How to solve the recurrence $T(n) = \frac{n}{2}T(\frac{n}{2}) + \log n$ I am trying to solve the recurrence below but I find myself stuck. 
$T(n) = \frac{n}{2}T(\frac{n}{2}) + \log n$
I have tried coming up with a guess by drawing a recurrence tree. What I have found is 
number of nodes at a level: $\frac{n}{2^{i}}$
running time at each node: $\log \frac{n}{2^{i}}$
total running time at each level: $\frac{n}{2^{i}}  \log \frac{n}{2^{i}}$
I try to sum this over through n > $\log n$ which is the height of the tree 
$$\begin{align}
\sum\limits_{i=0}^n \frac{n}{2^{i}}  \log \frac{n}{2^{i}}&=
n \sum\limits_{i=0}^n \frac{1}{2^{i}}  \log \frac{n}{2^{i}}\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}}  (\log n - \log 2^{i})\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}} \log n -  \sum\limits_{i=0}^n \frac{1}{2^{i}}  \log 2^{i}\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}} \log n -  \sum\limits_{i=0}^n \frac{ \log 2^{i}}{2^{i}}\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}} \log n -  d \sum\limits_{i=0}^n \frac{i}{2^{i}}\\
&=2 n  \log n - d \sum\limits_{i=0}^n \frac{i}{2^{i}}
\end{align}$$
I am still trying to figure out the sum of the second summation above but I somehow feel that it will be bigger than $2n \log n$ which makes my whole approach wrong. 
Is there another way to tackle this problem?  

Consider the equality
$$
1+x+x^2+...+x^n=\frac{x^{n + 1} -1}{x-1}
$$
Differentiate it by $x$, then multiply by $x$:
$$
x+2x^2+3x^3+...+n x^n=\frac{nx^{n+2} - (n + 1)x^{n+1} + x}{(x-1)^2}
$$
Now we can substitute $x=\frac{1}{2}$ and obtain
$$
\sum\limits_{i=0}^n\frac{i}{2^i}=n\left(\frac{1}{2}\right)^{n} - (n + 1)\left(\frac{1}{2}\right)^{n-1} + 2
$$
