So I have the following pseudo-code:Function(n):for (i = 4 to n^2):    for (j = 5 to floor(3ilog(i))):        // Some math that executes in constant timeSo far, I know that to find this i will be calculating$\sum_{i=4}^{n^2}\sum_{j=5}^{3i \log_{2}i}C$ where $C$ is a constant, but I am completely lost as to how to proceed past the first summation which, if I'm not mistaken, will give me $3C \cdot (\sum_{i=4}^{n^2}(i \log_{2}i) - 5(n^2 - 4))$, but from here I'm lost.  I don't need exact running time, but the asymptotic complexity.All help is greatly appreciated!  I realize that this might be a duplicate, but I haven't been able to find a nested for loop problem of this nature anywhere...
Running time of a nested loop with $\sum i \log i$ term
time complexity;algorithm analysis;asymptotics
$$F(n) = \sum\limits_{i = 4}^{n^2} \sum\limits_{j = 5}^{3i\log i}C = C\sum\limits_{i = 4}^{n^2}(3i\log i - 4) = 3C\sum\limits_{i = 4}^{n^2}i \log i - \Theta(n^2)$$For asymptotic analysis, we can start the summation from $i = 1$ instead of $4$.  Let$\begin{align}T(n) &= \sum\limits_{i = 1}^{n^2}i\log i = \sum\limits_{i = 1}^{n^2}\log {i^i} \\&= \log {1^1} + \log {2^2} + \log {3^3} + \dots + \log {(n^2)^{n^2}} \\&\le n^2 \cdot \log (n^2)^{n^2} \ \ \text{(there are \(n^2\) terms)}\\&= 2n^4 \log n \\&= O(n^4 \log n)\end{align}$Now, since $f(x) = x\log x$ is a continuous increasing function in $x \in [1, \infty)$, you can verify that :$\int\limits_{1}^{n^2}x\log x \ dx \le \sum\limits_{x = 1}^{n^2}x\log x \ \ $ (changing  $i$ to $x$ simply because $x$ looks better when integrating).   According to Wolfram, $\int\limits_{1}^{n^2}x\log x \ dx = \frac{1}{4}\left(n^4 \log \left({\frac{n^4}{e}}\right) + 1\right) = n^4 \log \left({\frac{n}{e}}\right) + \frac{1}{4}$Therefore, $\sum\limits_{x = 1}^{n^2}x\log x = \Omega(n^4\log n)$.Combining the upper and lower bounds, we have $T(n) = \Theta(n^4\log n)$Overall, $F(n) = \Theta(n^4\log n) - \Theta(n^2) = \Theta(n^4 \log n)$.So, your loops have complexity $\Theta(n^4 \log n)$.Remarks:1. You can show the lower and upper bounds in a similar way by observing that:$\int\limits_{1}^{n^2}x\log x \ dx \le \sum\limits_{x = 1}^{n^2}x\log x \le \int\limits_{0}^{n^2}(x+1)\log {(x+1)} \ dx$2. For the above inequalities, you can use the method of Riemann sum (use the left and the right Riemann sums to see the two bounds). Draw unit-width rectangles on an increasing function to see these bounds.3. I have done some implicit simplifications, such as just ignoring the constant $3C$ term, or starting the summation from $1$ instead of $4$ since we are dealing with $\Theta$. I have also made the common but slight abuse of notation when equating or subtracting $\Theta$ terms, without affecting the answer.