Challenging Integral: $\int_0^\infty\frac{\ln(2+x)\operatorname{Li}_2(-x)}{x(2+x)}dx$

Question

Prove that

$$\int_0^\infty\frac{\ln(2+x)\operatorname{Li}_2(-x)}{x(2+x)}dx=\frac32\operatorname{Li}_4\left(\frac12\right)-\frac{111}{32}\zeta(4)-\frac78\ln2\zeta(3)-\frac98\ln^22\zeta(2)+\frac1{16}\ln^42$$

where $\operatorname{Li}_r(x)=\sum_{n=1}^\infty\frac{x^n}{n^r}$ is the polylogarithm function and $\zeta$ is the Riemann zeta function.

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This problem is created by Cornel and can be found here. I managed to solve it but my solution turned out really long and I used results of many harmonic series, so is there an elegant way that spares us the tedious calculations? My solution will be posted soon in the answer section as its too long to be posted here. Thank you.

Answer 1

Starting with $x\mapsto \frac{2x}{1+x}$ we obtain

$$\mathcal{I}=\frac12\ln(2)\int_0^1\frac{\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx-\frac12\int_0^1\frac{\ln(1-x)\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx$$ $$\mathcal{I}=\frac12\ln(2)\mathcal{I}_1-\frac12\mathcal{I}_2\tag1$$

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From the solution above we have

$$\mathcal{I}_1=-\frac74\zeta(3)-\frac32\ln2\zeta(2)\tag2$$

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For $\mathcal{I}_2$, let $1-x\mapsto x$ then use the integral representation $\operatorname{Li}_2(z)=-\int_0^1\frac{z\ln(y)}{1-zy}dy$, we get

$$\mathcal{I}_2=2\int_0^1\int_0^1\frac{\ln(x)\ln(y)}{x-2xy+2y}dy=2\int_0^1\ln(y)\left(\int_0^1\frac{\ln(x)}{x-2xy+2y}dx\right)dy$$

$$=-2\int_0^1\frac{\ln(y)\operatorname{Li}_2\left(\frac{2y-1}{2y}\right)}{2y-1}dy=-\int_{-1}^1\frac{\ln\left(\frac{1+x}{2}\right)\operatorname{Li}_2\left(\frac{x}{1+x}\right)}{x}dx$$

using Landen's identity $-\operatorname{Li}_2\left(\frac{x}{1+x}\right)=\operatorname{Li}_2(-x)+\frac12\ln^2(1+x)$ yields

$$\small{\mathcal{I}_2=\int_{-1}^1\frac{\ln(1+x)\operatorname{Li}_2(-x)}{x}dx-\ln(2)\int_{-1}^1\frac{\operatorname{Li}_2(-x)}{x}dx-\frac12\ln(2)\int_{-1}^1\frac{\ln^2(1+x)}{x}dx+\frac12\int_{-1}^1\frac{\ln^3(1+x)}{x}dx}$$

$$=-\frac12\operatorname{Li}_2^2(-x)|_{-1}^1-\ln(2)\operatorname{Li}_3(-x)|_{-1}^1-\frac12\ln(2)\int_{-1}^1\frac{\ln^2(1+x)}{x}dx+\frac12\int_{-1}^1\frac{\ln^3(1+x)}{x}dx$$

$$=\frac{15}{16}\zeta(4)+\frac74\ln(2)\zeta(3)-\frac12\ln(2)\int_{-1}^1\frac{\ln^2(1+x)}{x}dx+\frac12\int_{-1}^1\frac{\ln^3(1+x)}{x}dx$$

Lets evaluate the last two integrals:

$$\int_{-1}^1\frac{\ln^2(1+x)}{x}dx=\int_{-1}^0\frac{\ln^2(1+x)}{x}dx+\int_{0}^1\frac{\ln^2(1+x)}{x}dx$$

where $\int_{-1}^0\frac{\ln^2(1+x)}{x}dx=-\int_{0}^1\frac{\ln^2(1-x)}{x}dx=-\int_{0}^1\frac{\ln^2(x)}{1-x}dx=-2\zeta(3)$ and $\int_{0}^1\frac{\ln^2(1+x)}{x}dx=\frac14\zeta(3)$

Similarly

$$\int_{-1}^1\frac{\ln^3(1+x)}{x}dx=\int_{-1}^0\frac{\ln^3(1+x)}{x}dx+\int_{0}^1\frac{\ln^3(1+x)}{x}dx$$

where $\int_{-1}^0\frac{\ln^3(1+x)}{x}dx=-\int_{0}^1\frac{\ln^3(1-x)}{x}dx=-\int_{0}^1\frac{\ln^3(x)}{1-x}dx=6\zeta(3)$ and $\int_0^1\frac{\ln^3(1+x)}{x}dx=6\zeta(4)-\frac{21}4\ln(2)\zeta(3)+\frac32\ln^2(2)\zeta(2)-6\operatorname{Li}_4(1/2)$

By grouping these results we get

$$\mathcal{I}_2=3\operatorname{Li}_4\left(\frac12\right)-\frac{111}{16}\zeta(4)-\frac34\ln^22\zeta(2)+\frac1{8}\ln^42\tag3$$

Plug $(2)$ and $(3)$ in $(1)$, the result of $\mathcal{I}$ follows.

Answer 2

First we start with subbing $ x=\frac{y}{2+y}\Longrightarrow y=\frac{2x}{1-x}\Longrightarrow dy=\frac{2}{(1-x)^2}$

$$\mathcal{I}=\int_0^\infty\frac{\ln(2+y)\operatorname{Li}_2(-y)}{y(2+y)}dy=\frac12\int_0^1\frac{\ln\left(\frac2{1-x}\right)\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx$$

$$=\frac12\ln2\int_0^1\frac{\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx-\frac12\int_0^1\frac{\ln(1-x)\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx$$

$$=\frac12\ln2\cdot\mathcal{I}_1+\frac12\mathcal{I}_2$$

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Evaluating the first integral $\mathcal{I}_1$:

By integration by parts we get

$$\mathcal{I}_1=\int_0^1\frac{\ln(x)\ln\left(\frac{1+x}{1-x}\right)}{x(1-x)}dx=\int_0^1\frac{\ln(x)\ln(1+x)}{x(1-x)}dx-\int_0^1\frac{\ln(x)\ln(1-x)}{x(1-x)}dx\\=\mathcal{A}-\mathcal{B}$$

$$\mathcal{A}=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\frac{x^{n-1}\ln(x)}{1-x}dx=-\sum_{n=1}^\infty \frac{(-1)^n}{n}[H_{n-1}^{(2)}-\zeta(2)]$$

$$=-\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{n}+\sum_{n-1}^\infty\frac{(-1)^n}{n^3}+\ln2\sum_{n=1}^\infty\frac{(-1)^n}{n}$$

$$=-\left(\frac12\ln2\zeta(2)-\zeta(3)\right)-\frac34\zeta(3)-\ln2\zeta(2)$$

$$=\frac14\zeta(3)-\frac32\ln2\zeta(2)$$

By symmetry, we have

$$\mathcal{B}=2\int_0^1\frac{\ln x\ln(1-x)}{x}dx=-\sum_{n=1}^\infty\frac1n\int_0^1x^{n-1}\ln xdx=2\sum_{n=1}^\infty\frac{1}{n^3}=2\zeta(3)$$

Collect $\mathcal{A}$ and $\mathcal{B}$ we get

$$\mathcal{I}_1=-\frac74\zeta(3)-\frac32\ln2\zeta(2)$$

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Evaluating the second integral $\mathcal{I}_2$:

Use Landen's identity $-\operatorname{Li}_2(z)=\operatorname{Li}_2\left(\frac{z}{z-1}\right)+\ln^2(1-z)$

Set $z=\frac{2x}{x-1}$ we get $-\operatorname{Li}2\left(\frac{2x}{x-1}\right)=\operatorname{Li}2\left(\frac{2x}{x+1}\right)+\frac12\ln^2\left(\frac{1-x}{1+x}\right)$

Then we can write

$$\mathcal{I}_2=\int_0^1\frac{\ln(1-x)\operatorname{Li}_2\left(\frac{2x}{x+1}\right)}{x}dx+\frac12\int_0^1\frac{\ln(1-x)\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx\\=\mathcal{J}+\frac12\mathcal{K}$$

Apply integration by parts for $\mathcal{J}$ we have

$$\mathcal{J}=-\frac52\zeta(4)+\int_0^1\frac{\operatorname{Li}_2(x)\ln\left(\frac{1+x}{1-x}\right)}{x(1+x)}dx$$

$$=-\frac52\zeta(4)+\int_0^1\frac{\operatorname{Li}_2(x)\ln(1+x)}{x(1+x)}dx-\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{x(1+x)}dx\\=-\frac52\zeta(4)+\mathcal{J}_1-\mathcal{J}_2$$

$$\mathcal{J}_1=\int_0^1\frac{\operatorname{Li}_2(x)\ln(1+x)}{x(1+x)}dx=-\sum_{n=1}^\infty (-1)^nH_n\int_0^1 x^{n-1}\operatorname{Li}_2(x)dx$$

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

You can find here:

$$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^2}{n^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{41}{16}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$

Substitute this result along with $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}=\frac12\ln^22-\frac12\zeta(2)$ we get

$$\boxed{\mathcal{J}_1=2\operatorname{Li}_4\left(\frac12\right)-\frac{21}{16}\zeta(4)+\frac74\ln2\zeta(3)-\ln^22\zeta(2)+\frac1{12}\ln^42}$$

For $\mathcal{J}_2$, break the denominator first

$$\mathcal{J}_2=\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{x}dx-\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{1+x}dx$$

$$=-\frac54\zeta(4)-\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{1+x}dx$$

Now set $1-x\to x$ then use the reflection formula $\operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)$

$$\mathcal{J}_2=-\frac54\zeta(4)-\int_0^1\frac{\ln x}{2-x}\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)dx$$

write $\frac1{2-x}=\sum_{n=1}^\infty \frac1{2^n}x^{n-1}$ to get

$$\mathcal{J}_2=-\frac54\zeta(4)-\sum_{n-1}^\infty \frac1{2^n}\int_0^1\ln x\ x^{n-1}\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)dx$$

$$=-\frac54\zeta(4)-\sum_{n-1}^\infty \frac1{2^n}\frac{\partial}{\partial n}\int_0^1x^{n-1}\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)dx, \quad IBP$$

$$=-\frac54\zeta(4)-\sum_{n-1}^\infty \frac1{2^n}\frac{\partial}{\partial n}\left(\frac{\zeta(2)}{n}-\frac{H_n^{(2)}}{n}\right)$$

$$=-\frac54\zeta(4)-\sum_{n=1}^\infty\frac{1}{2^n}\left(\frac{H_n^{(2)}}{n^2}+\frac{2H_n^{(3)}}{n}-\frac{\zeta(2)}{n^2}-\frac{2\zeta(3)}{n}\right)$$

The first sum is already proved here

$$\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42$$

For the second sum, we can just set $x=\frac12$ in the generating function to get

$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}=\operatorname{Li_4}\left(\frac12\right)-\frac{5}{16}\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac{1}{24}\ln^42$$

Collect these two results along with $\operatorname{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^22$ we get

$$\boxed{\mathcal{J}_2=-3\operatorname{Li_4}\left(\frac12\right)+\frac{9}{16}\zeta(4)+\frac14\ln^22\zeta(2)-\frac{1}{8}\ln^42}$$

Now collect the result of $\mathcal{J}_1$ and $\mathcal{J}_2$ we get

$$\boxed{\mathcal{J}=5\operatorname{Li_4}\left(\frac12\right)-\frac{35}{8}\zeta(4)+\frac74\ln2\zeta(3)-\frac54\ln^22\zeta(2)+\frac{5}{24}\ln^42}$$

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The integral $\mathcal{K}$ is already calculated here

$$\boxed{\mathcal{K}=-4\operatorname{Li}_4\left(\frac12\right)-\frac{41}{8}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac1{6}\ln^42}$$

Thus

$$\mathcal{I}_2=\mathcal{J}+\frac12\mathcal{K}=3\operatorname{Li}_4\left(\frac12\right)-\frac{111}{16}\zeta(4)-\frac34\ln^22\zeta(2)+\frac1{8}\ln^42$$

and finally

$$\mathcal{I}=\frac12\ln2\cdot\mathcal{I}_1+\frac12\mathcal{I}_2=\frac32\operatorname{Li}_4\left(\frac12\right)-\frac{111}{32}\zeta(4)-\frac78\ln2\zeta(3)-\frac98\ln^22\zeta(2)+\frac1{16}\ln^42$$

Answer 3

A great Solution by Sujee.

Starting with $x\mapsto \frac{2x}{1+x}$ we obtain

$$\mathcal{I}=\frac12\ln(2)\int_0^1\frac{\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx-\frac12\int_0^1\frac{\ln(1-x)\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx$$ $$\mathcal{I}=\frac12\ln(2)\mathcal{I}_1-\frac12\mathcal{I}_2\tag1$$

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From the solution above we have

$$\mathcal{I}_1=-\frac74\zeta(3)-\frac32\ln2\zeta(2)\tag2$$

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For $\mathcal{I}_1$, let $1-x\mapsto x$ then use the integral representation $\operatorname{Li}_2(z)=-\int_0^1\frac{z\ln(y)}{1-zy}dy$, we get

$$\mathcal{I}_2=2\int_0^1\int_0^1\frac{\ln(x)\ln(y)}{x-2xy+2y}dy=2\int_0^1\ln(y)\left(\int_0^1\frac{\ln(x)}{x-2xy+2y}dx\right)dy$$

$$=-2\int_0^1\frac{\ln(y)\operatorname{Li}_2\left(\frac{2y-1}{2y}\right)}{2y-1}dy=-\int_{-1}^1\frac{\ln\left(\frac{1+x}{2}\right)\operatorname{Li}_2\left(\frac{x}{1+x}\right)}{x}dx$$

using Landen's identity $-\operatorname{Li}_2\left(\frac{x}{1+x}\right)=\operatorname{Li}_2(-x)+\frac12\ln^2(1+x)$ yields

$$\small{\mathcal{I}_2=\int_{-1}^1\frac{\ln(1+x)\operatorname{Li}_2(-x)}{x}dx-\ln(2)\int_{-1}^1\frac{\operatorname{Li}_2(-x)}{x}dx-\frac12\ln(2)\int_{-1}^1\frac{\ln^2(1+x)}{x}dx+\frac12\int_{-1}^1\frac{\ln^3(1+x)}{x}dx}$$

$$=-\frac12\operatorname{Li}_2^2(-x)|_{-1}^1-\ln(2)\operatorname{Li}_3(-x)|_{-1}^1-\frac12\ln(2)\int_{-1}^1\frac{\ln^2(1+x)}{x}dx+\frac12\int_{-1}^1\frac{\ln^3(1+x)}{x}dx$$

$$=\frac{15}{16}\zeta(4)+\frac74\ln(2)\zeta(3)-\frac12\ln(2)\int_{-1}^1\frac{\ln^2(1+x)}{x}dx+\frac12\int_{-1}^1\frac{\ln^3(1+x)}{x}dx$$

Lets evaluate the last two integrals:

$$\int_{-1}^1\frac{\ln^2(1+x)}{x}dx=\int_{-1}^0\frac{\ln^2(1+x)}{x}dx+\int_{0}^1\frac{\ln^2(1+x)}{x}dx$$

where $\int_{-1}^0\frac{\ln^2(1+x)}{x}dx=-\int_{0}^1\frac{\ln^2(1-x)}{x}dx=-\int_{0}^1\frac{\ln^2(x)}{1-x}dx=-2\zeta(3)$ and $\int_{0}^1\frac{\ln^2(1+x)}{x}dx=\frac14\zeta(3)$

Similarly

$$\int_{-1}^1\frac{\ln^3(1+x)}{x}dx=\int_{-1}^0\frac{\ln^3(1+x)}{x}dx+\int_{0}^1\frac{\ln^3(1+x)}{x}dx$$

where $\int_{-1}^0\frac{\ln^3(1+x)}{x}dx=-\int_{0}^1\frac{\ln^3(1-x)}{x}dx=-\int_{0}^1\frac{\ln^3(x)}{1-x}dx=6\zeta(3)$ and $\int_0^1\frac{\ln^3(1+x)}{x}dx=6\zeta(4)-\frac{21}4\ln(2)\zeta(3)+\frac32\ln^2(2)\zeta(2)-6\operatorname{Li}_4(1/2)$

By grouping these results we get

$$\mathcal{I}_2=3\operatorname{Li}_4\left(\frac12\right)-\frac{111}{16}\zeta(4)-\frac34\ln^22\zeta(2)+\frac1{8}\ln^42\tag3$$

Plug $(2)$ in $(3)$ in $(1)$, the result of $\mathcal{I}$ follows.