Is there a closed form for $\sum_{j=1}^{n} j^2\log{j}$? Question
Is there a closed form for $\sum_{j=1}^{n} j^2\log{j} = 1\times0 + 2^2\times\log{2} + 3^2\log{3} + \dots + n^2\log{n}$? I'm trying to look for the simplest $\Theta$ notation.
Attempt
Let $g(n)$ be the function in $\Theta (g(n))$. My current take is the following,
$g(n) = 1\times0 + 2^2\times\log{2} + 3^2\log{3} + \dots + n^2\log{n}$
For $g(n)$ to be $O(g(n))$, we can simplify it this way,
$$g(n) = 1\times0 + 2^2\times\log{2} + 3^2\log{3} + \dots + n^2\log{n}$$
$$g(n) = n^2\times 0 + n^2\times\log{2} + \dots + n^2\log{n}$$
$$g(n) = n^2 (\log{1} + \log{2} + \dots + \log{n})$$
$$g(n) = n^2(\log{n!})$$
But $n^2$ seems to get in the way of $g(n)$ also being $\Omega(g(n))$, if we pick something with lower power, e.g. $n^{1.5}$, then $g(n)$ will fail to be $O(g(n))$, thus it's impossible to simplify further.
Is this conclusion correct?
 A: Because your function is monotone, it can be written as
$$\sum_{j=1}^nj^2\log j=\int_0^n t^2\log t\,dt+\theta(f(n)-f(0))$$
for some $-1\le\theta\le 1$. Then doing the integration, we get
$${1\over 3}\left(n^3\log(n^3)-1\right)+\theta(n^2\log n)=\Theta \left(n^3\log n\right)$$

In case you've never seen this result before, it's easy to see by using the left-hand and right-hand estimates on the integral, the former is less than the integral and the latter is greater.
To see this is simple, note
$$\sum_{j=0}^{n-1} j^2\log j<\int_1^nt^2\log t\,dt <\sum_{j=1}^n j^2\log j$$
And so the difference between the sum and the integral is bounded by $|f(n)-f(0)|$ as
$$0<\int_1^nt^2\log t\,dt -\sum_{j=0}^{n-1}j^2\log j\le f(n)-f(0)$$
ergo
$$\left|\int_1^nt^2\log t\,dt -\sum_{j=0}^{n-1}j^2\log j\right|\le |f(n)-f(0)|$$
which is the absolute value of the difference of the sums, so it must be equal to $\theta(f(n)-f(0))=\theta n^2\log n$, since those are the only numbers bounded by $|f(n)-f(0)|$.
A: Here is a closed form

$$ S =-\zeta'  \left( -2 \right) +\zeta_1' \left( -2,n+1 \right), $$

where $\zeta(x)$ and $\zeta(x,a)$ are the zeta and the Hurwitz zeta functions. Note that $\zeta_1'(s,a)$ is the derivative w.r.t. the first argument. Here is an example 

$$ \sum_{k=1}^{10} j^2\ln(j) \approx 776.2476538  $$

and with the formula is

$$ S \approx 776.2476539 .$$ 

Added: Here is a formula for a more general case in terms of the zeta and hurwitz zeta functions 

$$ \sum_{k=1}^{n} j^{s}\ln(j)^{m}= \lim_{\alpha \to s } \zeta^{(m)}_{\alpha}( - \alpha ) + \zeta^{(m)}_{\alpha} \left( -\alpha,n+1 \right)  $$

