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This morning I try to create interesting integrals involving harmonic numbers. See this Wikipedia. And look at, also if you need it, the definition of the Harmonic number $H_x$ using the digamma function from this MathWorld.

After I've asked to Wolfram Alpha online calculator this code

int -log(1-Harmonic(x))log(1+Harmonic(2x))dx, from x=0 to 1

I wondered this

Question. Can you justify a very good approximation of this $$\text{constant}=-\int_0^1\log\left(1-H_x\right)\log\left(1+H_{2x}\right)\,dx\tag{1}$$ using numerical analysis? Or well, can you provide me the expression of previous constant written as a series with a good rate of convergence? Many thanks.

You can to choose some of the two previous approaches. But if a different approach (yours) gets a very good approximation/information of the constant defined in $(1)$, feel free to add it as an answer.

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  • $\begingroup$ I added previous tags because these are good, but this problem is also about (integration). Many thanks. $\endgroup$
    – user243301
    Commented Sep 29, 2017 at 13:29
  • $\begingroup$ Why is this tagged asymptotics? It does not seem to be relevant. $\endgroup$
    – Antonio Vargas
    Commented Sep 29, 2017 at 17:16
  • $\begingroup$ Many thanks, following your experience now I remove this tag. And thanks for your attention and help @AntonioVargas $\endgroup$
    – user243301
    Commented Sep 29, 2017 at 19:44
  • $\begingroup$ I made a Taylor expansion of the integrand around $x=\frac 12$ up to fourth order ; it is just ugly (a bunch of polygamma functions) but, for sure, you can integrate termwise. The numerical value so obtained is $-0.847997$ to be compared to $-0.987235$ obtained by numerical integration. $\endgroup$
    – Claude Leibovici
    Commented Sep 30, 2017 at 9:21
  • $\begingroup$ Many thanks for your effort and contribution @ClaudeLeibovici $\endgroup$
    – user243301
    Commented Sep 30, 2017 at 9:50

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As said in comments, the expansion around $x=\frac 12$ is more than ugly. Moroever, the convergence does not look to ve very fast $$\left( \begin{array}{cc} n & \text{result} \\ 2 & 0.847997 \\ 4 & 0.901222 \\ 6 & 0.925063 \\ 8 & 0.938477 \\ 10 &0.947086 \end{array} \right)$$

A quick and dirty nonlinear regression $$y=\frac{0.619974+0.809459\, n}{1+0.820115\, n}$$ leads to an asymptotic value of $0.987006$.

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  • $\begingroup$ Any case, many thanks for your contribution. $\endgroup$
    – user243301
    Commented Oct 1, 2017 at 9:02
  • $\begingroup$ @user243301. You are very welcome ! My computer gave up trying $n=20$. This problem is awful (as I did approach it). May I ask where did you find it ? $\endgroup$
    – Claude Leibovici
    Commented Oct 1, 2017 at 9:05
  • $\begingroup$ Was a creation when I was inspired in PROBLEMA 124 La Gaceta de la RSME Vol. 13, nº 1, page 82 by Pablo Refolio (2010), with the purpose to write an integral about harmonic numbers $H_x$ or $H_{\lambda x}$. Look at from the web or La Gaceta de la Real Sociedad Matemática Española, and choose the links > english, >All issues, >Volume 13 (2010), >Issue 1 and finally Soluciones a los problemas 121 al 124 y 126. $\endgroup$
    – user243301
    Commented Oct 1, 2017 at 10:28
  • $\begingroup$ On the other hand this journal has very nice articles, but in spanish language, for example is it interesting the following article >english, >Sections, > The Column of Computational Mathematics, and in page 3, choose ¿Podemos fiarnos de los cálculos efectuados con ordenador? by Óscar Ciaurri and Juan Luis Varona (published in La Gaceta de la RSME, volume 9, issue 2 (2006)). $\endgroup$
    – user243301
    Commented Oct 1, 2017 at 10:28

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