Consider the following for $n\geq1$:

$$S_n = \sum_{i=1}^n i^{\ln(i)}$$

It seems to be hard to give an exact closed form formula for $S_n$.

It is straightforward however to see that $n^{\ln(n)} \leq S_n \leq n^{\ln(n)+1}$ but it is not clear to me if $S_n$ is asymptotic to some constant times the lower or upper bound or some function in between.

What is the asymptotic growth of $S_n$?

I would be interested in a $\Theta()$ answer if finding the precise constants is hard.

My motivation is mostly mathematical curiosity. I don't have a real application that isn't covered by the simple bounds given above.

(Much the same question asked previously on math.se.)

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    $\begingroup$ $n^{1+\ln n}/(2\ln n)$ $\endgroup$ – Brendan McKay Jun 1 '17 at 12:30
  • $\begingroup$ @BrendanMcKay That's very nice. How does one get that? $\endgroup$ – Lembik Jun 1 '17 at 12:32
  • $\begingroup$ Divide out the largest term. Approximate $(n-j)^{\ln(n-j)} / n^{\ln n}$, which is about $n^{-2j/n}$, and sum. I didn't rigorously handle the error terms but that wouldn't be difficult and it checks out numerically. $\endgroup$ – Brendan McKay Jun 1 '17 at 12:35
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    $\begingroup$ You can also use the identity $k^{\log k} = e^{(\log k)^2}$ and then approximate the sum using the integral $\int_1^n e^{(\log x)^2}dx$. This can be evaluated by expanding $e^{(\log x)^2}$ into its MacLaurin series and then change order of integration and summation. $\endgroup$ – Stanley Yao Xiao Jun 1 '17 at 12:39
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    $\begingroup$ (For the commenters) If you work out the asymptotics of the related integral don't you still have to show how this relates to the sum? That is at least if you want to get the constant right. $\endgroup$ – Anush Jun 1 '17 at 17:00

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