# Resistant integral $\int_0^1\left(\frac{\ln^2(1-x)\ln^2(1+x)}{1-x}-\frac{\ln^2(2)\ln^2(1-x)}{1-x}\right)\ dx$

Prove, without using harmonic series, that

$$I=\int_0^1\left(\frac{\ln^2(1-x)\ln^2(1+x)}{1-x}-\frac{\ln^2(2)\ln^2(1-x)}{1-x}\right)\ dx$$ $$=\frac18\zeta(5)-\frac12\ln2\zeta(4)+2\ln^22\zeta(3)-\frac23\ln^32\zeta(2)-2\zeta(2)\zeta(3)+\frac1{10}\ln^52+4\operatorname{Li}_5\left(\frac12\right)$$

This problem was proposed by Cornel and can be found here.

The main reason behind such constraint is that this integral can be simplified into $$S=\sum_{n=1}^\infty\frac{H_n}{n^42^n}$$ which was calculated here using real and complex methods. So evaluating $$I$$ without using harmonic series means we are providing a third solution to $$S$$.

I have already computed this integral ( will be posted soon) but I would like to see variant approaches.

Thanks.

In case the reader is curious about how this integral is related to $$\sum_{n=1}^\infty\frac{H_n}{n^42^n}$$, here is the steps

By integration by parts we have \begin{align} I&=\frac23\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{1+x}\ dx\overset{\color{red}{1-x\ \mapsto\ x}}{=}\frac13\int_0^1\frac{\ln^3x\ln(2-x)}{1-x/2}\ dx\\ &=\frac{\ln2}{3}\int_0^1\frac{\ln^3x}{1-x/2}\ dx+\frac13\int_0^1\frac{\ln^3x\ln(1-x/2)}{1-x/2}\ dx\\ &=\frac{\ln2}{3}\sum_{n=1}^\infty\frac{1}{2^{n-1}}\int_0^1x^{n-1}\ln^3x\ dx-\frac13\sum_{n=1}^\infty\frac{H_n}{2^n}\int_0^1x^n\ln^3x\ dx\\ &=\frac{\ln2}{3}\sum_{n=1}^\infty\frac{1}{2^{n-1}}\left(-\frac{6}{n^4}\right)-\frac13\sum_{n=1}^\infty\frac{H_n}{2^n}\left(-\frac{6}{(n+1)^4}\right)\\ &=-4\ln2\sum_{n=1}^\infty\frac{1}{n^42^n}+2\sum_{n=1}^\infty\frac{H_n}{(n+1)^42^n}\\ &=-4\ln2\operatorname{Li}_4\left(\frac12\right)+4\sum_{n=1}^\infty\frac{H_n}{n^42^n}-4\operatorname{Li}_5\left(\frac12\right) \end{align}

• What makes it “resistant”? Also, if you would like to see a variant, then we need to see what you already have. – gen-ℤ ready to perish Jul 21 '19 at 1:15
• you will come across tough integrals and you need some manipulations. I mentioned solution will be posted soon i am working on it. – Ali Shadhar Jul 21 '19 at 1:17

$$I=\int_0^1\frac{\ln^2(1-x)}{1-x}\left(\ln^2(1+x)-\ln^2(2)\right)\ dx\overset{IBP}{=}\frac23\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{1+x}\ dx$$ Using the algebraic identity $$\quad\displaystyle a^3b=\frac18(a+b)^4-\frac18(a-b)^4-ab^3$$

and by setting $$a=\ln(1-x)$$ and $$b=\ln(1+x)$$, we get

\begin{align} I=\frac1{12}\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{1+x}\ dx}_{\displaystyle I_1}-\frac1{12}\underbrace{\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{1+x}\ dx}_{\displaystyle I_2}-\frac23\underbrace{\int_0^1\frac{\ln(1-x)\ln^3(1+x)}{1+x}\ dx}_{\displaystyle I_3} \end{align}

The first integral was nicely done by Cornel and can be found in his book Almost Impossible Integral , Sums, and Series page $$80$$ and as follows:

\begin{align} I_1&=\int_0^1\frac{\ln^4(1-x^2)}{1+x}\ dx=\int_0^1(1-x)\frac{\ln^4(1-x^2)}{1-x^2}\ dx\overset{x^2=y}{=}\frac12\int_0^1\frac{1-\sqrt{y}}{\sqrt{y}}.\frac{\ln^4(1-y)}{1-y}\ dy\\ &\overset{IBP}{=}-\frac1{20}\int_0^1\frac{\ln^5(1-y)}{y^{3/2}}\ dy=-\frac{1}{20}\lim_{x\mapsto-1/2\\y\mapsto1}\frac{\partial^5}{\partial y^5}\text{B}(x,y)\\ &\boxed{I_1=\frac{16}5\ln^52-16\ln^32\zeta(2)+48\ln^22\zeta(3)-54\ln2\zeta(4)-24\zeta(2)\zeta(3)+72\zeta(5)} \end{align}

\begin{align} I_2=\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{1+x}\ dx\overset{\frac{1-x}{1+x}=y}{=}\int_0^1\frac{\ln^4x}{1+x}\ dx=\boxed{\frac{45}2\zeta(5)=I_2} \end{align}

\begin{align} I_3&=\int_0^1\frac{\ln(1-x)\ln^3(1+x)}{1+x}\ dx\overset{\frac1{1+x}=y}{=}-\int_{1/2}^1 \frac{\ln\left(\frac{2x-1}{x}\right)\ln^3x}{x}\ dx\\ &=\int_{1/2}^1\frac{\ln^4x}{x}\ dx-\int_{1/2}^1\frac{\ln(2x-1)\ln^3x}{x}\ dx, \quad \ln(2x-1)=\ln(1-2x)-i\pi\\ &=\frac15\ln^52-\int_{1/2}^1\frac{\ln(1-2x)\ln^3x}{x}\ dx-i\frac{\pi}{4}\ln^42\\ &=\frac15\ln^52+\sum_{n=1}^\infty\frac{2^n}{n}\int_{1/2}^1x^{n-1}\ln^3x\ dx-i\frac{\pi}{4}\ln^42\\ &=\frac15\ln^52+\sum_{n=1}^\infty\frac{2^n}{n}\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)-i\frac{\pi}{4}\ln^42\\ &=\frac15\ln^52+\ln^32\zeta(2)+3\ln^22\zeta(3)+6\ln2\zeta(4)+6\zeta(5)-6\operatorname{Li}_5(2)-i\frac{\pi}{4}\ln^42 \qquad\qquad\quad(1) \end{align} Using the polylogarithmic identity: $$\operatorname{Li}_5(x)=-\frac74\zeta(4)\ln(-x)-\frac16\zeta(2)\ln^3(-x)-\frac1{120}\ln^5(-x)+\operatorname{Li}_5(1/x)$$

Set $$x=2$$, we get

$$\operatorname{Li}_5(2)=2\ln2\zeta(4)+\frac13\ln^32\zeta(2)-\frac1{120}\ln^52+\operatorname{Li}_5\left(\frac12\right)-i\frac{\pi}{24}\ln^42\tag{2}$$

Plugging $$(2)$$ in $$(1)$$, we get $$\boxed{I_3=-6\operatorname{Li}_5\left(\frac12\right)+6\zeta(5)-6\ln2\zeta(4)+3\ln^22\zeta(3)-\ln^32\zeta(2)+\frac14\ln^52}$$

Combining the boxed results, we get the closed form of $$I$$.

• Reffering to this question: math.stackexchange.com/q/3293328/515527, you have solved one part of it. If we apply the same substitution: $\frac{1}{1+x}=y$ we get $$\int_0^1\frac{\ln^3(1+x)\ln(1-x)}{x}=\int_{1/2}^1 \frac{\ln\left(\frac{2x-1}{x}\right)\ln^3x}{x}dx-\int_{1/2}^1 \frac{\ln\left(\frac{2x-1}{x}\right)\ln^3x}{1-x}dx$$ The second one is still painful. – Zacky Jul 27 '19 at 22:44
• Good point. yes the first integral is manageable but the second one is a hell of integral and I think its more complicated than the original integral due to the limit $1/2$. Still, I will try it and see what I get. – Ali Shadhar Jul 27 '19 at 22:49
• It would be awesome if he can solve it without using the tough series I used. – Ali Shadhar Jul 27 '19 at 23:01
• This is what I got $$\int_0^1\frac{\ln^3(1+x)\ln(1-x)}{x}\ dx=3\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(H_n^3-H_nH_n^{(2)}-2\frac{H_n^2}{n}+2\frac{H_n}{n^2}\right)$$. Or we can relate it to the integral $\int_0^1\frac{\ln(1+x)\ln^3(1-x)}{x}\ dx$ using the algebraic identity $(a+b)^4-(a-b)^4=8a^3b+8ab^3$ where $\int_0^1 \frac{(a+b)^4}{x}$ and $\int_0^1 \frac{(a-b)^4}{x}$ are easy to do. – Ali Shadhar Jul 27 '19 at 23:49