon>0$, $\log x\le C_\epsilon x^\epsilon$ for $x\gg 1$, we have
$$ \log\log\frac1r \le C_\epsilon \left(\log\frac1r\right)^{\epsilon} \le C_\epsilon^{1+\epsilon}r^{-2\epsilon} $$
Choosing an $\epsilon$ sufficiently small, this gives
$$\left|\int_{\partial B_r} u\phi n^i d\Sigma\right| \le \tilde C_{n,\epsilon,\phi} r^{n-1-2\epsilon} \to 0 $$
for any $n>1$, as $r\to 0$.
