Apparently recurrences like this cannot be solved with the Master Theorem:
$T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{\log(n)}$
Because $n^{\log_b(a)} = n^1$ is not a polynomial multiple of $f(n) = \frac{n}{\log(n)}$ since $\frac{n}{\frac{n}{\log(n)}} = \log(n)$, which is less than $n^\epsilon$ for any positive $\epsilon$.
But when is this really even an issue?
For example why wouldn't the time complexity be $n \log n$ using case 2 of the Master Theorem? Unrolling the recursion:
$$T(n) = n\sum_{k=1}^{\log_2(n)} \frac{1}{\log_2(2^k)}$$
$$T(n) = n\sum_{k=1}^{\log_2(n)} \frac{1}{k}$$
Doesn't this suggest $\Theta(n \log n)$ even though it supposedly cannot be done with Master Theorem?