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My question is:

Can a closed-form expression for the following alternating quadratic Euler sum be found? Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$. $$S = \sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$$

What I have managed to do so far is to convert $S$ to two rather difficult integrals as follows.

Starting with the result $$\frac{H_{2n}}{2n} = -\int_0^1 x^{2n - 1} \ln (1 - x) \, dx \tag1$$ Multiplying (1) by $(-1)^n H_n/n$ then summing the result from $n = 1$ to $\infty$ gives $$S = -2 \int_0^1 \frac{\ln (1 - x)}{x} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n}. \tag2$$ From the following generating function for the harmonic numbers $$\sum_{n = 1}^\infty \frac{H_n x^n}{n} = \frac{1}{2} \ln^2 (1 - x) + \operatorname{Li}_2 (x),$$ replacing $x$ with $-x^2$ leads to $$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n} = \frac{1}{2} \ln^2 (1 + x^2) + \operatorname{Li}_2 (-x^2).$$ Substituting this result into (2) yields $$S = -2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx,$$ or, after integrating the first of the integrals by parts twice $$S = -\frac{5}{2} \zeta (4) + 4 \zeta (3) \ln 2 - 8 \int_0^1 \frac{x \operatorname{Li}_3 (x)}{1 + x^2} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx. \tag3$$

I have a slim hope the first of these integrals can be found (I cannot find it). As for the second of the integrals, it is proving to be a little difficult.

Can someone find each of the integrals appearing in (3)? Or perhaps an alternative approach to the sum will deliver the closed-form I seek, I am fine either way.


Update

Thanks to Ali Shather, the first of the integrals can be found. Here \begin{align} \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \ dx &=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{2n-1}\ln(1-x)\ dx\\ &= -\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n^3}\\ &=-4\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^3}\\ &=-4 \operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3}. \end{align} And using the result I calculated here, namely $$\operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3} = \frac{5}{8} \operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{195}{256} \zeta (4) + \frac{5}{192} \ln^4 2 - \frac{5}{32} \zeta (2) \ln^2 2 + \frac{35}{64} \zeta (3) \ln 2,$$ gives \begin{align} \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx &= -\frac{5}{2} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{195}{64} \zeta (4) - \frac{5}{48} \ln^4 2\\ & \qquad + \frac{5}{8} \zeta (2) \ln^2 2 - \frac{35}{16} \zeta (3) \ln 2. \end{align}

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    $\begingroup$ a little difficult seems to be a real understatement. I found quite good explicit approximations using for example $$\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}=\sum_{n = 1}^5 \frac{(-1)^n H_n H_{2n}}{n^2}+\sum_{n = 6}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$$ and expanding the last summand as series. Using expansions to $O(1/n^5)$, I obtained $-1.014450$ while the exact value is $-1.014452$ $\endgroup$ – Claude Leibovici Feb 19 '20 at 7:50
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    $\begingroup$ The first integral $$\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \ dx=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{2n-1}\ln(1-x)\ dx$$ $$=-\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n^3}=-4\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^3}=-4\Re \sum_{n=1}^\infty i^n\frac{H_n}{n^3}$$ and I think you already calculated this sum before. $\endgroup$ – Ali Shadhar Feb 19 '20 at 8:53
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    $\begingroup$ @Ali Shather Nice one. And yes, I have calculated this sum before. Now to the second integral which is an entirely different beast. $\endgroup$ – omegadot Feb 19 '20 at 8:58
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    $\begingroup$ After long calulations I find:$$\int_0^1 \frac{\ln (1 + x) \ln^2 (1 + x^2)}{x} \, dx,=-25\text{Li}_4(1/2) -\frac{229\ln 2}8\zeta(3) +\frac{2959 \pi^4}{11520} +\frac{91\pi^2\ln^2 2}{48} -\frac{51\ln^4 2}{16}+2{G^2}+\frac{3}{2}{\pi}G\ln{2}-\int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{1+x} \ dx+8\int_0^1 \frac{\ln (1 + x)\ln(1-x) \ln (1 + x^2)}{1+x} \ dx$$ $\endgroup$ – user178256 Feb 26 '20 at 10:50
  • $\begingroup$ @Edit profile and settings $\int_0^1 \frac{\ln (1 + x) \ln^2 (1 + x^2)}{x} \, dx+\int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{1+x} \, dx=-4 \pi \Im(\text{Li}_3(1+i))-7 \text{Li}_4\left(\frac{1}{2}\right)+\frac{5}{4} \zeta (3) \log (2)+\frac{641 \pi ^4}{3840}+\frac{7}{48} \log ^4(2)+\frac{5}{16} \pi ^2 \log ^2(2)+{2}{C^2}-\frac{1}{2} \pi C \log (2)$ $\endgroup$ – user178256 May 12 '20 at 8:30
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Using your integral representation, the sum equals to: $$\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}= -2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx$$ $$\small=-2 C^2+2 \pi C \log (2)-4 \pi \Im(\text{Li}_3(1+i))+3 \text{Li}_4\left(\frac{1}{2}\right)+\frac{21}{8} \zeta (3) \log (2)+\frac{487 \pi ^4}{5760}+\frac{\log ^4(2)}{8}+\frac{1}{8} \pi ^2 \log ^2(2)$$ For the second integral and its derivation, see here.

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  • $\begingroup$ Magical and thanks. $\endgroup$ – omegadot May 13 '20 at 5:29
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Remark: I have noticed too late that this integral has already been solved (in the update of omegadot).

Nevertheless I don't delete the contribution because, together with this information, it shows that the hypergeometric functions appearing here can be simplified appreciably which gives hope for other cases.

Original post

A closed expression of the integral

$$i = \int_0^1 \frac{x \operatorname{Li}_3(x)}{x^2+1}\tag{1}$$

can be found in terms of (sorry Ali) hypergeometric function as follows.

Partial integration gives

$$i=s_{0}-\int_0^1 \frac{\text{Li}_2(x) \log \left(x^2+1\right)}{2 x} \, dx\tag{2a}$$

where

$$s_0 = \frac{1}{2} \zeta (3) \log (2)\tag{2b}$$

Expanding the denominator of the integrand we find that $i=s_{0}+\sum a_{k}$ where

$$a_{k} =-\frac{1}{2} \int_0^1 \frac{(-1)^{k+1} x^{2 k-1} \text{Li}_2(x)}{k} \, dx=-\frac{(-1)^{k+1} \left(\pi ^2 k-3 H_{2 k}\right)}{24 k^3}\tag{3}$$

The two sums are

$$s_{1}=\frac{1}{24} \left(-\pi ^2\right) \sum _{k=1}^{\infty } \frac{(-1)^{k+1}}{k^2}=-\frac{\pi ^4}{288}\tag{4}$$

$$s_{2} = +\frac{1}{8} \sum _{k=1}^{\infty } \frac{(-1)^{k+1} H_{2 k}}{k^3}=\frac{1}{32} \left(-2 \,_P\tilde{F}_Q^{(\{0,0,0,0\},\{0,0,1\},0)}(\{1,1,1,1\},\{2,2,2\},-1)\\-\sqrt{\pi } \,_P\tilde{F}_Q^{(\{0,0,0,0,0\},\{0,0,0,1\},0)}\left(\left\{1,1,1,1,\frac{3}{2}\right\},\left\{2,2,2,\frac{3}{2}\right\},-1\right)\\+3 \zeta (3) (\gamma +\log (2))\right)\tag{5}$$

Where $\,_P\tilde{F}_Q$ is the regularized hypergeometric function. For more details see https://math.stackexchange.com/a/3544006/198592.

Two terms appear in $s_{2}$ due to the relation

$$H_{2 k}=\frac{1}{2} \left( H_{k-\frac{1}{2}}+ H_k \right)+\log (2)$$

The complete integral is then given by

$$i = s_{0}+s_{1}+s_{2}$$

The numeric check shows good agreement.

Discussion

I'm almost sure that the sum

$$\sum _{k=1}^{\infty } \frac{(-1)^{k+1} H_k}{k^3}$$

has a simpler expression, and so might

$$\sum _{k=1}^{\infty } \frac{(-1)^{k+1} H_{k-\frac{1}{2}}}{k^3}$$

and I'd be happy to replace the hypergeometric constructs.

No need to conjecture: omegadot has done it, see https://math.stackexchange.com/a/3290607/198592

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