4
$\begingroup$

In the solution to the interesting problem Evaluation of $\int_{0}^{1}\int_{0}^{1}\{\frac{1}{\,x}\}\{\frac{1}{x\,y}\}dx\,dy\,$ I found that the existence of a possible closed form depends on the asymptotic behaviour of the following three sums

$$\sigma_{a}(m) = \sum_{k=1}^m \frac{\log(k)}{k+1}$$

$$\sigma_{b}(m) = \sum_{k=1}^m k \log(k+1)\log(k)$$

$$\sigma_{c}(m) = \sum_{k=1}^m H_{k}\log(k)$$

To be more precise we need the asymptotic behaviour to the order of $1/m^3$ and possible logarithmic factors.

This is a challenge question.

Because of a lack of space here I'll show my solution attempts in a self answer. This reveals in more depth where I am stuck and gives a more detailed definition of the question.

$\endgroup$
5
$\begingroup$

Not only is this a challenge problem, it is a huge amount of grunt work. I'll solve only one, $\sigma_b(n),$ but the steps to find it are applicable to the others. The key observation is to break everything into sums with summands of the form $k^m\log{k}$ and $k^m\log^2{k}.$ Why? Because then we can use the asymptotic formula $$ \sum_{k=1}^n k^{-s}=\zeta{(s)}-\frac{1}{(s-1)n^{s-1}}+\frac{1}{2n^s}- \sum_{j=1}^\infty \frac{B_{2j}}{(2j)!}\frac{(s+2j-2)!}{(s-1)!}\frac{1}{n^{s+2j-1}}. $$ The expansion is derivable from the Euler-McLaurin formula. You can take derivatives of it but when evaluating at $s=1$ the appropriate limit must be taken. Because you want terms 'only' to $n^{-3},$ the infinite series can be truncated as follows, $$ \sum_{k=1}^n k^{-s}=\zeta{(s)}-\frac{1}{(s-1)n^{s-1}}+\frac{1}{2n^s}- \Big(\frac{1}{12}\frac{s}{n^{s+1}} + \frac{1}{720}\frac{s(s+1)(s+2)}{n^{s+3}} \Big). $$ We know we can stop here because the largest summand is $k\log^2(k),$ as seen from the expansion of $\sigma_b(n),$ $$ \sigma_b(n)=\sum_{k=2}^n k\,\log{k}\,\log{(k+1)} =\sum_{k=2}^n k\,\log{k} \big(\log{k}+\log{(1+1/k)}\big)$$ $$= \sum_{k=1}^n k\,\log^2{k} + \sum_{k=1}^n k\,\log{k}\Big(\frac{1}{k}-\frac{1}{2k^2}+\frac{1}{3k^3} +...\Big)$$ In the last step, the Taylor series for $\log(1+1/k)$ has been used, good since $k=2,3,...$, but then we turn around and set the starting index to 1 because $\log{(k=1)}$ =0. Now its just a matter of taking derivatives, limits, and keeping the expansion $up\,to\,order \,n^{-3}$. Below is the collection of results, with the shorthand that $L=\log{n}:$ $$ v_0=\sum_{k=1}^n k\,\log^2{k}=\frac{n^2}{2}\big(L^2-L+1/2) + \frac{n}{2}L^2+\frac{L^2}{12}+\frac{L}{6}+\zeta''(-1)-\frac{L}{360n^2}$$ $$ v_1=\sum_{k=1}^n \log{k}=n(L-1)+\frac{L}{2}-\zeta'(0)+\frac{1}{12n}+\frac{1}{360n^2}$$ $$v_2= \sum_{k=1}^n \frac{\log{k}}{k} =\frac{L^2}{2}+\gamma_1+\frac{L}{2n}+\frac{1}{12n^2}(1-L)$$ $$ v_3=\sum_{k=1}^n \frac{\log{k}}{k^2} =-\zeta'(2)-\frac{L+1}{n}+\frac{L}{2n^2}-\frac{1}{6n^3}(L-1/2)$$ $$ v_4=\sum_{k=1}^n \frac{\log{k}}{k^3} =-\zeta'(3)-\frac{L+1/2}{2n^2}+\frac{L}{2n^3}$$ $$ v_5=\sum_{k=1}^n \frac{\log{k}}{k^4} =-\zeta'(4)-\frac{L+1/3}{3n^3}$$ $$ v_6=\sum_{k=1}^n \frac{\log{k}}{k^5} =-\zeta'(5) ...$$ In figuring $v_2$ a limit had to be made and the first Stieltjes constant occurred. It's obvious now that you the add the properly weighted pieces up. For shorthand, let $\tilde{v}_k=v_k$ but without the constant term. Therefore, $$\sigma_b(n)=v_0+v_1-\frac{v_2}{2}+ \frac{\tilde{v}_3}{3}-\frac{\tilde{v}_4}{4}+\frac{\tilde{v}_5}{5} -\sum_{n=2}^\infty \frac{\zeta'(n)}{n+1}(-1)^n $$ Not only is there the $\gamma_1$ and $\zeta''(-1)$ ,($\zeta'(0)$ is known) this analysis introduces the new constant $$ \kappa:=-\sum_{n=2}^\infty \frac{\zeta'(n)}{n+1}(-1)^n =0.27331079196...$$ Let $\kappa^*=\kappa+\zeta''(-1)-\gamma_1/2+\log{(2\pi)}/2.$ The the final result can be stated as $$ \sigma_b(n) = \frac{n^2}{2}\big(L^2-L+\frac{1}{2}\big)+n\big(\frac{L^2}{2}+L-1\big)-\frac{L^2}{6}+\frac{2}{3}L +\kappa^* - \frac{1}{n}\big(\frac{7L}{12}+\frac{1}{4}\big) + $$ $$+\frac{1}{n^2}\big(\frac{119}{360}L + \frac{17}{720}\big) + \frac{1}{n^3}\big(\frac{-89}{360}L+\frac{1}{180}\big) + O(L/n^4).$$ For a numerical test I let $n=10$ and obtained 6 digits agreement. For $\sigma_c(n),$ write $H_k=\log{k}+\gamma+\frac{1}{2k} +...$ and many of the results in this answer can be reused. The $\sigma_c(n)$ should be even easier, with only a single log power, and writing $1/(1+k)= 1/(k(1+1/k))=1/k(1-1/k+...).$

$\endgroup$
  • $\begingroup$ @ skbmoore Impressive derivation of an important result. I'll study the method in detail. $\endgroup$ – Dr. Wolfgang Hintze Aug 23 '18 at 19:11
  • $\begingroup$ @ skbmoore Your answer was really enlightling. I have understood the perfectly laid out method and was able to calculate the necessary asymptotics to present the first closed form in my original answer to math.stackexchange.com/questions/2879699/…. I am having difficulties with $\sigma(c)$ as the infinite sum defining a "constant" does not converge. $\endgroup$ – Dr. Wolfgang Hintze Aug 24 '18 at 22:00
  • $\begingroup$ Try this, Dr. Hintze, and tell me if it works. The idea is to get that 1/n weighting in the infinite sum. $\sum_k H_k\log{k}=\sum_k \log{k} (H_{k+1}-1/(k+1)).$ 2nd term is $\sigma_a(n),$ which apparently you are fine with calculating. The asymptotic expansion for $H_{k+1} = \log(k+1)+\gamma+1/(2(k+1))+...$ Now that you have $\sum_k \log(1+k)\log(k),$ treat it like I did in my answer. The problem is that those terms implied by the ellipses, as well as the one before it, 1/(2(k+1)), need expanding in primitives of 1/k^m. Don't have the resources available weekends to try it; good luck. $\endgroup$ – skbmoore Aug 25 '18 at 22:47
  • $\begingroup$ @ skbmoore Thanks for the proposal. Before starting this I am still checking the methods for simpler cases. BTW I have posted a related question in 2017 but received no answer (math.stackexchange.com/questions/2584246/…). $\endgroup$ – Dr. Wolfgang Hintze Aug 26 '18 at 20:21
  • $\begingroup$ @ skbmoore Thanks for the proposal but I think I have found the solution: the constant is just defined as $\lim_{n\to\infty}\left( \sigma_{c}(n) - (\text{leading terms})(n)\right)$. I had already mentioned this limit as (10) but didn't recognize it and was instead still looking for a definition similar to that of the other two cases. Now this raises the usual question if this Limit can be expressed by known constants (has a closed form). $\endgroup$ – Dr. Wolfgang Hintze Aug 26 '18 at 23:11
1
$\begingroup$

EDIT 31.08.18 / 25.08.18

This is a completely reworked version which includes the results obtained so far. Taken together these can be considered as the complete solution to my question.

It's a pleasure to mention that the answer of skbmoore for $\sigma_{b}$ was a breakthrough for me methodically, and cut the Gordian knot.

The asymptotic expansions of all three sums $\sigma_{a,b,c}$ have been calculated, and corresponding (new?) constants $\kappa_{a,b,c}$ have been found. Two of the constants are defined by convergent sums, $\kappa_{c}$ is a divergent sum which has been given a meaning in form of a double integral in another interesting contribution of skbmoore. I have calculated single integral representation for the other two constants $\kappa_{a,b}$.

Method

I have adopted now a similar method in which the CAS Mathematica is very helpful.

The "generating function" for calculating finite series of the log-power type $k^p \log(k)^q$ ($p$ and $q$ integer, $q\ge 0$) is the sum

$$\nu(n,x) = \sum_{k=1}^n k^x = H_{n,-x}\tag{1}$$

where $H_{n,m}=H_{n}^{(m)}$ is the generalised harmonic number (defined by this equation).

For instance we have

$$\sum_{k=1}^n k \log(k) = \frac{\partial }{\partial x}\nu(n,x) |x\to 1, \sum_{k=1}^n \frac{ \log(k)^2}{k} = \frac{\partial^2}{\partial x^2}\nu(n,x) |x\to -1$$

and so on.

In what follows we shall calculate the asymptotic behaviour for large $n$ of sums using the asymptotic expression of the generating function. Mathematica gives up to order $1/n^5$

$$\nu_{a}(n,x)\simeq \zeta (-x)\\+n^x\left(\frac{n}{x+1}+\frac{1}{2}+\frac{x}{12 n}+\frac{-x^3+3 x^2-2 x}{720 n^3}+\frac{(x-4) (x-3) (x-2) (x-1) x}{30240 n^5}\right) \tag{2}$$

Results

By reduction to log-power series I have confirmed the result of skbmoore for $\sigma_{b}$ and have calculated $\sigma_{a}$. For $\sigma_{c}$ I shall provide a result below which, however, needs to be discussed.

In what follows we set $L = \log(n)$.

1) asymptotic behaviour of $\sigma_{a}$

Here we start writing

$$\frac{\log(k)}{k+1}=\frac{\log(k)}{k \left(1+\frac{1}{k}\right)}= \frac{Log(k)}{k}-\frac{\log(k)}{k^2}+\frac{\log(k)}{k^3}-\frac{\log(k)}{k^4}+\frac{\log(k)}{k^5}-+...\tag{3}$$

This is in fact an asymptotic expansion of $1/(1+k)$ about $k=\infty$ but it is valid already for $k\gt 1$, and, what is even more important, the expansion is convergent.

Now we calculate the asymptotics of these log-power integrals up to the fifth power and add the results which gives finally

$$\sigma_{a}(n) = \left(\frac{1}{2}L^2\right)+\left(\gamma _1-\kappa_{a}\right) +\left(\frac{1}{n}(\frac{3}{2}L+1) -\frac{1}{n^2}(\frac{13}{12}L +\frac{1}{6}) +\frac{1}{n^3}(L+\frac{1}{36}) +O(L/n^4)\right)\tag{4}$$

Where $\gamma _1$ is Stieltjes gamma 1, and the new constant is defined by

$$\kappa_a = - \sum_{k=2}^\infty (-1)^k \zeta'(k)\simeq 0.788531 \tag{5}$$

Originally, the expansion (3) up to the fifth degree resulted in this expression instead of $\kappa_{a}$

$$\zeta '(2)-\zeta '(3)+\zeta '(4)-\zeta '(5)$$

which was then extended by adding more orders in (3) and finally lead to $\kappa_{a}$

It is interesting that there is an integral representation of $\kappa_{a}$.

Euler's formula for the $\zeta$-function is

$$\zeta (k)=\frac{1}{\Gamma (k)}\int_0^{\infty } \frac{t^{k-1}}{e^t-1} \, dt$$

Differentiating with respect to $k$ under the integral gives for the integrand

$$\zeta'(k) = \frac{t^{k-1} (\log (t)-\psi ^{(0)}(k))}{\left(e^t-1\right) \Gamma (k)}$$

Now luckily the sum $-\sum_{k=2}^\infty (-1)^k \zeta'(k)$ can be performed explicitly so that we obtain the integral representation

$$\kappa_{a} = \int_{0}^\infty \frac{e^{-t} \left(\gamma e^t+\log (-t)+e^t \log (t)-\log (t)+\Gamma (0,-t)\right)}{e^t-1}\tag{6}$$

where $\Gamma (a,z)$ is the incomplete $\Gamma$-function. The integral is numerically in agreement with the sum. The imaginary part from $\log(-t)$ is cancelled by that of $\Gamma (0,-t)$.

2) Calculation of $\sigma_{b}$

This was first done in the answer of skbmoore and checked by myself. For reference I repeat those results here.

The asymptotic expansion is

$$ \sigma_b(n) =\big(\kappa_{b}+\zeta''(-1)-\gamma_1/2+\log{(2\pi)}/2\big)\\+ \frac{n^2}{2}\big(L^2-L+\frac{1}{2}\big)+n\big(\frac{L^2}{2}+L-1\big)-\frac{L^2}{6}+\frac{2}{3}L \\- \frac{1}{n}\big(\frac{7L}{12}+\frac{1}{4}\big) + \frac{1}{n^2}\big(\frac{119}{360}L + \frac{17}{720}\big) + \frac{1}{n^3}\big(\frac{-89}{360}L+\frac{1}{180}\big) + O(L/n^4).$$

The constant is given by the convergent sum

$$\kappa_{b} = -\sum_{k=2}^\infty (-1)^k \frac{1}{1+k} \zeta'(k)\simeq 0.27331079196...$$

The integral representation is

$$\kappa_{b} = \int_{0}^\infty f_{b}(t)\,dt$$

with

$$f_{b}(t) = \frac{1}{2 \left(e^t-1\right) t^2}\\\Big(e^{-t} \left(e^t \left(2 \text{Ei}(-t)+\gamma \left(t^2-4\right)+\left(t^2-4\right) \log (t)-2 t+2\right)+2 (t+1) \text{Ei}(t)-2\right)\Big)$$

Here $\text{Ei}$ is the exponential integral function defined by

$$\text{Ei(z)} = -P\int_{-z}^{\infty } \frac{\exp (-t)}{t} \, dt$$

where the principal value has to be taken.

To derive $f_{b}(t)$ we have to calculate $-\sum_{k=2}^\infty (-1)^k \frac{1}{1+k}\zeta'(k)$ with \zeta'(k) given above. We generate the denominator by $\frac{1}{1+k} = \int_0^1 x^k \,dx$, perform the $k$-sum, and then do the $x$-integral (which Mathematica can do as an indefinte integral of which we take limits). The result is $f_{b}(t)$.

3) Calculation of $\sigma_{c}$

Here the summand is $\log(k) H_{k}$. Proceeding as in $\sigma_{a}$ we have to look for the asymptotics of the cofactor of the $\log$.

We have

$$H_k\simeq \log ^2(k)+\log (k) \left(\gamma+\frac{1}{2 k}-\frac{1}{12 k^2} +\frac{1}{120 k^4} -\frac{1}{252 k^6}\right)\tag{7}$$

Employing the Bernoulli numbers $B_{k}$ this can also be formally written as

$$H_k\simeq \log ^2(k)+\log (k) \left(\gamma +\frac{1}{2 k}+\sum _{m=1}^\infty \frac{B_{2 m}}{(2 m) k^{2 m}}\right)$$

The sum is divergent, hence it must be terminated after a finite number of terms or treated differently.

Proceedings with the log-power integrals as before we arrive at

$$\sigma_{c}(n) = \left(\kappa_{c}\right)+\left(\frac{3 \gamma _1}{2}+\frac{\gamma ^2}{2}-\frac{\pi ^2}{24}-\frac{1}{2} \log ^2(2 \pi )+\frac{1}{2} \gamma \log (2 \pi )\right)\\+\left(L^2 n+\frac{3 L^2}{4}+(\gamma -2) (L-1) n+\frac{\gamma L}{2}\right)+O(L/n)\tag{8}$$

The first bracket was originally with the expansion (7) up to 6th order given by

$$\frac{1}{2} B(2) \zeta '(2)+\frac{1}{4} B(4) \zeta '(4)+\frac{1}{6} B(6) \zeta '(6)$$

Extending the expansion (7) would lead to a suspected constant of the form

$$\kappa_{c} {\dot=} \sum _{m=1}^\infty \frac{\zeta'(2m) B_{2 m}}{(2 m)} \tag{9}$$

In contrast to that in $\sigma_{a}$ and $\sigma_{b}$ the series is not convergent, and the question arises: what is the constant, how many terms do we have to take?

Fortunately, just in time someone else rephrased my problem (Constant term in Stirling type formula for $\sum^N_{n=1} H_n \cdot \ln(n)$) and skbmoore in an answer ingeniously found a valid interpretation of the divergent sum in terms of this integral:

$$\kappa_{c,i}=\int_0^\infty \frac{dt/t}{e^t-1}\Big(\, \log{t}\big(\frac{t}{e^t-1}-1+t/2\big) - \Psi(t)\Big) = -0.077596...\tag{9a}$$

here

$$\Psi(t)=-\int_0^t \frac{\log{(1-u/t)}}{e^u-1}\Big(1-\frac{u\,e^u}{e^u-1}\Big)\,du- \gamma\Big(\frac{t}{e^t-1} - 1\Big)+ \big(1-\frac{\gamma}{2}\big)t + \log{\big(\frac{t}{e^t-1}\big) }\tag{9b}$$

Hence the expression is in fact a double integral.

Alternatively, as stated already in my version of 25.08.18, the overall constant can be found by this limit

$$(\text{overall constant = $\kappa_{c}$ + 2nd bracket of (8)})\\=\lim_{n\to\infty}\big( \sigma_{c}(n) -\text{ (leading terms = 3rd bracket of (8))}\big)\tag{10}$$

Original post as of 23.08.18

This self answer shows the attemps I made to solve the problems.

1) $\sigma_{a}(m)=\sum_{k=1}^m \frac{\log(k)}{k+1}$

Writing $\frac{1}{1+k}=\int_0^1 x^k \,dx$, $\log(k) = (\frac{\partial }{\partial t}k^t)|t\to 0$

and defining

$$\mu (x,t,m) = \sum _{k=1}^m x^k t^k \tag{a1} $$

we have

$$\sigma_{a}(m) =\frac{\partial}{\partial t}\left( \int_{0}^1 \mu(x,t,m) \,dx \right)| t \to 0 \tag{a2}$$.

The "kernel" function $\mu$ can be expressed by standard functions

$$\mu(x,t,m) = \text{Li}_{-t}(x)-x^{m+1} \Phi (x,-t,m+1)$$

Here

$$\Phi(z,s,a) = \sum_{k=0}^\infty \frac{x^k}{(k+a)^s}$$

is the Lerch transcendent and

$$\text{Li}_{\alpha}(x) = \sum_{k=1}^\infty z^k/k^\alpha$$

is the polylog function.

The strategy now could be to investigate the asymptotic behaviour of the kernel. But here I am stuck.

2) $\sigma_{b}(m) = \sum_{k=1}^m k \log(k) \log(k+1)$

I have tried some Abelian partial summation but with no avail. In the meantime an answer to this sum was given.

3) $\sigma_{c}(m) = \sum_{k=1}^m H_{k} \log(k)$

This is strongly related to 1).

Observing the definition

$$H_{k} = \int_{0}^1 \frac{1-x^k}{1-x}\,dx$$

and $\sum_{k=1}^m log(k) = \log(m!)$ we find

$$\sigma_{c}(m)=\int_0^1 \frac{\,dx}{1-x}\left(\log(m!) -\frac{\partial}{\partial t} \mu(x,t,m) \right)|t\to 0\tag{c1}$$

where $\mu$ is defined in $(a2)$

$\endgroup$
  • $\begingroup$ Interesting thread. Definitely worth a more detailed study. (+1) $\endgroup$ – Markus Scheuer Aug 31 '18 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.