How to evaluate: $\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$

Question

$$\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x)\,\mathrm dx=1.03693\ldots$$

This number looks like $\zeta(5)$ value.

We expand the terms

$$\int_0^1\frac{\frac{\pi^2}{6}}{1-x}\cdot \ln^2(x) \, \mathrm dx-\int_0^1 \frac{\operatorname{Li}_2(1-x)}{1-x} \cdot \ln^2(x) \, \mathrm dx$$

$$2\zeta(2)\zeta(3)-\int_0^1 \frac{\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$$

this last integral it is very complicate to compute...

Can any user please help to show that whether this value is equal to $\zeta(5)$ or not?

Answer 1

We can have a nice generalization for

$$I_n=\int_0^1\frac{\zeta(n)-\operatorname{Li}_n(1-x)}{1-x}\ln^2x\ dx=\int_0^1\frac{\zeta(n)-\operatorname{Li}_n(x)}{x}\ln^2(1-x)\ dx$$

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From $$\ln^2(1-x)=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}x^k$$

It follows that

$$I_n=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\int_0^1x^{k-1}(\zeta(n)-\operatorname{Li}_n(x))\ dx$$ By integration by parts we have

$$\int_0^1x^{k-1}\operatorname{Li}_n(x)\ dx=(-1)^{n-1}\frac{H_k}{k^n}-\sum_{i=1}^{n-1}(-1)^i\frac{\zeta(n-i+1)}{k^i}$$

$$\Longrightarrow I_n=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{\zeta(n)}{k}+(-1)^{n}\frac{H_k}{k^n}+\sum_{i=\color{red}{1}}^{n-1}(-1)^i\frac{\zeta(n-i+1)}{k^i}\right)$$

or

$$I_n=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\sum_{i=\color{red}{2}}^{n-1}(-1)^i\frac{\zeta(n-i+1)}{k^i}+(-1)^{n}\frac{H_k}{k^n}\right),\quad n=2,3,...$$

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Applications

$$I_2=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{H_k}{k^2}\right)$$

$$ I_3=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{\zeta(2)}{k^2}-\frac{H_k}{k^3}\right)$$

$$I_4=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{\zeta(3)}{k^2}-\frac{\zeta(2)}{k^3}+\frac{H_k}{k^4}\right)$$

$$I_5=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{\zeta(4)}{k^2}-\frac{\zeta(3)}{k^3}+\frac{\zeta(2)}{k^4}-\frac{H_k}{k^5}\right)$$

Answer 2

Start with $1-x\mapsto x$ $$I=\int_0^1\frac{\operatorname{Li}_2(1-x)\ln^2x}{1-x}dx=\int_0^1\frac{\operatorname{Li}_2(x)\ln^2(1-x)}{x}dx$$

By Cauchy product we have

$$\ln(1-x)\operatorname{Li}_2(x)=-\sum_{n=1}^\infty\left(2\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac3{n^3}\right)x^n$$

multiply both sides by $\frac{\ln(1-x)}{x}$ then integrate from $x=0$ to $x=1$ and use the fact that $-\int_0^1x^{n-1}\ln(1-x)\ dx=\frac{H_n}{n}$ we get

$$I=\sum_{n=1}^\infty\left(2\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac3{n^3}\right)\left(\frac{H_n}{n}\right)$$

$$=2\sum_{n=1}^\infty\frac{H_n^2}{n^3}+\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}-3\sum_{n=1}^\infty\frac{H_n}{n^4}$$

For the first sum, its evaluated here

$$\sum_{n=1}^\infty\frac{H_n^2}{n^3}=\frac72\zeta(5)-\zeta(2)\zeta(3)$$

The second sum, can be found here

$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}=\zeta(5)+\zeta(2)\zeta(3)$$

and the well-known result

$$\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$

Combine the three sums to get

$$\boxed{I=2\zeta(2)\zeta(3)-\zeta(5)}$$

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A different approach is to apply integration by parts to your original integral before breaking the inegrand.

Answer 3

Here is an easier way, let $1-x\mapsto x$

$$I=\int_0^1\frac{\zeta(2)-\operatorname{Li}_2(1-x)}{1-x}\ln^2x\ dx=\int_0^1\frac{\zeta(2)-\operatorname{Li}_2(x)}{x}\ln^2(1-x)\ dx$$

Now use $\ln^2(1-x)=2\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n$ we get

$$I=2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\int_0^1 x^{n-1}(\zeta(2)-\operatorname{Li}_2(x))\ dx$$

$$=2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\left(\frac{\zeta(2)}{n}-\frac{\zeta(2)}{n}+\frac{H_n}{n^2}\right)$$

$$=2\sum_{n=1}^\infty\frac{H_n^2}{n^3}-2\sum_{n=1}^\infty\frac{H_n}{n^4}$$

These two sums are mentioned in my previous solution above, collecting them gives $ I=\zeta(5)$

Answer 4

Here is an alternative approach through the harmonic sum thicket.

Starting with Euler's reflexion formula for the dilogarithm function, namely $$\operatorname{Li}_2 (x) + \operatorname{Li}_2 (1 - x) = \zeta (2) - \ln x \ln (1- x),$$ your integral $I$ can be rewritten as $$I = \int_0^1 \frac{\ln^3 x \ln (1 - x)}{1 - x} \, dx + \int_0^1 \frac{\ln^2 x \operatorname{Li}_2 (x)}{1 - x} \, dx = I_1 + I_2.$$

For the first integral, from the generating function for the harmonic numbers $H_n$ we have $$\frac{\ln (1 - x)}{1 - x} = - \sum_{n = 1}^\infty H_n x^n.$$ Thus \begin{align} I_1 &= -\sum_{n = 1}^\infty H_n \int_0^1 x^n \ln^3 x \, dx\\ &= -\sum_{n = 1}^\infty H_n \frac{d^3}{ds^3} \left [\int_0^1 x^{n + s} \, dx \right ]_{s = 0}\\ &= -\sum_{n = 1}^\infty H_n \frac{d^3}{ds^3} \left [\frac{1}{n + s + 1} \right ]_{s = 0}\\ &= 6 \sum_{n = 1}^\infty \frac{H_n}{(n + 1)^4} = 6 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^4}, \end{align} after reindexing $n \mapsto n - 1$. Since $$H_n = H_{n - 1} + \frac{1}{n},$$ this leads to $$I_1 = 6 \sum_{n = 1}^\infty \frac{H_n}{n^4} - 6 \sum_{n = 1}^\infty \frac{1}{n^2} = 6 \sum_{n = 1}^\infty \frac{H_n}{n^4} - 6 \zeta (5).$$

For the second integral, from the generating function for the generalised harmonic numbers of order two $H^{(2)}_n$ we have $$\frac{\operatorname{Li}_2 (x)}{1 - x} = \sum_{n = 1}^\infty H^{(2)}_n x^n.$$ Thus \begin{align} I_2 &= \sum_{n = 1}^\infty H^{(2)}_n \int_0^1 x^n \ln^2 x \, dx\\ &= \sum_{n = 1}^\infty H^{(2)}_n \frac{d^2}{ds^2} \left [\int_0^1 x^{n + s} \, dx \right ]_{s = 0}\\ &= \sum_{n = 1}^\infty H^{(2)}_n \frac{d^2}{ds^2} \left [\frac{1}{n + s + 1} \right ]_{s = 0}\\ &= 2\sum_{n = 1}^\infty \frac{H^{(2)}_n}{(n + 1)^3} = 2 \sum_{n = 2}^\infty \frac{H^{(2)}_{n-1}}{n^3}, \end{align} after reindexing $n \mapsto n - 1$. Since $$H^{(2)}_n = H^{(2)}_{n - 1} + \frac{1}{n^2},$$ we have $$I_2 = 2 \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} - 2 \sum_{n = 1}^\infty \frac{1}{n^5} = 2 \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} - 2 \zeta (5).$$

Returning to our integral, we have $$I = 6 \sum_{n = 1}^\infty \frac{H_n}{n^4} + 2 \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} - 8 \zeta (5).$$ Since $$\sum_{n = 1}^\infty \frac{H_n}{n^4} = 3 \zeta (5) - \zeta (2) \zeta (3),$$ and $$\sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} = 3 \zeta (2) \zeta (3) - \frac{9}{2} \zeta (5),$$ it is immedate that $I = \zeta (5)$, as desired.

Answer 5

A solution that is not using harmonic series.

\begin{align}J&=\int_0^1 \frac{\Big(\operatorname{Li}_2(1)-\operatorname{Li}_2(1-x)\Big)\ln^2(x)}{1-x} \, \mathrm dx\\ &\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)\Big(\operatorname{Li}_2(1)-\operatorname{Li}_2(1-x)\Big)\right]_0^1+\int_0^1 \left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)\frac{\ln x}{1-x}dx\\ &=2\zeta(3)\zeta(2)+\int_0^1 \left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)\frac{\ln x}{1-x}dx\\ &\overset{\text{IBP}}=-\int_0^1 \left(\int_0^x \frac{\ln t}{1-t}dt\right)\frac{\ln^2 x}{1-x}dx\\ &=-\int_0^1 \int_0^1 \frac{x\ln(tx)\ln^2 x}{(1-tx)(1-x)}\,dtdx\\ &=\int_0^1 \int_0^1\left(\frac{\ln(tx)\ln^2 x}{(1-t)(1-tx)}-\frac{\ln(tx)\ln^2 x}{(1-t)(1-x)}\right)dt dx\\ &=2\zeta(2)\zeta(3)+\int_0^1 \int_0^1\left(\frac{\ln^3(tx)-2\ln t\ln^2(tx)+\ln^2 t\ln(tx)}{(1-t)(1-tx)}-\frac{\ln^3 x}{(1-t)(1-x)}\right)dt dx\\ &=2\zeta(2)\zeta(3)-\int_0^1 \frac{1}{1-t}\left(\int_t^1 \frac{\ln^3 u}{1-u}du\right)dt+\int_0^1\frac{1}{t}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)dt\\&-\int_0^1 \frac{2\ln t}{t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt -2\underbrace{\int_0^1 \frac{\ln t}{1-t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt}_{J-2\zeta(2)\zeta(3)}+\\&\int_0^1\frac{\ln^2 t}{t}\left(\int_0^t \frac{\ln u}{1-u}du\right)dt+\underbrace{\int_0^1\frac{\ln^2 t}{1-t}\left(\int_0^t \frac{\ln u}{1-u}du\right)dt}_{-J}\\ &=2\zeta(2)\zeta(3)-\int_0^1 \frac{1}{1-t}\left(\int_t^1 \frac{\ln^3 u}{1-u}du\right)dt+\int_0^1\frac{1}{t}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)dt\\&-\int_0^1 \frac{2\ln t}{t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt -2\Big(J-2\zeta(2)\zeta(3)\Big)+\int_0^1\frac{\ln^2 t}{t}\left(\int_0^t \frac{\ln u}{1-u}du\right)dt-J\\ &\overset{\text{IBP}}=6\zeta(2)\zeta(3)-3J+\int_0^1 \frac{\ln(1-t)\ln^3 t}{1-t}dt-\frac{1}{3}\int_0^1 \frac{\ln^4 t}{1-t}dt\\ &=6\zeta(2)\zeta(3)-3J+\int_0^1 \frac{\ln(1-t)\ln^3 t}{1-t}dt-8\zeta(5) \end{align} On the other hand, \begin{align}K&=\int_0^1 \frac{\ln(1-t)\ln^3 t}{1-t}dt,C=\int_0^1 \frac{\ln^3 t}{1-t}dt\\ &\overset{\text{IBP}}=\left[\left(\int_0^t\frac{\ln^3 u}{1-u}du-C\right)\ln(1-t)\right]_0^1+\int_0^1 \frac{1}{1-t}\left(\int_0^t\frac{\ln^3 u}{1-u}du-C\right)dt\\ &=\int_0^1\int_0^1\left(\frac{t\ln^3 (tu)}{(1-tu)(1-t)}-\frac{C}{1-t}\right)dt du\\ &=\int_0^1\int_0^1\left(\frac{\ln^3 (tu)}{(1-t)(1-u)}-\frac{\ln^3 (tu)}{(1-u)(1-tu)}-\frac{C}{1-t}\right)dt du\\ &=-12\zeta(2)\zeta(3)+\int_0^1 \frac{1}{1-u}\left(\int_u^1\frac{\ln^3 t}{1-t}dt\right)du-\int_0^1 \frac{1}{u}\left(\int_0^u\frac{\ln^3 t}{1-t}dt\right)du+\\&\int_0^1 \int_0^1 \left(\frac{\ln^3 u}{(1-t)(1-u)}-\frac{C}{1-t}\right)dt du\\ &\overset{\text{IBP}}=-12\zeta(2)\zeta(3)-J+\int_0^1\frac{\ln^4 u}{1-u}du+\underbrace{\int_0^1 \int_0^1 \left(\frac{\ln^3 u}{(1-t)(1-u)}-\frac{C}{1-t}\right)dt du}_{L}\\ &0\leq A<1\\ L(A)&=\int_0^A \int_0^A \left(\frac{\ln^3 u}{(1-t)(1-u)}-\frac{C}{1-t}\right)dt du\\ &=\ln(1-A)\int_0^A \left(C-\frac{\ln^3 u}{1-u}\right)du\\ &=\ln(1-A)\int_A^1\frac{\ln x}{1-x}dx-(1-A)\ln(1-A)C\\ L&=\lim_{A\rightarrow 1}L(A)\\&=0\\ K&=\boxed{-6\zeta(2)\zeta(3)+12\zeta(5)}\\ \end{align} Therefore,

$\boxed{\displaystyle J=\zeta(5)}$

I assume that, \begin{align}\int_0^1 \frac{\ln t}{1-t}dt&=-\zeta(2)=-\frac{1}{6}\pi^2\\ \int_0^1 \frac{\ln^2 t}{1-t}dt&=2\zeta(3)\\ \int_0^1 \frac{\ln^4 t}{1-t}dt&=24\zeta(5)\\ \end{align}