Solving a complicated recurrence relation
Following Timot's suggestion, let $S(m) = T(2^m)$, so that $S$ satisfies the recurrence
$$ S(m) = \begin{cases} 2S(m/2) + \tfrac{m}{\log m} &amp; \text{ if } m &gt; 2 \\ 1 &amp; \text{ otherwise.} \end{cases} $$
In fact, we'll analyze the recurrence with a base case of $m = 1$ instead, and assume the logarithm is base $2$. We have
$$
\begin{align*}
S(2^k) &amp;= \frac{2^k}{k} + 2 \frac{2^{k-1}}{k-1} + 4 \frac{2^{k-2}}{k-2} + \cdots + 2^{k-1} \frac{2^1}{1} + 2^k \\ &amp;=
2^k \left[ \frac{1}{k} + \frac{1}{k-1} + \cdots + \frac{1}{1} + 1 \right] = \Theta(2^k\log k).
\end{align*}
$$
Therefore $S(m) = \Theta(m\log\log m)$ and $T(n) = \Theta(\log n\log\log\log n)$.