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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]

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As you wrote, it seems to be hard to give an exact formula for this. $$S_n = \sum_{i=1}^n i^{\log(i)}$$

Computing a few values reproduced below for $\log(S_n)$ $$\left( \begin{array}{cc} n & \log(S_n) \\ 100 & 23.5591 \\ 200 & 30.9636 \\ 300 & 35.7524 \\ 400 & 39.3537 \\ 500 & 42.2632 \\ 600 & 44.7156 \\ 700 & 46.8417 \\ 800 & 48.7224 \\ 900 & 50.4113 \\ 1000 & 51.9458 \end{array} \right)$$ and plotting them, it seems that they look like $$\log(S_n)=a\, n^b+c$$ Performing a nonlinear regression, the following is obtained $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} \\ a & 28.7896 & 0.77787 \\ b & 0.164932 & 0.00226 \\ c & -37.9895 & 1.03575 \\ \end{array}$$ corresponding to a quite good fit of the data ($R^2$ almost equal to $1$) .

Extrapolating is always dangerous : using this model, $\log(S_{2000})=62.8620$ while the correct value should be $62.6013$ which is not too bad (I hope).

Edit

Inspired by OmG's answer, we could use $$\int_1^n x^{\log(x)}\,dx=n^{\log (n)+1} F\left(\log (n)+\frac{1}{2}\right)-F\left(\frac{1}{2}\right)$$ where appears the Dawson integral.

Update

We have the double inequality $$n^{\log(n)} < S_n <n^{1+\log(n)}$$ Considering $T_n=\frac{S_n}{n^{\log(n)}}$ and curve fitting $T_n$ for the range $10^2 < n <10^4$,an approximation could be $$T_n=0.133133\,n^{0.89683}+3.14905$$For example, using $n=10^4$, this approximation gives $S_n=3.59432\times 10^{39}$ while the exact value should be $S_n=3.59626\times 10^{39}$

Similarly,for infinitely large values of $x$ (see here), $$F(x) = \sum_{k=0}^{\infty} \frac{(2k-1)!!}{2^{k+1} x^{2k+1}} = \frac{1}{2 x} + \frac{1}{4 x^3} + \frac{3}{8 x^5}+O\left(\frac 1{x^7} \right) $$ Limiting the the first term, this would make $$\int_1^n x^{\log(x)}\,dx\approx \frac{n^{1+\log (n)}}{2 \log (n)+1}$$ For example, using $n=10^4$, this last approximation gives for the integral $=3.57353\times 10^{39}$ while its exact value should be $3.59279\times 10^{39}$. For sure, adding more terms will make the approximation better.

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  • $\begingroup$ Is $F(\log(n) + 1/2)$ greater than or less than $1$? $\endgroup$ – user66307 May 31 '17 at 10:14
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    $\begingroup$ @Lembik. It is less than $1$ but $n F\left(\log (n)+\frac{1}{2}\right)$ looks like a straight line with a small positive slope. For $n=1000$, this term is "just" $\approx 68.13$ $\endgroup$ – Claude Leibovici May 31 '17 at 10:18
  • $\begingroup$ What is $F(1/2)$ approximately? $\endgroup$ – user66307 May 31 '17 at 10:21
  • $\begingroup$ @Lembik. Almost $0.4244$ $\endgroup$ – Claude Leibovici May 31 '17 at 10:22
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    $\begingroup$ nice (+1), maybe we can employ euler mac-laurin to make this rigorous? $\endgroup$ – tired May 31 '17 at 19:22
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An upper bound could be like the following:

$$\int x^{ln(x)} = \frac{\sqrt{\pi} erfi(\log{x}+0.5)}{2e^{0.25}}$$

$$\sum_{i=1}^n i^{ln(i)}\leq \int_1^n x^{ln(x)}dx \leq \frac{\sqrt{\pi} erfi(\log{n}+0.5)}{2e^{0.25}}$$

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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 $$I_n+1=n^{\log (n)+1} F\left(\log (n)+\frac{1}{2}\right)-F\left(\frac{1}{2}\right)+1$$ $$J_n+1= (n+1)^{\log (n+1)+1} F\left(\log (n+1)+\frac{1}{2}\right)-2^{1+\log (2)} F\left(\frac{1}{2}+\log (2)\right)+1$$

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