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.


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}

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ Magical and thanks. $\endgroup$ – omegadot May 13 '20 at 5:29

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

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


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

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.


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


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.