18
$\begingroup$

Is it possible to evaluate this integral in closed form? $$ \int_0^1 \frac{du}{u}\text{Li}_2(u)^2\log u \stackrel{?}{=} -\frac{\zeta(6)}{3}.$$ I found the possible closed form using an integer relation algorithm.

I found several other possible forms for similar integrals, including $$ \int_0^1 \frac{du}{u}\text{Li}_2(u)^2(\log u)^2 \stackrel{?}{=} -20\zeta(7)+12\zeta(2)\zeta(5).$$

There doesn't seem to be an equivalent form when the integrand contains $(\log u)^3$, at least not just in terms of $\zeta$.

Does anybody know a trick for evaluating these integrals?

Update. The derivation of the closed form for the second integral follows easily along the ideas O.L. used in the answer for the first integral.

Introduce the functions $$ I(a,b,c) = \int_0^1 \frac{du}{u}(\log u)^c \text{Li}_a(u)\text{Li}_b(u) $$ and $$ S(a,b,c) = \sum_{n,m\geq1} \frac{1}{n^am^b(n+m)^c}. $$ Using integration by parts, the expansion of polylogarithms from their power series definition and also that $$ \int_0^1 (\log u)^s u^{t-1}\,du = \frac{(-1)^s s!}{t^{s+1}},$$ check that $$ I(2,2,2) = -\frac23 I(1,2,3) = 4S(1,2,4). $$

Now use binomial theorem and the fact that $S(a,b,c)=S(b,a,c)$ to write $$ 6S(1,2,4) + 2S(3,0,4) = 3S(1,2,4) + 3S(2,1,4)+S(0,3,4)+S(3,0,4) = S(3,3,1). $$ Now, using Mathematica, $$ S(3,3,1) = \sum_{n,m\geq1}\frac{1}{n^3m^3(n+m)} = \sum_{m\geq1}\frac{H_m}{m^6} - \frac{\zeta(2)}{m^5} + \frac{\zeta(3)}{m^4}, $$ and $$ \sum_{m\geq1}\frac{H_m}{m^6} = -\zeta(4)\zeta(3)-\zeta(2)\zeta(5)+4\zeta(7), $$ so $$ S(3,3,1) = 4\zeta(7)-2\zeta(2)\zeta(5). $$

Also, $$ S(0,3,4) = \zeta(3)\zeta(4) - \sum_{m\geq1} \frac{H_{n,4}}{m^3} = -17\zeta(7)+10\zeta(2)\zeta(5)+\zeta(3)\zeta(4), $$ from which it follows that $$ I(2,2,2) = \frac23\left(S(3,3,1)-2S(0,3,4)\right) = -20\zeta(7)+12\zeta(2)\zeta(5). $$

$\endgroup$
9
  • $\begingroup$ I derived this integral representations for Euler's sum. Try if you can relate your integral to one of these sums. $\endgroup$ Commented Jun 23, 2013 at 7:58
  • $\begingroup$ Check these integral techniques which they may be useful for your purpose. $\endgroup$ Commented Jun 23, 2013 at 8:01
  • $\begingroup$ Trying to evaluate some Euler sums is actually why I found these sorts of integrals in the first place, although these two don't come exactly from Euler sums as far as I can tell. $\endgroup$
    – Kirill
    Commented Jun 23, 2013 at 21:12
  • $\begingroup$ In general you will find that: $\displaystyle\int _{0}^{1}\!{\frac {{\text{ Li}_a} \left( u \right) {\text{ Li}_b} \left( u \right) \left(\log u \right) ^{c}}{u}}{d u}= \left( -1 \right) ^{c}\Gamma \left( c+1 \right) \sum _{k=1}^{ \infty } \left( \sum _{j=1}^{\infty }{\frac {1}{ \left( j+k \right) ^{ c+1}{j}^{a}{k}^{b}}} \right)$, so $I(a,b,c)= \left( -1 \right) ^{c}\Gamma \left( c+1 \right) S(a,b,c+1) $. Proof: $u=e^{-y}$, series expand polylogs, $\Gamma \left( c+1 \right) =\int _{0}^{\infty }\!{y}^{c}{{\rm e}^{-y}} {dy}$, $\mathbb{R}(c)>0$. $\endgroup$ Commented Jun 24, 2013 at 16:56
  • $\begingroup$ Sorry: $\mathbb{R}(c)\ge-1$. $\endgroup$ Commented Jun 24, 2013 at 17:02

5 Answers 5

10
+100
$\begingroup$

Now it is a proof.


Let us integrate once by parts to replace the (first) integral by $$I=\int_0^1\frac{\ln u\,\mathrm{Li}_2(u)^2du}{u}=\int_0^1\frac{\ln^2 u\ln(1-u)}{u}\mathrm{Li}_2(u)\,du.$$ Next replace $\mathrm{Li}_2(u)=\sum_{m=1}^{\infty}u^m/m^2$ and $\ln(1-u)=-\sum_{n=1}^{\infty}u^{n}/n$ by the corresponding Taylor series. Exchanging the order of summation and integration, evaluate the integrals with respect to $u$. This can be done using that $$\int_0^1 u^{s-1}\ln^2u\,du=\frac{2}{s^3}.$$ So $I$ can be written as a double series $$I=-2\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n(m+n)^3}.$$ Now let us introduce the following sums: \begin{align} &S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^3(m+n)^3},\\ &S_2=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{n^3(m+n)^3},\\ &S_3=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n(m+n)^3},\\ &S_4=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{mn^2(m+n)^3}. \end{align} It is obvious that $S_1=S_2$ and $S_3=S_4$. What is more funny (but still obvious to prove) is that $$S_1+S_2+3S_3+3S_4=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^3n^3}=\zeta(3)^2.\tag{1}$$ Therefore, if we manage to compute $S_1=S_2$, we will be able to compute $I$. But $$S_1=-\sum_{m=1}^{\infty}\frac{\psi''(1+m)}{2m^3}=\frac12\left(\zeta(3)^2-\frac{\pi^6}{945}\right).\tag{2}$$ Here the first equality follows from the recursion relation $\psi''(z+1)-\psi''(z)={2}/{z^3}$ and telescoping argument, whereas the second was obtained using Mathematica.

Now combining (1), (2) and the fact that $I=-(S_3+S_4)$, we find $$I=-\frac13\left[\left(S_1+S_2+3S_3+3S_4\right)-2S_1\right]=-\frac{1}{3}\times\frac{\pi^6}{945}=-\frac{\zeta(6)}{3}.$$

$\endgroup$
6
  • $\begingroup$ Very nice!${}{}{}$ $\endgroup$
    – Kirill
    Commented Jun 23, 2013 at 21:28
  • $\begingroup$ @Kirill Thanks! I am now thinking about the second integral. Following the same approach, the main obstacle is that Mathematica cannot compute the analog of the formula (2). I actually think that understanding how to obtain (2) by hands would be sufficient to prove the 2nd integral. $\endgroup$ Commented Jun 23, 2013 at 21:34
  • $\begingroup$ The reason I originally considered these integrals is that $$ \sum_{m\geq1}H_{m,r}z^m = \frac{\text{Li}_r(z)}{1-z}, $$ where $H_{m,r+1}=\psi_{r}(1+m)-\psi_{r}(1)$. So evaluating $(2)$ is equivalent to evaluating a integrals of the form that I gave in the question. (I know this doesn't help much.) $\endgroup$
    – Kirill
    Commented Jun 23, 2013 at 21:39
  • $\begingroup$ I guess I was hoping to evaluate these integrals without converting them back to Euler sums, using some other technique. (Although I had no way of evaluating them at all.) $\endgroup$
    – Kirill
    Commented Jun 23, 2013 at 21:43
  • $\begingroup$ @O.L.: This technique is similar to the one I used to derive "other forms of solution 1" in this problem(+1). $\endgroup$ Commented Jun 23, 2013 at 22:36
3
$\begingroup$

By Cauchy product we have

$$\operatorname{Li}_2^2(x)=\sum_{n=1}^\infty\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)x^n$$

Multiply both sides by $\frac{\ln^2x}{x}$ then integrate from $x=0$ to $1$ and use the fact that $\int_0^1 x^{n-1}\ln^2xdx=\frac{2}{n^3}$

we get

$$\int_0^1\frac{\operatorname{Li}_2^2(x)\ln^2x}{x}dx=8\sum_{n=1}^\infty \frac{H_n}{n^6}+4\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5}-12\zeta(7)$$

By Euler identity we have $$\sum_{n=1}^\infty \frac{H_n}{n^6}=4\zeta(7)-\zeta(2)\zeta(5)-\zeta(3)\zeta(4)$$ and in my solution here I managed to prove $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^5}=-10\zeta(7)+5\zeta(2)\zeta(5)+2\zeta(3)\zeta(4)$$

By collecting these results we get

$$\int_0^1\frac{\operatorname{Li}_2^2(x)\ln^2x}{x}dx=-20\zeta(7)+12\zeta(2)\zeta(5)$$

$\endgroup$
2
$\begingroup$

I've decided to publish my work so far - I do not promise a solution, but I've made some progress that others may find interesting and/or helpful.

$$\text{Let } I_{n,k}=\int_{0}^{1}\frac{\text{Li}_{k}(u)}{u}\log(u)^{n}du$$ Integrating by parts gives $$I_{n,k}=\left[\text{Li}_{k+1}(u)\log(u)^{n}\right]_{u=0}^{u=1}-\int_{0}^{1}\frac{\text{Li}_{k+1}(u)}{u}n\log(u)^{n-1}du$$ $$\text{Hence, }I_{n,k}=-nI_{n-1,k+1} \implies I_{n,k}=(-1)^{r}\frac{n!}{(n-r)!}I_{n-r,k+r}$$ Taking $r=n$ gives $I_{n,k}=(-1)^{n}n!I_{0,n+k}$. $$\text{But obviously } I_{0,n+k}=\int_{0}^{1}\frac{\text{Li}_{n+k}(u)}{u}du=\text{Li}_{n+k+1}(1)-\text{Li}_{n+k+1}(0)=\zeta(n+k+1)$$

$$\text{Now consider }J_{n,k,l}=\int_{0}^{1}\frac{\text{Li}_{k}(u)}{u}\text{Li}_{l}(u)\log(u)^{n}du$$ Integrating by parts again, $$J_{n,k,l}=\left[\text{Li}_{k+1}(u)\text{Li}_{l}(u)\log(u)^{n}\right]_{0}^{1}-\int_{0}^{1}\frac{\text{Li}_{l-1}(u)}{u}\text{Li}_{k+1}(u)\log(u)^{n}-\int_{0}^{1}\frac{n\log(u)^{n-1}}{u}\text{Li}_{k+1}(u)\text{Li}_{l}(u) du$$ So $J_{n,k,l}=-J_{n,k+1,l-1}-nJ_{n-1,k+1,l}$; continuing in the spirit of the first part suggests that we ought to try to increase the first and second indices, while decreasing the third. If we can succeed in this, we have found a closed form.

$\endgroup$
3
  • $\begingroup$ Thank you. When I was looking at this myself, I also found some nice closed forms for $J_{n,k,l}$ for small $n$, $k$ and $l$; $J_{1,2k,0}$ is especially simple because of integration by parts. $\endgroup$
    – Kirill
    Commented Jun 20, 2013 at 1:19
  • $\begingroup$ Since the differentiation and integration of $\text{Li}_{k}^{m}$ introduces other indices, I think analysing $J$ may be the more sensible of the two options. I'll try to put in a bit more work on paper to make more progress, and it does seem that the closed form you seek exists. The goal, it seems, is to increase both $k$ and $l$ while decreasing $l$. $\endgroup$ Commented Jun 20, 2013 at 16:42
  • $\begingroup$ That should say "increase $k$ and $l$ while decreasing $n$" $\endgroup$ Commented Jun 23, 2013 at 21:12
1
$\begingroup$

\begin{align}J&=\int_0^1 \frac{\ln^2 x\left(\text{Li}_2(x)\right)^2}{x}dx\\ &\overset{\text{IBP}}=\frac{1}{3}\Big[\left(\text{Li}_2(x)\right)^2\ln^3 x\Big]_0^1+\frac{2}{3}\int_0^1 \frac{\text{Li}_2(x)\ln(1-x)\ln^3 x}{x}dx\\ &\overset{\text{IBP}}=\frac{1}{6}\Big[\text{Li}_2(x)\ln(1-x)\ln^4 x\Big]_0^1+\frac{1}{6}\int_0^1 \ln^4 x\left(\frac{\ln^2(1-x)}{x}+\frac{\text{Li}_2(x)}{1-x}\right)dx\\ &\overset{\text{IBP}}=\left(\frac{1}{30}\Big[\ln^5 x\ln^2(1-x)\Big]_0^1+\frac{1}{15}\int_0^1 \frac{\ln(1-x)\ln^5 x}{1-x}dx\right)+\frac{1}{6}\int_0^1 \frac{\text{Li}_2(x)\ln^4 x}{1-x}dx\\ &=\frac{1}{15}\underbrace{\int_0^1 \frac{\ln(1-x)\ln^5 x}{1-x}dx}_{J_1}+\frac{1}{6}\underbrace{\int_0^1 \frac{\text{Li}_2(x)\ln^4 x}{1-x}dx}_{J_2}\\ C&=\int_0^1\frac{\ln^5 t}{1-t}dt\\ J_1&\overset{\text{IBP}}=\left[\left(\int_0^x\frac{\ln^5 t}{1-t}dt-C\right)\ln(1-x)\right]+\int_0^1 \frac{1}{1-x}\left(\int_0^x\frac{\ln^5 t}{1-t}dt-C\right)dx\\ &=\int_0^1 \int_0^1 \left(\frac{x\ln^5(tx)}{(1-t)(1-x)}-\frac{C}{1-x}\right)dt dx\\ &=\int_0^1 \int_0^1 \left(\frac{\ln^5(tx)}{(1-t)(1-x)}-\frac{\ln^5(tx)}{(1-t)(1-tx)}-\frac{C}{1-x}\right)dt dx\\ &=-240\zeta(2)\zeta(5)-240\zeta(3)\zeta(4)+\int_0^1 \frac{1}{1-t}\left(\int_t^1 \frac{\ln^5 u}{1-u}du\right)dt-\\&\int_0^1 \frac{1}{t}\left(\int_0^t \frac{\ln^5 u}{1-u}du\right)dt+\int_0^1 \int_0^1 \left(\frac{\ln^5 t}{(1-t)(1-x)}-\frac{C}{1-x}\right)dt dx\\ &=-240\zeta(2)\zeta(5)-240\zeta(3)\zeta(4)-J_1+\int_0^1 \frac{\ln^6 t}{1-t}dt+\\&\underbrace{\int_0^1 \int_0^1 \left(\frac{\ln^5 t}{(1-t)(1-x)}-\frac{C}{1-x}\right)dt dx}_{K}\\ &0\leq A<1\\ K(A)&=\int_0^A \int_0^A \left(\frac{\ln^5 t}{(1-t)(1-x)}-\frac{C}{1-x}\right)dt dx\\ &=\ln(1-A)\left(AC-\int_0^A \frac{\ln^5 t}{1-t}dt\right)\\ &=-(1-A)\ln(1-A)C+\ln(1-A)\int_A^1 \frac{\ln^5 t}{1-t}dt\\ K&=\lim_{A\rightarrow 1}K(A)\\ &=0 \end{align} Therefore,

$\boxed{\displaystyle J_1=360\zeta(7)-120\zeta(2)\zeta(5)-120\zeta(3)\zeta(4)}$ \begin{align*}J_2&\overset{\text{IBP}}=\left[\left(\int_0^x\frac{\ln^4 t}{1-t}dt\right)\text{Li}_2(x)\right]_0^1+\int_0^1 \frac{\ln(1-x)}{x} \left(\int_0^x\frac{\ln^4 t}{1-t}dt\right)dx\\ &\overset{\text{IBP}}=24\zeta(5)\zeta(2)+\left[\left(\int_0^x\frac{\ln^4 t}{1-t}dt\right)\ln(1-x)\ln x\right]_0^1-\\&\int_0^1 \ln x\left(\frac{\ln^4 x\ln(1-x)}{1-x}-\frac{1}{1-x}\left(\int_0^x\frac{\ln^4 t}{1-t}dt\right)\right)dx\\ &=24\zeta(5)\zeta(2)-J_1+\underbrace{\int_0^1 \frac{\ln x}{1-x}\left(\int_0^x\frac{\ln^4 t}{1-t}dt\right)dx}_{J_3}\\ J_3&\overset{\text{IBP}}=-24\zeta(5)\zeta(2)-\int_0^1 \frac{\ln^4 x}{1-x}\left(\int_0^x\frac{\ln t}{1-t}dt\right)dx\\ &=-24\zeta(5)\zeta(2)-\int_0^1\int_0^1 \frac{x\ln^4 x\ln(tx)}{(1-x)(1-tx)}dt dx\\ &=-24\zeta(5)\zeta(2)-\int_0^1\int_0^1 \left(\frac{\ln^4 x\ln(tx)}{(1-t)(1-x)}-\frac{\ln^4 x\ln(tx)}{(1-t)(1-tx)}\right)dt dx\\ &=-\int_0^1 \frac{1}{1-t}\left(\int_t^1 \frac{\ln^5 u}{1-u}du\right)dt+\int_0^1 \frac{1}{t}\left(\int_0^t \frac{\ln^5 u}{1-u}du\right)dt-4\underbrace{\int_0^1 \frac{\ln t}{1-t}\left(\int_0^t \frac{\ln^4 u}{1-u}du\right)dt}_{J_1+J_2-24\zeta(2)\zeta(5)}-\\&4\int_0^1 \frac{\ln t}{t}\left(\int_0^t \frac{\ln^4 u}{1-u}du\right)dt+6\int_0^1 \frac{\ln^2 t}{1-t}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)dt+6\int_0^1 \frac{\ln^2 t}{t}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)dt-\\&4\int_0^1 \frac{\ln^3 t}{1-t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt-4\int_0^1 \frac{\ln^3 t}{t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt+\underbrace{\int_0^1 \frac{\ln^4 t}{1-t}\left(\int_0^t \frac{\ln u}{1-u}du\right)dt}_{-24\zeta(2)\zeta(5)-J_3}+\\& \int_0^1 \frac{\ln^4 t}{t}\left(\int_0^t \frac{\ln u}{1-u}du\right)dt\\ &\overset{\text{IBP}}=\underbrace{\Big(360\zeta(7)-120\zeta(2)\zeta(5)-120\zeta(3)\zeta(4)\Big)}_{J_1}- 720\zeta(7)-\\&4\Big(\big(360\zeta(7)-120\zeta(2)\zeta(5)-120\zeta(3)\zeta(4)\big)+J_2-24\zeta(2)\zeta(5)\Big)+4\times \frac{1}{2}\times 720\zeta(7)+\\&6\int_0^1 \frac{\ln^2 t}{1-t}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)dt-6\times \frac{1}{3}\times 720\zeta(7)-4\times -6\zeta(4)\times 2\zeta(3)+\\&4\int_0^1 \frac{\ln^2 t}{1-t}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)dt+4\times\frac{1}{4}\times 720\zeta(7)+\Big(-24\zeta(2)\zeta(5)-J_3\Big)-\frac{1}{5}\times 720\zeta(7)\\ &=-1224\zeta(7)+432\zeta(2)\zeta(5)+408\zeta(3)\zeta(4)-4J_2-J_3+10\int_0^1 \frac{\ln^2 t}{1-t}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)dt\\ J_3&=-612\zeta(7)+216\zeta(2)\zeta(5)+204\zeta(3)\zeta(4)-2J_2+5\underbrace{\int_0^1 \frac{\ln^2 t}{1-t}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)dt}_{J_4}\\ \end{align*} \begin{align*} J_4&=\int_0^1\int_0^1 \frac{t\ln^2 t\ln^3(tu)}{(1-t)(1-tu)}dtdu\\ &=\int_0^1\int_0^1\left(\frac{\ln^2 t\ln^3(tu)}{(1-t)(1-u)}-\frac{\ln^2 t\ln^3(tu)}{(1-u)(1-tu)}\right)dt du\\ &=-48\zeta(3)\zeta(4)-72\zeta(2)\zeta(5)+\\&\int_0^1\int_0^1\left(\frac{\ln^5 t}{(1-t)(1-u)}-\frac{\ln^2 u\ln^3(tu)-2\ln u\ln^4(tu)+\ln^5(tu)}{(1-u)(1-tu)}\right)dt du\\ &=-48\zeta(3)\zeta(4)-72\zeta(2)\zeta(5)+\int_0^1 \frac{1}{1-u}\left(\int_u^1 \frac{\ln^5 t}{1-t}dt\right)du-\\&\int_0^1 \frac{1}{u}\left(\int_0^t\frac{\ln^5 t}{1-t}dt\right)du-\underbrace{\int_0^1 \frac{\ln^2 u}{1-u}\left(\int_0^t\frac{\ln^3 t}{1-t}dt\right)du}_{J_4}-\\&\int_0^1 \frac{\ln^2 u}{u}\left(\int_0^t\frac{\ln^3 t}{1-t}dt\right)du+2\underbrace{\int_0^1 \frac{\ln u}{1-u}\left(\int_0^t\frac{\ln^4 t}{1-t}dt\right)du}_{J_3}+\\&2\int_0^1 \frac{\ln u}{u}\left(\int_0^t\frac{\ln^4 t}{1-t}dt\right)du\\ J_4&=-60\zeta(7)+24\zeta(2)\zeta(5)+36\zeta(3)\zeta(4)+J_3\\ J_3&=228\zeta(7)-84\zeta(2)\zeta(5)-96\zeta(3)\zeta(4)+\frac{1}{2}J_2\\ J_2&=\boxed{-264\zeta(7)+120\zeta(2)\zeta(5)+48\zeta(3)\zeta(4)}\\ \end{align*} Therefore, \begin{align*}\boxed{J=12\zeta(2)\zeta(5)-20\zeta(7)}\end{align*}

$\endgroup$
0
$\begingroup$

You might find Two New Zeta Constants: Fractal String, Continued Fraction, and Hypergeometric Aspects of the Riemann Zeta Function interesting . It relates hypergeometric functions, polylogarithms, and the zeta function

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .