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We know a closed-form of the first two powers of the integral of trilogarithm function between $0$ and $1$. From the result here we know that

$$I_1=\int_0^1 \operatorname{Li}_3(x)\,dx = \zeta(3)-\frac{\pi^2}{6}+1.$$

From here we also know that

$$I_2=\int_0^1 \operatorname{Li}_3^2(x)\,dx = 20-8\zeta(2)-10\zeta(3)+\frac{15}{2}\zeta(4)-2\zeta(2)\zeta(3)+\zeta^2(3).$$

Is there a closed-form of

$$I_3=\int_0^1 \operatorname{Li}_3^3(x)\,dx$$ and $$I_4=\int_0^1 \operatorname{Li}_3^4(x)\,dx\,?$$

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    $\begingroup$ I suspect that this is going to increase in complexity like the general formulae for first, second, third and fourth order polynomials do. Not that it will stop at 4 but that it is going to be big and look like a random rational combination of zeta functions $\endgroup$ – Ali Caglayan Oct 20 '14 at 0:32
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    $\begingroup$ I summon Cleo... $\endgroup$ – Start wearing purple Oct 20 '14 at 0:35
  • $\begingroup$ You can follow the same arguments from the paper to evaluate the integral(use integrals by parts and create a recurrence relation), but it takes a lot of time to do it even for $I_3$. It would be interesting if someone came up with a different idea. $\endgroup$ – karvens Oct 21 '14 at 1:27
  • $\begingroup$ @user153012 Do you need proof or would the answer suffice? $\endgroup$ – Andrew Oct 26 '14 at 19:15
  • $\begingroup$ @Andrew Feel free to post you result. $\endgroup$ – user153012 Oct 26 '14 at 20:24
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Numerically, $$ I_3 = 27 \zeta (5) \zeta (2)-3 \zeta (3)^2 \zeta (2)+156 \zeta (3) \zeta (2)-\tfrac{153}{8} \zeta (7)-90 \zeta (5)+\zeta (3)^3+21 \zeta (3)^2-660 \zeta (3)-420 \zeta (2)-315 \zeta (4)-\tfrac{477}{4} \zeta (6)+1680 $$ and $I_4$ doesn't seem to have a similar closed form.

The same approach that I used in this answer to decompose the integral as a sum of multiple zeta values works here as well because of the integral form $$ \mathrm{Li}_3(x) = \int_0^x \frac{dt}{t} \int_0^t \frac{du}{u} \int_0^u \frac{du}{1-u}, $$ so it's possible to write $\int_0^1 \mathrm{Li}_3(x)^4\,dx$ as an iterated 13-fold integral, but that looks a little tedious to apply it by hand.

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  • $\begingroup$ Thank you for your answer Kirill! I hope at the end of the analytically solution by @David H we will see the same expression for $I_3$ as yours. $\endgroup$ – user153012 Oct 27 '14 at 6:17
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(In Progress)

For $I_3$ at least, an alternative to evaluating via repeated integration by parts is to expand the integral as a series.

For $|x|<1$ we have the power series representation of the trilogarithm,

$$\operatorname{Li}_{3}{\left(x\right)}=\sum_{k=1}^{\infty}\frac{x^k}{k^3}\tag{1}.$$

Expanding one of the trilogarithm factors with series $(1)$, and then pulling the sum outside the integral, we have the following series representation of $I_3$:

$$\begin{align} I_3 &=\int_{0}^{1}\operatorname{Li}_{3}^{3}{\left(x\right)}\,\mathrm{d}x\\ &=\int_{0}^{1}\sum_{k=1}^{\infty}\frac{x^k}{k^3}\operatorname{Li}_{3}^{2}{\left(x\right)}\,\mathrm{d}x\\ &=\sum_{k=1}^{\infty}\frac{1}{k^3}\int_{0}^{1}x^k\operatorname{Li}_{3}^{2}{\left(x\right)}\,\mathrm{d}x\\ &=\sum_{k=1}^{\infty}\frac{1}{k^3}J{\left(k,3,3\right)}\tag{2},\\ \end{align}$$

where in the last line of $(2)$ we've substituted the notation from Freitas' paper for integrals of the form,

$$J{\left(m,p,q\right)}:=\int_{0}^{1}x^m\operatorname{Li}_{p}{\left(x\right)}\operatorname{Li}_{q}{\left(x\right)}.$$

Lemma 3.5 from Freitas states that for $p,q\ge 2$ and $m\ge 0$, the following recurrence relation holds:

$$\begin{cases} J{\left(m,p,q\right)}=\frac{\zeta{(p)}\zeta{(q)}}{m+1}-\frac{1}{m+1}\left[J{\left(m,p-1,q\right)}+J{\left(m,p,q-1\right)}\right];\\ J{\left(m,1,q\right)}=\frac{\zeta{(q)}}{m+1}-\frac{1}{m+1}\left[mJ_{0}{(m,q)}+J_{0}{(m,q-1)}\right]\\ ~~~~~~~~~~~~~~~~~~~~~+\frac{1}{m+1}\left[mJ{\left(m-1,1,q\right)}+J{\left(m-1,1,q-1\right)}-J{\left(m,1,q-1\right)}\right];\\ J_{0}{(m,q)}=\frac{\zeta{(q)}}{m+1}-\frac{1}{m+1}J_{0}{(m,q-1)};\tag{3}\\ \end{cases}$$

where $J_{0}{(m,q)}:=\int_{0}^{1}x^m\operatorname{Li}_{q}{\left(x\right)}\,\mathrm{d}x$.

Thus, we have:

$$\begin{align} J{\left(k,3,3\right)} &=\frac{\zeta^{2}{(3)}}{k+1}-\frac{1}{k+1}\left[J{\left(k,2,3\right)}+J{\left(k,3,2\right)}\right]\\ &=\frac{\zeta^{2}{(3)}}{k+1}-\frac{2}{k+1}J{\left(k,2,3\right)}\\ &=\frac{\zeta^{2}{(3)}}{k+1}-\frac{2}{k+1}\left(\frac{\zeta{(2)}\zeta{(3)}}{k+1}-\frac{1}{k+1}\left[J{\left(k,1,3\right)}+J{\left(k,2,2\right)}\right]\right)\\ &=\frac{\zeta^{2}{(3)}}{k+1}-\frac{2\zeta{(2)}\zeta{(3)}}{(k+1)^2}+\frac{2}{(k+1)^2}J{\left(k,1,3\right)}+\frac{2}{(k+1)^2}J{\left(k,2,2\right)}\\ &=\frac{\zeta^{2}{(3)}}{k+1}-\frac{2\zeta{(2)}\zeta{(3)}}{(k+1)^2}+\frac{2}{(k+1)^2}J{\left(k,1,3\right)}\\ &~~~~~ +\frac{2}{(k+1)^2}\left(\frac{\zeta^{2}{(2)}}{k+1}-\frac{2}{k+1}J{\left(k,1,2\right)}\right)\\ &=\frac{\zeta^{2}{(3)}}{k+1}-\frac{2\zeta{(2)}\zeta{(3)}}{(k+1)^2}+\frac{2\zeta^{2}{(2)}}{(k+1)^3}+\frac{2}{(k+1)^2}J{\left(k,1,3\right)}-\frac{4}{(k+1)^3}J{\left(k,1,2\right)}.\\ \end{align}$$

Hence, we can write $I_3$ as a sum of five series:

$$\begin{align} I_3 &=\sum_{k=1}^{\infty}\frac{J{\left(k,3,3\right)}}{k^3}\\ &=\sum_{k=1}^{\infty}\frac{\zeta^{2}{(3)}}{k^3(k+1)}-\sum_{k=1}^{\infty}\frac{2\zeta{(2)}\zeta{(3)}}{k^3(k+1)^2}+\sum_{k=1}^{\infty}\frac{2\zeta^{2}{(2)}}{k^3(k+1)^3}+\sum_{k=1}^{\infty}\frac{2\,J{\left(k,1,3\right)}}{k^3(k+1)^2}\\ &~~~~~ -\sum_{k=1}^{\infty}\frac{4\,J{\left(k,1,2\right)}}{k^3(k+1)^3}\\ &=\zeta^{2}{(3)}\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)}-2\zeta{(2)}\zeta{(3)}\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)^2}+2\zeta^{2}{(2)}\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)^3}\\ &~~~~~ +2\sum_{k=1}^{\infty}\frac{J{\left(k,1,3\right)}}{k^3(k+1)^2}-4\sum_{k=1}^{\infty}\frac{J{\left(k,1,2\right)}}{k^3(k+1)^3}.\tag{4}\\ \end{align}$$

The first three sums are routine calculations, so I'll simply quote their values and relegate their derivations to an appendix:

$$\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)}=1-\zeta{(2)}+\zeta{(3)};$$

$$\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)^2}=4-3\zeta{(2)}+\zeta{(3)};$$

$$\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)^3}=10-6\zeta{(2)}.$$

After using the three sums above to simply $(4)$, we're left with a representation of $I_3$ as a sum of products of zeta values plus the remaining two series:

$$I_3=20\zeta^2{(2)}-8\zeta{(2)}\zeta{(3)}+\zeta^2{(3)}-12\zeta^3{(2)}+6\zeta^2{(2)}\zeta{(3)}-3\zeta{(2)}\zeta^2{(3)}+\zeta^3{(3)}\\ +2\sum_{k=1}^{\infty}\frac{J{\left(k,1,3\right)}}{k^3(k+1)^2}-4\sum_{k=1}^{\infty}\frac{J{\left(k,1,2\right)}}{k^3(k+1)^3}.\tag{5}$$

The remaining two series are trickier because they involve the integrals $J{\left(k,1,2\right)}$ and $J{\left(k,1,3\right)}$, and unfortunately, for integrals of type $J{\left(k,1,q\right)}$, where $k\ge 0$ and $q\ge 2$, the recurrence relation $(3)$ relates these integrals to similar integrals of lower order in both $k$ and $q$ instead of in $q$ alone (which would be much easier to take advantage of). So it seems these integrals will have to evaluated directly, and the best way to accomplish that is probably to begin by integrating by parts.

Note that $\operatorname{Li}_{1}{\left(x\right)}=-\ln{\left(1-x\right)}$. In preparation for integrating by parts, we'll need the following indefinite integral:

$$\begin{align} -\int x^k\ln{\left(1-x\right)}\,\mathrm{d}x &=-\frac{x^{k+1}}{k+1}\ln{\left(1-x\right)}-\int\frac{x^{k+1}}{(k+1)(1-x)}\,\mathrm{d}x\\ &=-\frac{x^{k+1}}{k+1}\ln{\left(1-x\right)}+\frac{1}{k+1}\int\frac{x^{k+1}}{x-1}\,\mathrm{d}x\\ &=-\frac{x^{k+1}}{k+1}\ln{\left(1-x\right)}+\frac{1}{k+1}\int\frac{x^{k+1}-1+1}{x-1}\,\mathrm{d}x\\ &=-\frac{x^{k+1}}{k+1}\ln{\left(1-x\right)}-\frac{1}{k+1}\int\frac{1}{1-x}\,\mathrm{d}x+\frac{1}{k+1}\int\frac{x^{k+1}-1}{x-1}\,\mathrm{d}x\\ &=-\frac{x^{k+1}}{k+1}\ln{\left(1-x\right)}+\frac{1}{k+1}\ln{\left(1-x\right)}+\frac{1}{k+1}\int\sum_{j=0}^{k}x^j\,\mathrm{d}x\\ &=\frac{1-x^{k+1}}{k+1}\ln{\left(1-x\right)}+\frac{1}{k+1}\sum_{j=0}^{k}\int x^j\,\mathrm{d}x\\ &=\frac{1-x^{k+1}}{k+1}\ln{\left(1-x\right)}+\frac{1}{1+k}\sum_{j=0}^{k}\frac{x^{j+1}}{j+1}+\color{grey}{constant}\\ &=\frac{1-x^{k+1}}{k+1}\ln{\left(1-x\right)}+\frac{1}{1+k}\sum_{j=1}^{k+1}\frac{x^{j}}{j}+\color{grey}{constant}.\\ \end{align}$$

We'll also need the derivative,

$$\frac{\partial}{\partial x}\operatorname{Li}_{q}{\left(x\right)}=\frac{\operatorname{Li}_{q-1}{\left(x\right)}}{x}.$$

Thus, after integration by parts, we have for positive integers $k,q\in\mathbb{Z}^+$ where $q\ge 2$:

$$\begin{align} J{\left(k,1,q\right)} &=\int_{0}^{1}x^k\operatorname{Li}_{1}{\left(x\right)}\operatorname{Li}_{q}{\left(x\right)}\\ &=-\int_{0}^{1}x^k\ln{\left(1-x\right)}\operatorname{Li}_{q}{\left(x\right)}\\ &=\left[\left(\frac{1-x^{k+1}}{k+1}\ln{\left(1-x\right)}+\frac{1}{1+k}\sum_{j=1}^{k+1}\frac{x^{j}}{j}\right)\operatorname{Li}_{q}{\left(x\right)}\right]_{0}^{1}\\ &~~~~~ -\int_{0}^{1}\left(\frac{1-x^{k+1}}{k+1}\ln{\left(1-x\right)}+\frac{1}{1+k}\sum_{j=1}^{k+1}\frac{x^{j}}{j}\right)\frac{\operatorname{Li}_{q-1}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=\frac{\operatorname{Li}_{q}{\left(1\right)}}{1+k}\sum_{j=1}^{k+1}\frac{1}{j}\\ &~~~~~ -\int_{0}^{1}\left(\frac{1-x^{k+1}}{k+1}\ln{\left(1-x\right)}+\frac{1}{1+k}\sum_{j=1}^{k+1}\frac{x^{j}}{j}\right)\frac{\operatorname{Li}_{q-1}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=\frac{\zeta{\left(q\right)}H_{k+1}}{1+k}\\ &~~~~~ -\int_{0}^{1}\left(\frac{1-x^{k+1}}{k+1}\ln{\left(1-x\right)}+\frac{1}{1+k}\sum_{j=1}^{k+1}\frac{x^{j}}{j}\right)\frac{\operatorname{Li}_{q-1}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=\frac{\zeta{\left(q\right)}H_{k+1}}{1+k}-\int_{0}^{1}\frac{1-x^{k+1}}{k+1}\ln{\left(1-x\right)}\frac{\operatorname{Li}_{q-1}{\left(x\right)}}{x}\,\mathrm{d}x\\ &~~~~~ -\int_{0}^{1}\frac{1}{1+k}\sum_{j=1}^{k+1}\frac{x^{j-1}}{j}\operatorname{Li}_{q-1}{\left(x\right)}\,\mathrm{d}x\\ &=\frac{\zeta{\left(q\right)}H_{k+1}}{1+k}-\frac{1}{k+1}\int_{0}^{1}\left(1-x^{k+1}\right)\ln{\left(1-x\right)}\frac{\operatorname{Li}_{q-1}{\left(x\right)}}{x}\,\mathrm{d}x\\ &~~~~~ -\frac{1}{k+1}\int_{0}^{1}\sum_{j=0}^{k}\frac{x^{j}}{j+1}\operatorname{Li}_{q-1}{\left(x\right)}\,\mathrm{d}x\\ &=\frac{\zeta{\left(q\right)}H_{k+1}}{1+k}-\frac{1}{k+1}\int_{0}^{1}\frac{\ln{\left(1-x\right)}\operatorname{Li}_{q-1}{\left(x\right)}}{x}\,\mathrm{d}x\\ &~~~~~ +\frac{1}{k+1}\int_{0}^{1}x^{k}\ln{\left(1-x\right)}\operatorname{Li}_{q-1}{\left(x\right)}\,\mathrm{d}x\\ &~~~~~ -\frac{1}{k+1}\sum_{j=0}^{k}\frac{1}{j+1}\int_{0}^{1}x^{j}\operatorname{Li}_{q-1}{\left(x\right)}\,\mathrm{d}x\\ &=\frac{\zeta{\left(q\right)}H_{k+1}}{1+k}+\frac{1}{k+1}\int_{0}^{1}\frac{\operatorname{Li}_{1}{\left(x\right)}\operatorname{Li}_{q-1}{\left(x\right)}}{x}\,\mathrm{d}x\\ &~~~~~ -\frac{1}{k+1}\int_{0}^{1}x^{k}\operatorname{Li}_{1}{\left(x\right)}\operatorname{Li}_{q-1}{\left(x\right)}\,\mathrm{d}x\\ &~~~~~ -\frac{1}{k+1}\sum_{j=0}^{k}\frac{1}{j+1}\int_{0}^{1}x^{j}\operatorname{Li}_{q-1}{\left(x\right)}\,\mathrm{d}x\\ &=\small{\frac{\zeta{\left(q\right)}H_{k+1}}{1+k}+\frac{J{\left(-1,1,q-1\right)}}{k+1}-\frac{J{\left(k,1,q-1\right)}}{k+1}-\frac{1}{k+1}\sum_{j=0}^{k}\frac{J_{0}{\left(j,q-1\right)}}{j+1}}.\tag{6}\\ \end{align}$$

The recurrence relation we have developed in the last line of $(6)$ above is much more useful for our purposes than what we get from $(3)$.

Letting $q=3$ in $(6)$ gives us the relation,

$$J{\left(k,1,3\right)}=\frac{\zeta{(3)}H_{k+1}}{1+k}+\frac{J{\left(-1,1,2\right)}}{k+1}-\frac{J{\left(k,1,2\right)}}{k+1}-\frac{1}{k+1}\sum_{j=0}^{k}\frac{J_{0}{\left(j,2\right)}}{j+1}.\tag{7}$$

The integral represented by the numerator of the second term of $(7)$ is easy to evaluate:

$$\begin{align} J{\left(-1,1,2\right)} &=\int_{0}^{1}\frac{\operatorname{Li}_{1}{\left(x\right)}\operatorname{Li}_{2}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=-\int_{0}^{1}\frac{\ln{\left(1-x\right)}\operatorname{Li}_{2}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=\left[\frac{\operatorname{Li}_{2}^2{\left(x\right)}}{2}\right]_{0}^{1}\\ &=\frac12\zeta^2{(2)}=\frac54\zeta{(4)}=\frac{\pi^4}{72}. \end{align}$$

The integral represented by the numerator of the summand in the last term of $(7)$ can be evaluated as well:

$$\begin{align} J_{0}{\left(j,2\right)} &=\int_{0}^{1}x^j\operatorname{Li}_{2}{\left(x\right)}\,\mathrm{d}x\\ &=\left[\frac{x^{j+1}}{j+1}\operatorname{Li}_{2}{\left(x\right)}\right]_{0}^{1}+\int_{0}^{1}\frac{x^{j}\ln{\left(1-x\right)}}{j+1}\,\mathrm{d}x\\ &=\frac{\operatorname{Li}_{2}{\left(1\right)}}{j+1}+\frac{1}{j+1}\int_{0}^{1}x^{j}\ln{\left(1-x\right)}\,\mathrm{d}x\\ &=\frac{\zeta{(2)}}{j+1}+\frac{1}{j+1}\left[\frac{x^{j+1}-1}{j+1}\ln{\left(1-x\right)}-\frac{1}{j+1}\sum_{\ell=1}^{j+1}\frac{x^{\ell}}{\ell}\right]_{0}^{1}\\ &=\frac{\zeta{(2)}}{j+1}-\frac{1}{(j+1)^2}\sum_{\ell=1}^{j+1}\frac{1}{\ell}\\ &=\frac{\zeta{(2)}}{j+1}-\frac{H_{j+1}}{(j+1)^2}.\\ \end{align}$$

Thus, we have the finite sum:

$$\begin{align} \sum_{j=0}^{k}\frac{J_{0}{\left(j,2\right)}}{j+1} &=\sum_{j=0}^{k}\frac{\zeta{(2)}}{(j+1)^2}-\sum_{j=0}^{k}\frac{H_{j+1}}{(j+1)^3}\\ &=\zeta{(2)}\sum_{j=1}^{k+1}\frac{1}{j^2}-\sum_{j=1}^{k+1}\frac{H_{j}}{j^3}\\ &=\zeta{(2)}H_{k+1,2}-\sum_{j=1}^{k+1}\frac{H_{j}}{j^3}.\\ \end{align}$$

Hence, equation $(7)$ simplifies to:

$$J{\left(k,1,3\right)}=\frac{\zeta{(3)}H_{k+1}}{1+k}+\frac12\frac{\zeta^2{(2)}}{k+1}-\frac{J{\left(k,1,2\right)}}{k+1}-\frac{\zeta{(2)}H_{k+1,2}}{k+1}+\frac{1}{k+1}\sum_{j=1}^{k+1}\frac{H_{j}}{j^3}.$$

Dividing through by $k^3(k+1)^2$ and summing over $k\ge1$ gives us the first unevaluated series of $(5)$:

$$\begin{align} \sum_{k=1}^{\infty}\frac{J{\left(k,1,3\right)}}{k^3(k+1)^2} &=\frac12\zeta^2{(2)}\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)^3}+\zeta{(3)}\sum_{k=1}^{\infty}\frac{H_{k+1}}{k^3(k+1)^3}-\sum_{k=1}^{\infty}\frac{J{\left(k,1,2\right)}}{k^3(k+1)^3}\\ &~~~~~ -\zeta{(2)}\sum_{k=1}^{\infty}\frac{H_{k+1,2}}{k^3(k+1)^3}+\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)^3}\sum_{j=1}^{k+1}\frac{H_{j}}{j^3}\\ &=\frac12\zeta^2{(2)}\left[10-6\zeta{(2)}\right]+\zeta{(3)}\sum_{k=1}^{\infty}\frac{H_{k+1}}{k^3(k+1)^3}-\sum_{k=1}^{\infty}\frac{J{\left(k,1,2\right)}}{k^3(k+1)^3}\\ &~~~~~ -\zeta{(2)}\sum_{k=1}^{\infty}\frac{H_{k+1,2}}{k^3(k+1)^3}+\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)^3}\sum_{j=1}^{k+1}\frac{H_{j}}{j^3}\\ &=5\zeta^2{(2)}-3\zeta^3{(2)}+\zeta{(3)}\sum_{k=1}^{\infty}\frac{H_{k+1}}{k^3(k+1)^3}-\sum_{k=1}^{\infty}\frac{J{\left(k,1,2\right)}}{k^3(k+1)^3}\\ &~~~~~ -\zeta{(2)}\sum_{k=1}^{\infty}\frac{H_{k+1,2}}{k^3(k+1)^3}+\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)^3}\sum_{j=1}^{k+1}\frac{H_{j}}{j^3}.\\ \end{align}$$

To evaluate the series $\sum_{k=1}^{\infty}\frac{H_{k+1}}{k^3(k+1)^3}$, we'll need the partial fraction expansions,

$$\begin{align} \frac{1}{k^3(k+1)^3} &=\frac{1}{k^3}-\frac{3}{k^2}+\frac{6}{k}-\frac{1}{(k+1)^3}-\frac{3}{(k+1)^2}-\frac{6}{k+1}\\ &=\frac{1}{k^3}-\frac{1}{(k+1)^3}-\frac{3}{k^2}-\frac{3}{(k+1)^2}+\frac{6}{k(k+1)},\\ \end{align}$$

and,

$$\begin{align} \frac{1-3k}{k^3(k+1)} &=\frac{1}{k^3}-\frac{4}{k^2}+\frac{4}{k}-\frac{4}{k+1}. \end{align}$$

We'll also need the series sum,

$$\begin{align} \sum_{k=1}^{\infty}\frac{H_{k+1}}{k(k+1)} &=\sum_{k=1}^{\infty}\left(\frac{H_{k+1}}{k}-\frac{H_{k+1}}{k+1}\right)\\ &=\sum_{k=1}^{\infty}\left(\frac{H_{k}}{k}+\frac{1}{k(k+1)}-\frac{H_{k+1}}{k+1}\right)\\ &=\sum_{k=1}^{\infty}\left(\frac{H_{k}}{k}+\frac{1}{k}-\frac{H_{k+1}}{k+1}-\frac{1}{k+1}\right)\\ &=2.\\ \end{align}$$

Thus,

$$\begin{align} \frac{H_{k+1}}{k^3(k+1)^3} &=\frac{H_{k+1}}{k^3}-\frac{H_{k+1}}{(k+1)^3}-\frac{3H_{k+1}}{k^2}-\frac{3H_{k+1}}{(k+1)^2}+\frac{6H_{k+1}}{k(k+1)}\\ &=\frac{H_{k}}{k^3}+\frac{1}{k^3(k+1)}-\frac{H_{k+1}}{(k+1)^3}-\frac{3H_{k}}{k^2}-\frac{3}{k^2(k+1)}-\frac{3H_{k+1}}{(k+1)^2}+\frac{6H_{k+1}}{k(k+1)}\\ &=\frac{1}{k^3(k+1)}-\frac{3}{k^2(k+1)}+\frac{H_{k}}{k^3}-\frac{H_{k+1}}{(k+1)^3}-\frac{3H_{k}}{k^2}-\frac{3H_{k+1}}{(k+1)^2}+\frac{6H_{k+1}}{k(k+1)}\\ &=\frac{1-3k}{k^3(k+1)}+\frac{H_{k}}{k^3}-\frac{H_{k+1}}{(k+1)^3}-\frac{3H_{k}}{k^2}-\frac{3H_{k+1}}{(k+1)^2}+\frac{6H_{k+1}}{k(k+1)}.\\ \end{align}$$

And thus,

$$\begin{align} \sum_{k=1}^{\infty}\frac{H_{k+1}}{k^3(k+1)^3} &=\sum_{k=1}^{\infty}\frac{1-3k}{k^3(k+1)}+\sum_{k=1}^{\infty}\left[\frac{H_{k}}{k^3}-\frac{H_{k+1}}{(k+1)^3}\right]-\sum_{k=1}^{\infty}\frac{3H_{k}}{k^2}-\sum_{k=1}^{\infty}\frac{3H_{k+1}}{(k+1)^2}\\ &~~~~~ +\sum_{k=1}^{\infty}\frac{6H_{k+1}}{k(k+1)}\\ &=\sum_{k=1}^{\infty}\frac{1-3k}{k^3(k+1)}+1-3\sum_{k=1}^{\infty}\frac{H_{k}}{k^2}-3\sum_{k=2}^{\infty}\frac{H_{k}}{k^2}+6\sum_{k=1}^{\infty}\frac{H_{k+1}}{k(k+1)}\\ &=4+\sum_{k=1}^{\infty}\frac{1-3k}{k^3(k+1)}-6\sum_{k=1}^{\infty}\frac{H_{k}}{k^2}+6\sum_{k=1}^{\infty}\frac{H_{k+1}}{k(k+1)}\\ &=4\sum_{k=1}^{\infty}\frac{1}{k^3}-4\sum_{k=1}^{\infty}\frac{1}{k^2}+4\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)-6\sum_{k=1}^{\infty}\frac{H_{k}}{k^2}+6\sum_{k=1}^{\infty}\frac{H_{k+1}}{k(k+1)}\\ &=8-4\zeta{(2)}+\zeta{(3)}-6\sum_{k=1}^{\infty}\frac{H_{k}}{k^2}+6\sum_{k=1}^{\infty}\frac{H_{k+1}}{k(k+1)}\\ &=8-4\zeta{(2)}+\zeta{(3)}-6\cdot 2\zeta{(3)}+6\cdot 2\\ &=20-4\zeta{(2)}-11\zeta{(3)}.\\ \end{align}$$

Note that in the second-to-last line above we've made use of the Euler sum, $\sum_{k=1}^{\infty}\frac{H_{k}}{k^2}=2\zeta{(3)}$ (for proof, see appendix).

Hence,

$$\begin{align} \sum_{k=1}^{\infty}\frac{J{\left(k,1,3\right)}}{k^3(k+1)^2} &=20\zeta{(3)}+5\zeta^2{(2)}-4\zeta{(2)}\zeta{(3)}-11\zeta^2{(3)}-3\zeta^3{(2)}-\sum_{k=1}^{\infty}\frac{J{\left(k,1,2\right)}}{k^3(k+1)^3}\\ &~~~~~ -\zeta{(2)}\sum_{k=1}^{\infty}\frac{H_{k+1,2}}{k^3(k+1)^3}+\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)^3}\sum_{j=1}^{k+1}\frac{H_{j}}{j^3}.\\ \end{align}$$

And hence,

$$\begin{align} I_3 &=40\zeta{(3)}+30\zeta^2{(2)}-16\zeta{(2)}\zeta{(3)}-21\zeta^2{(3)}-18\zeta^3{(2)}+6\zeta^2{(2)}\zeta{(3)}-3\zeta{(2)}\zeta^2{(3)}\\ &~~~~~ +\zeta^3{(3)}-6\sum_{k=1}^{\infty}\frac{J{\left(k,1,2\right)}}{k^3(k+1)^3}-2\zeta{(2)}\sum_{k=1}^{\infty}\frac{H_{k+1,2}}{k^3(k+1)^3}+2\sum_{k=1}^{\infty}\frac{1}{k^3(k+1)^3}\sum_{j=1}^{k+1}\frac{H_{j}}{j^3}.\\ \end{align}$$

...

TBC

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    $\begingroup$ Thanks David H for the answer! $\endgroup$ – user153012 Oct 26 '14 at 1:30
  • $\begingroup$ $\sum_{k=1}^{\infty}\frac{H_{k}}{k^2}=2\zeta{(3)}$ comes directly from Euler's theorem 2.2 at page 20 in this paper. $\endgroup$ – user153012 Nov 2 '14 at 21:07
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Here's my approach. To reduce clutter let

$$f_m(x) = Li_m(x)$$

and let

$$f_M(x) = f_{m_1}(x)\dots f_{m_n}(x)$$

where $M$ has $n$ entries. Then let

$$I_k(M) = \int f_M(x) dx$$

where we can interpret this as an indefinite or definite integral. Consider the following derivative

$$(x^{k+1}f_M)' = (k+1)x^k f_M + x^{k+1} \sum_{k=1}^n f_{m_k}'\prod_{i\neq k}f_{m_i} = (k+1)x^k f_M + x^{k} \sum_{k=1}^n f_{m_k-1}\prod_{i\neq k}f_{m_i}$$

Integrating with respect to $x$ (and ignoring the extra constant) yields

$$x^{k+1}f_M = (k+1)I_k(M) + \sum_{i=1}^n I_k(M)_i$$

where the notation $I_k(M)_i$ means we subtract $1$ from the $i$-th entry in $M$. This works for every $k+1 \neq 0$. For the case where $k+1 = 0$ we have

$$f_M' = \sum_{k=1}^n f_{m_k}'\prod_{i\neq k}f_{m_i} = \frac{1}{x}\sum_{k=1}^n f_{m_k-1}\prod_{i\neq k}f_{m_i}$$

Integrating yields the formula

$$f_M = \sum_{i=1}^n I_{-1}(M)_i$$

The integrals you seek are $I_0(3,3,3)$ and $I_0(3,3,3,3)$. The difficulty is in the number of terms and not necessarily the size of each term. We can write the recurrence relation

$$I_0(M) = x f_M - \sum_{i=1}^n I_0(M)_i$$

These identities should also hold for the definite version of the integrals.

Anyway, we can start with $I_0(3,3,3)$ and reduce it to $I_0(2,3,3)$. Then we reduce it further to $I_0(1,3,3)$ and $I_0(2,2,3)$ and so on. The question is where do we stop. One should ideally stop when one can compute all the quantities. If not, then one should go further. In the end I think stopping when the smallest entry is 1 might be a good strategy. We reduce

$$(3,3,3) \to (2,3,3) \to (1,3,3),(2,2,3)$$

and then

$$(2,2,3) \to (1,2,3),(2,2,2)$$

The definite version of $I_0(2,2,2)$ corresponds to $\mathcal{I}_3$ from Kirill's linked answer so we have

$$I_0(3,3,3) = 6xf_2^2 f_3- 3xf_2 f_3^2 + x f_3^3 - 6 I_0(2,2,2) - 12 I_0(1,2,3) + 3 I_0(1,3,3)$$

Letting $x \to 1$ and $x \to 0$ we find

$$I_0(3,3,3) = 6 \zeta(2)^2 \zeta(3) - 3 \zeta(2) \zeta(3)^2 + \zeta(3)^3 - 6 \mathcal{I_3} - 12 I_0(1,2,3) + 3 I_0(1,3,3)\\ = 540 - 108 \zeta(2) - 216 \zeta(3) + 72 \zeta(2) \zeta(3) + 6 \zeta(2)^2 \zeta(3) - 3 \zeta(3) \zeta(3)^2 + \zeta(3)^3 - 171 \zeta(4) - 36 \zeta(5) - \tfrac{105}{4} \zeta(6) - 12 I_0(1, 2, 3) + 3 I_0(1, 3, 3)$$

Therefore the problem reduces to calculating The quantities

$$I_0(1, 3, 3) = \int_0^1 Li_1(x)Li_3(x)^2 dx = - \int_0^1 \log(1-x)Li_3(x)^2 dx$$

$$I_0(1, 2, 3) = \int_0^1 Li_1(x)Li_2(x)Li_3(x) dx = -\int_0^1 \log(1-x)Li_2(x)Li_3(x) dx$$

I'm not sure how much of a reduction this is because the number of terms has not been decreased. However, from Kirill's answer, it seems that it'll make his approach easier since one would need fewer iterated integrals.

EDIT: As Kirill mentioned, the procedure reduces the overall weight $|M| = m_1+\dots+m_n$ which could be useful on its own. If we allowed the appearance of $0$'s in $M$ we get divergent integrals, so one would have to handle them in some more special way. If we just went ahead with it we'd get

$$I_0(3,3,3) = 90 f_1^3 x+18 \left(f_3-5 f_2\right) f_1^2 x+3 \left(12 f_2^2-6 f_3f_2+f_3^2\right) f_1 x-6f_2^3 x+f_3^3 x-3 f_2 f_3^2 x+6 f_2^2 f_3 x + R$$

where $R$ consists of

$$R = -270 I_0(0, 1, 1) + 180 I_0(0, 1, 2)- 36 I_0(0, 1, 3) - 36 I_0(0, 2, 2) + 18 I_0(0, 2,3) - 3 I_0(0, 3, 3)$$

We've reduced the weight from 9 to 6, 5, 4, 3 and 2. One could even go a little further and think about making the weight 0, although I'm not sure if that would be a good way to go about this. In any case, we can try this and we obtain a larger sum with $M$'s $(-6,3,3)$, $(-5,2,3)$, $(-4,1,3)$, $(-4,2,2)$, $ (-3,0,3)$, $(-3,1,2)$, $(-2,-1,3)$, $(-2,0,2)$, $(-2,1,1)$, $(-1,-1,2)$, $(-1,0,1)$ and $(0,0,0)$.

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  • $\begingroup$ (+1) Very systematic. Regarding your last comment about the number of terms not decreasing: your procedure reduces the weight $|M|=\sum m_i$, which is quite useful on its own. $\endgroup$ – Kirill Nov 2 '14 at 15:17
  • $\begingroup$ That is true. From 9 to 7 and 6 in this case. In the other integral we'd get from 12 to 10, 9 and 8. If we allowed going down to 0 on one of the terms we get divergent integrals but if one can handle it somehow we'd get even lower weights. In the first case we would go from 9 to 2, 3, 4, 5 and 6. $\endgroup$ – Alexander Vlasev Nov 2 '14 at 20:17

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