Let $t(n):=\begin{cases} \frac{2+\text{log}n}{1+\text{log}n}t(\lfloor\frac{n}{2}\rfloor) + log ((n!)^{\text{log} n}) \hspace{1cm} \text{if}\hspace{0.5cm} n>1 \\ 1 \hspace{0.5cm} \text{if}\hspace{0.2cm} n=1 \end{cases}$
We need to prove that $t(n) \in O(n²)$, thus $t(n) \leq c*n²$
I tried to play around with the master theorem (since $a,b > 1$)
so $a=\frac{2+\text{log} n}{1+ \text{log} n}$, $b=2$, $f(n)=\text{log}((n!)^{\text{log} n})=\text{log}n(\text{log}(n!))$
I have difficulties with the asymptotics of the $f(n)$ due to all the logarithms, help would be much appreciated.
Hint: \begin{align} \log(n!) &= \log(1 \cdot 2 \cdots (n-1) \cdot n) \\ &= \log(1) + \log(2) + \ldots + \log(n-1) + \log(n) \\ &\le \int_1^{n+1} \log(x) ~ dx\\ &= C\left. (x \log x - x) \right|_1^{n+1}\\ &= C\left(\left[ (n+1) \log (n+1) - (n+1)\right]- \left[ 1 \log (1) - 1\right] \right)\\ &= C\left(\left[ (n+1) \log (n+1) - n-1)\right]+1\right) \\ &= C\left( (n+1) \log (n+1) - n\right)\\ &\le C \left[ (n+1) \log (n+1) \right] \end{align}