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.


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 $1/(1+k)= 1/(k(1+1/k))=1/k(1-1/k+...).$

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  • $\begingroup$ @ 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

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 $\kappa_{a,b}$.


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 $1/n^5$

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


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

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


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

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:

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


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

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

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


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

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 $(a2)$

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

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