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Given a series of the type:

$$Q(s,n) = \ln(1)^s + \ln(2)^s + \ln(3)^s + \cdots+ \ln(n)^s $$

How does one evaluate it?

Something I noticed was:

$$Q(1,n) = \ln(1) + \ln(2) + \ln(3)+ \cdots+\ln(n) = \ln(1\cdot 2\cdot 3 \cdots n) = \ln(n!) $$

I also noticed that:

$$\int^{n}_{1}\ln(x)^s\, dx\quad\sim\quad\sum^{n}_{i = 1}\ln(i)^s$$

But I am really interested in an exact formula or at least one whose difference from the actual value progressively decreases as opposed to merely whose ratio from the actual progressively decreases.

Sidharth Ghoshal
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    I've solved this using the idea of an "indefinite summation"-formula and with an ansatz which should mimic that how the Bernoulli-polynomials were found for sums of like powers. It's too long to write it down here, but perhaps you find it instructive to read http://go.helms-net.de/math/divers/BernoulliForLogSums.pdf – Gottfried Helms Jul 06 '13 at 20:04
  • the sum is divergent , regularizatio is needed so $ (-1)^{s}\zeta ^{(s)} (0) $ – Jose Garcia Jul 06 '13 at 20:06
  • I would just like the n'th value not an infinite – Sidharth Ghoshal Jul 06 '13 at 20:07
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    @frogeyedpeas: Please note that \ln produces the correct upright notation $\ln$, whereas ln just means "the product of two variables named $l$ and $n$". – Zev Chonoles Jul 06 '13 at 20:30
  • oh, I'll keep that in mind next time :) – Sidharth Ghoshal Jul 06 '13 at 20:31
  • Related: http://math.stackexchange.com/questions/207455 asks about the exponential of your Q(s,n)-function (but with no answer), and in http://math.stackexchange.com/questions/279401 I looked at the convergence-radius of the power-series solution for $t_1(x)$ in my answer – Gottfried Helms Jul 07 '13 at 05:50

3 Answers3

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Here is a short excerpt of the discussion to which I've linked in my first comment.

For $s=1$ (which is somehow nearly trivial) we can define the function $$ t_1(x)=-\zeta '(0)-\ln(\Gamma(\exp(x)))$$ which gives for instance $$ t_1(\ln(2)) - t_1(\ln(4)) = \ln(2)+\ln(3) $$ and in general $$ t_1(\ln(a)) - t_1(\ln(b)) = \sum _{k=a}^{b-1} \ln(k) $$ The key is, that the artificial-looking version of $t_1(x)$ gives the infinite series $$ t_1(\ln(x)) = \sum _{k=x}^\infty \ln(x) = \ln(x) + \ln(x+1) + \ldots $$

The coefficients of the power series of $t_1(x)$ can easily be given for instance using Pari/GP 

t_1(x) + O(x^8)
%1321 = 0.91893853 + 0.57721566*x - 0.53385920*x^2 - 0.32557879*x^3 
      - 0.12527414*x^4 - 0.033725651*x^5 - 0.0068593536*x^6 - 0.0011726081*x^7
      + O(x^8)

where the coefficients can be described exactly by compositions of Stirling numbers 2nd kind and $\zeta()$-values at positive integer arguments, and where moreover $\zeta(1)$ is replaced by the Euler-$\gamma $ (which, btw, indicates that we have somehow the Ramanujan-like zeta-renormalization at work here)

The first answer is then $$ Q(1,n) = t_1(\ln(1)) - t_1(\ln(n+1)) $$

For $s=2$ $$ t_2(\ln(x)) = \sum_{k=x}^{\infty} \ln(x)^2 $$

such that analoguously $$ Q(2,n) = t_2(\ln(1)) - t_2(\ln(n+1)) = \sum_{k=1}^n \ln(k)^2 $$

I don't have an exact representation for the power series in terms of zetas and Euler-gamma; here is an approximation, where the constant term is $\zeta''(0)$ (the generation scheme allows arbitrary precision depending on the possible size of involved matrices): 

t_2(x) = -2.006356455908585 - 0.1456316909673534*x + 0.6345699670487060*x^2 
        - 0.3868588771980126*x^3 - 0.2407113770463571*x^4 - 0.09916202534448954*x^5
        - 0.02847303775799426*x^6 - 0.005923792714748150*x^7 - 0.0009884022636657563*x^8
        - 0.0001620035246035620*x^9 - 0.00002414672567100699*x^10 
        - 0.000001216451660450317*x^11 + 0.0000001409130267444575*x^12
        - 0.0000001437552825860954*x^13 - 0.00000003587528042872192*x^14 
        + 0.00000001359539422026695*x^15 + O(x^16)

and $$Q(2,n) = - \sum_{k=1}^\infty c_k \cdot \ln(n+1)^k $$ where $c_k$ are the coefficients of the power series and the index $k$ begins at $1$ such that the constant term is skipped.

The numbers and the generation-scheme (even for the higher $s$) can be taken from the discussion to which I've linked in my first comment.

Gottfried Helms
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I hope you don't mind if I use $\log$ as you use $\ln$ (it is more standard in analytic number theory).

Since $\log$ is monotonic increasing: $$\int_{1}^{n}\log^s x \ dx < Q(s,n) < \int_{1}^{n+1}\log^s x \ dx$$ (using left/right endpoints and $Q(s,1)=0$).

This shows that $Q(s,n)=\int_1^n\log^s x + O(\log^s n)$, which is already a pretty good asymptotic formula; this error term is massively dwarfed by the main term (even though it is not a decreasing function as you have asked for - this requirement might be too strict).

Evaluating integrals, we obtain: $$Q(s,n)=\int_{1}^{n}\log^s x \ dx+O(\log^s n)=n\left(\sum_{k=0}^{s}(-1)^{s-k}\frac{s!}{k!}\log^k n\right)+O(\log^s n)\qquad (*)$$ when $s$ is an integer, and an analogous expression in terms of the incomplete gamma function otherwise.

Suffice it to say, as far as you are probably concerned, $$Q(s,n)=n\log^s(n)+O\left[n\log^{s-1}(n)\right]$$The full version is $(*)$.

pre-kidney
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  • Can this be transformed to a series expansion of some type? for example $Q(s,n) = a_1n\ln^s(n) + a_2n\ln^{s-1}(n)... a_sn\ln(n) + a_{s+1}n + a_{s+2} + a_{s+3}n/\ln(n) + a_{s+4}n/\ln(n)^2...$ – Sidharth Ghoshal Jul 06 '13 at 23:05
  • Your latex got a little messed up so I'm not sure what you mean, but I think what you are getting at is covered by the integral I evaluated. Terms like $n\log^k(n)$ with $k<0$ don't appear, because they are dwarfed by the $O(\log^s n)$ term already. And the $k>=0$ case appear explicitly in the integral evaluation $(*)$. – pre-kidney Jul 06 '13 at 23:08
  • I corrected the latex The idea was that I had a series that went from $\log^k(n) to \log^0(n)$ and then all the negative powers of $\log(n)$ each multiplied by some coefficient and the value of n – Sidharth Ghoshal Jul 06 '13 at 23:10
  • I edited my answer to clarify your question. See $(*)$, which gives the series expansion you desire. – pre-kidney Jul 06 '13 at 23:10
  • Just gave it a try. printing the results by Q followed by the explicite summation I got this:$$ Q(1,5)= 3.04718956217 \quad ( 4.78749174278) \ Q(1,15)=25.6207530165 \quad (27.8992713838) \ Q(2,15)=58.7615323423 \quad ( 60.4520312916) \ Q(2,115)= 1727.81941430 \quad ( 1737.07712499) $$ – Gottfried Helms Nov 07 '14 at 13:12
  • Seems like a pretty decent approximation. Also did you try plugging in terms from the series I labelled (*)? – pre-kidney Dec 15 '14 at 07:13
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You may use Euler-Maclaurin formula to get $Q(x,s)=\sum_{n\leq x}(\log n)^s$. That would be $x(\log x)^s-s\int_{1}^{x}(\log t)^{s-1} \ dt+O((\log x)^s)$. It should be a fine approximation for your work!

Kunnysan
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  • You should expand your answer. Using just the first 3 terms of the Euler-Macclaurin formula, you can show that the error is less than $\frac{s}{12n}(\log(n))^{s-1}$ for $s\gt 1$. Unfortunately the error bound doesn't start going down until $n \gt e^{s-1}$. – Apprentice Queue Jul 07 '13 at 18:16
  • @Apprentice Queue: That's a good observation. But, are you sure that the error term will be $O((\log n)^{s-1}/n)$, I mean does that go to zero? – Kunnysan Jul 07 '13 at 18:23
  • yes it does approach zero because if you calculate the error term or Euler-Maclaurin, you will see it is bounded by the first derivative (ie $s(\log(n))^{s-1}/n$). – Apprentice Queue Jul 07 '13 at 22:10