Solving $T(n) = 2T(\frac{n}{2}) + n\log(n)$ without master theorem
For direct result let's use direct summation assuming $n=2^k$:
$$T(n) = 2T\left(\frac{n}{2}\right) + n\log(n)=2\left[ 2T\left(\frac{n}{2^2}\right) +\frac{n}{2}\log \frac{n}{2}\right]+ n\log(n) = \\
=2^2T\left(\frac{n}{2^2}\right)+n\log \frac{n}{2}+ n\log(n) =\\
=\cdots=\\
=2^kT\left(1\right)+n\log \frac{n}{2^{k-1}}+ \cdots +n\log \frac{n}{2}+n\log(n)=\\
=2^kT\left(1\right) +n\log 2 + \cdots +n\log 2^k=\\
=nT\left(1\right) +n\log 2 (1 + \cdots + k) =nT\left(1\right) + \frac{k(k+1)}{2} n\log 2 \in \Theta (n \log^2 n)$$
where, of course, $k=\log_2 n$. Hope you can obtain from here everything what you wanted.
At end let me note also, that $n \sim n\log(n)$ and other equivalences, which you are using, are not correct, if we understand standard $\lim \frac{f}{g}=1$, for symbol &quot;$\sim$&quot;. Correct is, for example, $n = o(n\log(n))$ etc.