Summation with Ceilinged Logarithmic Function According to Johann Blieberger's paper - "Discrete Loops and Worst Case Performance" (1994):
$$
\sum_{i = 1}^{n}\left \lceil \log_2{(i)} \right \rceil = n\left \lceil \log_2{(n)} \right \rceil - 2^{\left \lceil \log_2{(n)} \right \rceil} + 1
$$
Now, I was wondering if someone knows what the following may equal?
$$
\sum_{i = 1}^{n}i\left \lceil \log_2{(i)} \right \rceil = ?
$$

Based on @vonbrand's answer, using integration by parts is a good approximation:

Concretely, 
$$
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = n.n(\left \lceil \log_a{(n)} \right \rceil - 1) - 1(\left \lceil \log_a{(1)} \right \rceil - 1) - \int_{1}^{n}i(\left \lceil \log_a{(i)} \right \rceil - 1) di
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = n^2(\left \lceil \log_a{(n)} \right \rceil - 1) + 1 - \left [ \frac{i^2(2\left \lceil \log_a{(i)} \right \rceil - 3)}{4}\right ]_{1}^{n}
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = n^2(\left \lceil \log_a{(n)} \right \rceil - 1) + 1 - \left [ \frac{n^2(2\left \lceil \log_a{(n)} \right \rceil - 3)}{4} - \frac{1(2\left \lceil \log_a{(1)} \right \rceil - 3)}{4}\right]
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = n^2(\left \lceil \log_a{(n)} \right \rceil - 1) + 1 - \left [ \frac{n^2(2\left \lceil \log_a{(n)} \right \rceil - 3)}{4} + \frac{3}{4}\right]
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = n^2\left \lceil \log_a{(n)} \right \rceil - n^2 + \frac{1}{4} - \frac{n^2\left \lceil \log_a{(n)} \right \rceil}{2} +\frac{3n^2}{4}
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = \frac{n^2\left \lceil \log_a{(n)} \right \rceil}{2} - \frac{n^2}{4} + \frac{1}{4} 
\\\\
\text{Therefore, } \sum_{i = 1}^{n}i\left \lceil \log_a(i) \right \rceil \approx \left \lfloor  \frac{n^2\left \lceil \log_a{(n)} \right \rceil}{2} - \frac{n^2}{4} + \frac{1}{4} \right \rfloor
$$
