Bounds for $\sum_{i=1}^n i^{\ln(i)}$ Consider for $n\geq1$:
$$S_n = \sum_{i=1}^n i^{\ln(i)}$$
It seems to be hard to give an exact formula for this.

For large $n$, are there good bounds one can get for $S_n$?

Ideally I would like bounds written in elementary terms so I can understand their asymptotics.

Update
It is straightforward to see that $n^{\ln(n)} \leq S_n \leq n^{\ln(n)+1}$.  However, is $S_n$ asymptotic to some constant times the lower or upper bound or something in between?
[Cross-posted to https://mathoverflow.net/questions/271193/asymptotic-growth-of-sum-i-1n-i-lni]
 A: An upper bound could be like the following:
$$\int x^{ln(x)} = \frac{\sqrt{\pi} erfi(\log{x}+0.5)}{2e^{0.25}}$$
$$\sum_{i=1}^n i^{ln(i)}\leq \int_1^n x^{ln(x)}dx \leq \frac{\sqrt{\pi} erfi(\log{n}+0.5)}{2e^{0.25}}$$
A: I prefer to add another answer just related to the bounds.
$$S_n = \sum_{i=1}^n i^{\ln(i)}=1+\sum_{i=2}^n i^{\ln(i)}$$ Since $x^{\log(x)}$ is increasing for $x \geq 1$ then $$I_n=\int_{x=1}^nx^{\log(x)} \leq\sum_{i=2}^n i^{\ln(i)}\leq J_n=\int_{x=2}^{n+1}x^{\log(x)}\,dx$$ whith make
$$I_n+1 \leq S_n\leq J_n+1$$ (this holds for any $n >2$) using 
$$I_n+1=n^{\log (n)+1} F\left(\log (n)+\frac{1}{2}\right)-F\left(\frac{1}{2}\right)+1$$ $$J_n+1= (n+1)^{\log (n+1)+1} F\left(\log (n+1)+\frac{1}{2}\right)-2^{1+\log (2)}
   F\left(\frac{1}{2}+\log (2)\right)+1$$
A: Write $S_n := \sum_{ i =1}^n e^{\log(i)^2}$. For $k$ that is small compared to $n$, consider the contribution to the sum from the term with $i = n-k$. Here we have
\begin{align*}
\log(n-k)^2 = \log(n)^2 - 2 \frac{\log n }{n}k + O(k^2/n^2).
\end{align*}
In particular, we anticipate that for large $n$
\begin{align*}
S_n = (1 + o(1)) e^{ \log(n)^2 } \sum_{ k \geq 1} e^{ - 2 \frac{\log n}{n} k }.
\end{align*}
Now using the fact that for small $\varepsilon$, $\sum_{ k \geq 1} e^{ - \varepsilon k } \approx 1/\varepsilon$, it seems reasonable to guess that
\begin{align*}
S_n = (1 + o(1)) e^{ \log(n)^2 } \frac{n}{2 \log n }.
\end{align*}
as $n \to \infty$.
