Solution of $T(n)=2T(n/2) + n\log(\log n)$ I am struggling to solve this equation:
$$T(n)=2T(n/2) + n\log(\log n).$$ I concluded that the Master Theorem does not apply in this situation so I tried to successively substitute the terms in order to solve this equation but cannot proceed. Could somebody tell me what would be the best way to solve this?
Thanks.

We are given that $$T(n) = 2 T(n/2) + n \log_2(\log_2(n))$$ (Note that if we take $\log$ to the base $2$ or to the base $e$, the only difference is in the coefficient of the leading order term).
Let $n = 2^k$. Call $T(n) = g(k)$. Then we have
\begin{align}
g(k) & = 2g(k-1) + 2^k \log_2(k)\\
& = 2 (2g(k-2) + 2^{k-1} \log_2(k-1)) + 2^k \log_2(k)\\
& = 4 g(k-2) + 2^k \log_2(k-1) + 2^k \log_2(k)\\
& = 4 (2g(k-3) + 2^{k-2} \log_2(k-2))+ 2^k \log_2(k-1) + 2^k \log_2(k)\\
& = 8 g(k-3) + 2^k \log_2(k-2) + 2^k \log_2(k-1) + 2^k \log_2(k)
\end{align}
Hence, in general, 
\begin{align}
g(k) & = 2^k g(0) + 2^k \left( \log_2(1) + \log_2(2) + \log_2(3) + \cdots + \log_2(k)\right)\\
& = 2^k g(0) + 2^k \log_2(k!)
\end{align}
Now from Stirling's formula, we have that $$k! \sim C \sqrt{k} \left(\dfrac{k}{e} \right)^k$$
Hence, $$\log_2(k!) = k \log_2(k) - k \log_2(e) + \dfrac12 \log_2(k) + \mathcal{O}(1)$$
\begin{align}
g(k)& = 2^k g(0) + 2^k \left( k \log_2(k) -k \log_2(e) + \dfrac12 \log_2(k) + \mathcal{O}(1) \right)\\
& = 2^k k \log_2(k) - k2^k \log_2(e) + 2^{k-1} \log_2(k) + \mathcal{O} \left(2^k \right)
\end{align}
Now plug in $k = \log_2(n)$, to get
\begin{align}
T(n) & = n \log_2(n) \log_2(\log_2 n) - n \log_2(n) \log_2(e) + \dfrac{n}2 \log_2(\log_2 n) + \mathcal{O}(n)
\end{align}
