TITLE: Solving recurrences of the form $T(n)=aT(n/a)+Θ(nlgn)$
QUESTION [6 upvotes]: On pages 95 and 96 of the third edition of the CLRS book, we find
the following, which applies here since $a=b$ is all
it takes to block the application of the Master Theorem: "Although
$n\lg n$ is asymptotically larger than n, it is not polynomially
larger because the ratio $\tfrac{n\lg n}{n}=\lg n$ is asymptotically
less than $n^{\epsilon}$ for any positive constant $\epsilon$. Consequently,
the recurrence falls into the gap between case 2 and case 3." For a solution, the authors send us to exercise 4.6.2 on page 106:
"Show that if $f\left(n\right)=\Theta\left(n^{\log_{b}a}\lg^{k}n\right)$,
where $k\geq0$, then the master recurrence has solution $T\left(n\right)=\Theta\left(n^{\log_{b}a}\lg^{k+1}n\right)$.
For simplicity, confine your analysis to exact powers of b."
(Here $\lg^k n$ is CLRS's notation for $(\log_2 n)^k$.)
This is where I am starting to have problems...

REPLY [2 votes]: This is amenable to analysis by the Akra-Bazzi method. The calculations go like this, using the notation of the Wikipedia article:
\begin{eqnarray*}
k & = & 1 \\
a_1 & = & a \\
b_1 & = & 1/a \\
g(x) & = & x \log x \\
h(x) & = & 0 \\
p & = & 1 \Leftarrow a_1 b_1=1 \\
T(x) & = & \Theta \left ( x \left ( 1 + \int_1^x \frac{u \log u}{u^{p+1}} du \right ) \right ) \\
& = & \Theta \left ( x \left ( 1 + \Theta \left (\log(x)^2 \right ) \right ) \right ) \\
& = & \Theta \left ( x \log(x)^2 \right )
\end{eqnarray*}
This may not help you that much, as you may not be able to prove that the Akra-Bazzi method works.