I am looking for a closed form for: $$\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x.$$ I am assuming integration by parts multiple times but I can't get anywhere with it. Any help/ hint would be greatly appreciated. Thanks!
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$$\int_0^1\frac{\text{Li}_3(-x)\text{Li}_2(x)}{x}dx=\sum_{n=1}^\infty\frac{(-1)^n}{n^3}\int_0^1 x^{n-1}\text{Li}_2(x)dx$$
$$=\sum_{n=1}^\infty\frac{(-1)^n}{n^3}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)$$
$$=\zeta(2)\text{Li}_4(-1)-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$$
$$=-\frac{49}{28}\zeta(6)-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$$
Unfortunately there is no known closed form for the latter sum.
Ali Shadhar
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Do you think we will be able to find a closed form for the latter sum any time soon? – Jun 14 '20 at 03:38
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And is this the case for many of the alternating harmonic sums involving an odd exponent? For example $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n^3}$? These seem particularly interesting... – Jun 14 '20 at 03:39
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What I know of that there is no closed form for $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n^a}$ for $a=5,7,...$. – Ali Shadhar Jun 14 '20 at 03:56
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So there is a closed form for $a=3$? If so, has it been found? – Jun 14 '20 at 05:01
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Yep, you can find it evaluated here https://math.stackexchange.com/questions/457371/alternating-harmonic-sum-sum-k-geq-1-frac-1kk3h-k/3219475#3219475 – Ali Shadhar Jun 14 '20 at 08:58
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A conjectural evaluation for the remaining alternating sum is given in equation (19) of https://arxiv.org/abs/1908.04770, namely $$ \sum_{n=1}^{\infty} \frac{(-1)^n H_{n-1}}{n^5} = \frac{1}{13} \left( \frac{1}{3}\operatorname{Li}_6\left(-\frac{1}{8}\right) - 162 \operatorname{Li}_6\left(-\frac{1}{2}\right) - 126 \operatorname{Li}_6\left(\frac{1}{2}\right) \right) -\frac{1787}{624} \zeta(6) + \frac{3}{8} \zeta(3)^2 \\ +\frac{31}{16}\zeta(5) \log(2) - \frac{15}{26} \zeta(4) \log^2(2) + \frac{3}{104} \zeta(2) \log^4(2) - \frac{1}{208} \log^6(2) $$ Subtract this from $-\frac{9}{16} \zeta(6)$ to get the answer for the original question. Note the relation $$ \int_0^1 \frac{\operatorname{Li}_3(-x)\operatorname{Li}_2(x)}{x} \mathrm{d} x = -\frac{9}{16} \zeta(6) - \sum_{n=1}^{\infty} \frac{(-1)^n H_{n-1}}{n^5} = -\frac{49}{32} \zeta(6) - \sum_{n=1}^{\infty} \frac{(-1)^n H_{n}}{n^5} $$ where the denominator in the coefficient of $\zeta(6)$ in the above answer is corrected, the $-\frac{49}{28} \zeta(6)$ there should read $-\frac{49}{32}\zeta(6)$.
guest
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Wow! This is the first conjecture I have seen of that sum. Is there any note on how it was achieved? – Jul 12 '20 at 14:19
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If the conjecture is indeed correct, then many integrals on this site involving $\sum_{n\geqslant 1} \frac{H_n}{n^52^n}$ may be computed. – Jul 12 '20 at 14:24
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Can you mention the page number and equation number of this sum please? – Ali Shadhar Jul 15 '20 at 10:05
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Also, the only conjectural part is the coefficient of Zeta(6). That is, it is a theorem that the identity holds for some rational factor in front of Zeta(6) (all other coefficients are proven). The coefficient of Zeta(6) was fixed numerically. – guest Jul 30 '20 at 21:13
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The proof uses quite abstract machinery (motivic periods). It would be great if anyone would find an elementary proof! – guest Aug 01 '20 at 13:03