I have an exercise on my book, where's written:
Let $\Omega =B_1(0)$ and $$f(x,y)=|\log(\frac{1}{\sqrt{x^2+y^2}})|^k$$ for $k \in (0, \frac{1}{2})$. Prove that the $H^1$ norm squared is $$||f||_{H^1}^2 = 2 \pi \int_0^1 |\log(s)|^{2k}s ds + 2k^2 \pi^2 \int_0^1 \frac{1}{s}|\log(s)|^{2k -2}ds$$
- The first term is just the $L^2$ norm of $f$, and I have been able to show it. My big big problem is the second term, which should correspond to $$\sum_{\alpha \leq 1} \int_{B_1(0)} (D^{\alpha}f)^2$$
Question: in this particular case, is the above summation just $$\int \int_{B_1(0)} (\partial_x f)^2 + (\partial_y f)^2 dxdy$$?
If so, I have problems in obtaining the expected result. For instance, if I compute the $L^2$ norm of the first partial I obtain the following (wolframapha computation of the derivative)
$$\int_{B_1(0)} (\partial_x f)^2 =k^2 \int_0^{2 \pi} \cos^2(\theta) d\theta \int_0^1 \frac{1}{s} |\log(s)|^{2k-2} = k^2 \pi \int_0^1 \frac{1}{s} |\log(s)|^{2k-2}ds$$
The other integral for $\partial_y f$ is analogous, and hence my second addendum would be
$$\color{blue}{2} k^2 \pi \int_0^1 \frac{1}{s} |\log(s)|^{2k-2}ds$$
I can't figure out how to get that $\color{red}{\pi^2}$ factor.