sum of logarithms I have to solve find the value of
$$\sum_{k=1}^{n/2} k\log k$$
as a part of question. 
How should I proceed on this ?
 A: Got it.  The constant in Moron's answer is $C = \log A$, where $A$ is the Glaisher-Kinkelin constant.  Thus $C = \frac{1}{12} - \zeta'(-1)$.  
The expression $H(n) = \prod_{k=1}^n k^k$ is called the hyperfactorial, and it has the known asymptotic expansion
$$H(n) = A e^{-n^2/4} n^{n(n+1)/2+1/12} \left(1 + \frac{1}{720n^2} - \frac{1433}{7257600n^4} + \cdots \right).$$
Taking logs and using the fact that $\log (1 + x) = O(x)$ yields an asymptotic expression for the OP's sum
$$\sum_{k=1}^n k \log k = C - \frac{n^2}{4} + \frac{n(n+1)}{2} \log n + \frac{\log n}{12} + O \left(\frac{1}{n^2}\right),$$ 
the same as the one Aryabhata obtained with Euler-Maclaurin summation.

Added: Finding an asymptotic formula for the hyperfactorial is Problem 9.28 in Concrete Mathematics (2nd ed.).  The answer they give uses Euler-Maclaurin, just as Aryabhata's answer does.  They also mention that a derivation of the value of $C$ is in N. G. de Bruijn's Asymptotic Methods in Analysis, $\S$3.7.  
A: Here is an asymptotic expression using EulerMcLaurin Summation.
$$ \sum _{k=1}^{n} k \log k = \int_{1}^{n} x \log x\ \text{d}x + (n\log n)/2 + C' + (\log n + 1)/12+ \mathcal{O}(1/n^2)$$
$$ = n^2(2 \log n - 1)/4 + (n\log n)/2 + (\log n)/12 + C + \mathcal{O}(1/n^2)$$
for some constant $C$. 
