Solving $\int_0^1\frac{\log^4(1-x^2)}{1+x}dx$

Question

This question is similar to my last question.
It's also from the same book, Problem 1.8(iii). $$I=\int_0^1\dfrac{\log^4(1-x^2)}{1+x}dx$$ Wolfram Alpha can't give the closed form, it is only giving a numerical expression.
According to the book $$I=\dfrac{16}{5}\log^52-16\log^32\zeta(2)+48\log^22\zeta(3)-54\log2\zeta(4)-24\zeta(2)\zeta(3)+72\zeta(5)$$ I tried taking substitutions like $y=1-x^2$. It didn't take me anywhere. Is there some way to solve this integral?

Answer 1

Too long for a comment

One very elegant way is to exploit The Master Theorem of Series given in (Almost) Impossible Integrals, Sums, and Series (2019). So, we have that

$$\int_0^1 \frac{\log^4(1-x^2)}{1+x} \textrm{d}x$$ $$\small =\frac{1}{4}\lim_{n\to -1/2 }\frac{H_n^5+10H_n^3 H_n^{(2)}+15 H_n (H_n^{(2)})^2+20 H_n^2 H_n^{(3)}+20 H_n^{(2)} H_n^{(3)}+30 H_n H_n^{(4)}+24 H_n^{(5)}}{5 n},$$

and done. One would usually like to use the Polygamma function notation (here one may observe the special cases with the argument $1/2$ are well-known).

Who cares to learn to use The Master Theorem of Series may find examples with its use and power both in (Almost) Impossible Integrals, Sums, and Series (2019) and More (Almost) Impossible Integrals, Sums, and Series (2023).

End of story

Answer 2

Complete solution.

No harmonic sums, no Beta function are used.

The "obvious" part, \begin{align}J&=\int_0^1\frac{\log^4(1-x^2)}{1+x}dx\\ &=2\int_0^1\frac{\log^4(1-x)}{1+x}dx+12\int_0^1\frac{\log^2(1-x)\log^2(1+x)}{1+x}dx+2\underbrace{\int_0^1\frac{\log^4(1+x)}{1+x}dx}_{=\frac{1}{5}\ln^5 2}-\\& \underbrace{\int_0^1\frac{\log^4(\frac{1-x}{1+x})}{1+x}dx}_{u=\frac{1-x}{1+x}}\\ &=2\underbrace{\int_0^1\frac{\log^4(1-x)}{1+x}dx}_{=J_1}+12\underbrace{\int_0^1\frac{\log^2(1-x)\log^2(1+x)}{1+x}dx}_{=J_2}+\frac{2}{5}\ln^5 2-\underbrace{\int_0^1 \frac{\ln^4 u}{1+u}du}_{=J_3}\\ J_3&=\int_0^1\frac{\ln^4 u}{1-u}du-\underbrace{\int_0^1\frac{2u\ln^4 u}{1-u^2}du}_{z=u^2}=\left(1-\frac{1}{16}\right)\int_0^1\frac{\ln^4 u}{1-u}du=\boxed{\frac{45}{2}\zeta(5)}\\ J_1&\overset{u=1-x}=\frac{1}{2}\int_0^1 \frac{\ln^4 u}{1-\frac{u}{2}}du=24\text{Li}_5\left(\frac{1}{2}\right)\\ \end{align}

The hardest part is to compute, \begin{align}J_2=\int_0^1\frac{\log^2(1-x)\log^2(1+x)}{1+x}dx\end{align}

NB:I assume, \begin{align}a\neq 0,\left|a\right|\leq 1,n\geq 1,\text{integer},\int_0^1\frac{\ln^n x}{1-ax}dx=\frac{(-1)^nn!}{a}\text{Li}_{n+1}(a)\end{align}

First addendum. \begin{align}K_n&=\int_0^1 \frac{\ln^n (1+x)}{x}dx\\ &\overset{u=\frac{1}{1+x}}=(-1)^n\int_{\frac{1}{2}}^1\frac{\ln^n u}{u(1-u)}du=\frac{\left(\ln 2\right)^{n+1}}{1+n}+(-1)^n\int_{\frac{1}{2}}^1\frac{\ln^n u}{1-u}du\\ &=\frac{\left(\ln 2\right)^{n+1}}{1+n}+(-1)^n\int_0^1\frac{\ln^n u}{1-u}du-(-1)^n\underbrace{\int_0^{\frac{1}{2}}\frac{\ln^n u}{1-u}du}_{z=2u}\\ \end{align}

For $n\geq 1$, integer, \begin{align}A_n&=\int_0^{\frac{1}{2}} \frac{\ln^n x}{1-x}dx\\ &\overset{z=2x}=\frac{1}{2}\int_0^1\frac{\ln^n\left(\frac{z}{2}\right)}{1-\frac{z}{2}}dz=\frac{1}{2}\int_0^1\frac{\sum_{k=0}^n\binom{n}{k}(-1)^k(\ln 2)^k(\ln z)^{n-k}}{1-\frac{z}{2}}dz\\ &=(-1)^n(\ln 2)^{n+1}+(-1)^n\underbrace{\sum_{k=0}^{n-1}\binom{n}{k}(n-k)!(\ln 2)^k\text{Li}_{n+1-k}\left(\frac{1}{2}\right)}_{l=n-k}\\&=\boxed{(-1)^n(\ln 2)^{n+1}+(-1)^n\sum_{l=1}^n\binom{n}{l}l!(\ln 2)^{n-l}\text{Li}_{l+1}\left(\frac{1}{2}\right)}\\ K_n&=\int_0^1 \frac{\ln^n (1+x)}{x}dx\\ &\overset{u=\frac{1}{1+x}}=(-1)^n\int_{\frac{1}{2}}^1\frac{\ln^n u}{u(1-u)}du=\frac{\left(\ln 2\right)^{n+1}}{1+n}+(-1)^n\int_{\frac{1}{2}}^1\frac{\ln^n u}{1-u}du\\ &=\frac{\left(\ln 2\right)^{n+1}}{1+n}+(-1)^n\int_0^1\frac{\ln^n u}{1-u}du-(-1)^n\int_0^{\frac{1}{2}}\frac{\ln^n u}{1-u}du\\ &=\frac{\left(\ln 2\right)^{n+1}}{1+n}+n!\zeta(n+1)-(-1)^nA_n\\&=\boxed{\frac{-n\left(\ln 2\right)^{n+1}}{1+n}+n!\zeta(n+1)-\sum_{l=1}^n\binom{n}{l}l!(\ln 2)^{n-l}\text{Li}_{l+1}\left(\frac{1}{2}\right)} \end{align} \begin{align}J_2&\overset{u=\frac{1-x}{1+x}}=\int_0^1\frac{\ln^2\left(\frac{2}{1+u}\right)\left(\ln^2\left(\frac{2}{1+u}\right)+\ln^2 u+2\ln\left(\frac{2}{1+u}\right)\ln u\right)}{1+u}du\\ &=\underbrace{\int_0^1 \frac{\ln^4\left(\frac{2}{1+u}\right)}{1+u}du}_{z=\frac{1-u}{1+u}}-2\underbrace{\int_0^1 \frac{\ln^3 (1+u)\ln u}{1+u}du}_{\text{IBP}}+\underbrace{\int_0^1 \frac{\ln^2 (1+u)\ln^2 u}{1+u}du}_{=J_4}+6\times\\&\ln 2\underbrace{\int_0^1 \frac{\ln^2 (1+u)\ln u}{1+u}du}_{\text{IBP}}- 2\ln 2\underbrace{\int_0^1 \frac{\ln(1+u)\ln^2 u}{1+u}du}_{=J_5}-6\ln^2 2\underbrace{\int_0^1 \frac{\ln (1+u)\ln u}{1+u}du}_{\text{IBP}}+\\&\ln^2 2\int_0^1 \frac{\ln^2 u}{1+u}du+2\ln^3 2\int_0^1 \frac{\ln u}{1+u}du\\ &=\frac{\ln^5 2}{5}+\frac{K_4}{2}+J_4-2K_3\ln 2-2J_5\ln 2+3K_2\ln^2 2+\frac{3\zeta(3)\ln^2 2}{2}-\frac{\pi^2\ln^3 2}{6} \end{align} \begin{align} J_5&=-\frac{1}{3}\left(\underbrace{\int_0^1\frac{\ln^3\left(\frac{u}{1+u}\right)}{1+u}du}_{z=\frac{u}{1+u}}-\int_0^1\frac{\ln^3 u}{1+u}du+\int_0^1\frac{\ln^3(1+u)}{1+u}du-3\underbrace{\int_0^1\frac{\ln^2(1+u)\ln u}{1+u}du}_{\text{IBP}}\right)\\ &=-\frac{A_3}{3}-\frac{7\pi^4}{360}-\frac{\ln^4 2}{12}-\frac{K_3}{3} \end{align} Therefore, \begin{align}J_2=3K_2\ln^2 2-\frac{4K_3\ln 2}{3}+\frac{K_4}{2}+\frac{2A_3\ln 2}{3}+\frac{11\ln^5 2}{30}-\frac{\pi^2\ln^3 2}{6}+\frac{3\zeta(3)\ln^2 2}{2}+\frac{7\pi^4\ln 2}{180}+J_4\end{align} Since, \begin{align}K_2&=-\frac{2\ln^3 2}{3}+2\zeta(3)-2\text{Li}_2\left(\frac{1}{2}\right)\ln 2-2\text{Li}_3\left(\frac{1}{2}\right)=\frac{\zeta(3)}{4}\\ A_3&=-\ln^4 2-3\text{Li}_2\left(\frac{1}{2}\right)\ln^2 2-6\text{Li}_3\left(\frac{1}{2}\right)\ln 2-6\text{Li}_4\left(\frac{1}{2}\right)\\ &=-\frac{\ln^4 2}{2}+\frac{\pi^2\ln^2 2}{4}-\frac{21\zeta(3)\ln 2}{4}-6\text{Li}_4\left(\frac{1}{2}\right)\\ K_3&=-\frac{\ln^4 2}{4}+\frac{\pi^2\ln^2 2}{4}-\frac{21\zeta(3)\ln 2}{4}+\frac{\pi^4}{15}-6\text{Li}_4\left(\frac{1}{2}\right)\\ K_4&=-\frac{4\ln^5 2}{5}+\frac{2\pi^2\ln^3 2}{3}-\frac{21\zeta(3)\ln^2 2}{2}-24\text{Li}_4\left(\frac{1}{2}\right)\ln 2+24\zeta(5)-24\text{Li}_5\left(\frac{1}{2}\right) \end{align} then, \begin{align}J_2=-\frac{\ln^5 2}{30}+\frac{\zeta(3)\ln^2 2}{2}-\frac{\pi^4\ln 2}{20}-8\text{Li}_4\left(\frac{1}{2}\right)\ln 2+12\zeta(5)-12\text{Li}_5\left(\frac{1}{2}\right)+J_4\end{align} I assume, \begin{align}\text{Li}_2\left(\frac{1}{2}\right)&=\frac{\pi^2}{12}-\frac{\ln^2 2}{2},\text{Li}_3\left(\frac{1}{2}\right)=\frac{7\zeta(3)}{8}+\frac{\ln^3 2}{6}-\frac{\pi^2 \ln 2}{12}\\ \end{align}

To terminate the computation i have to compute, \begin{align}J_4=\int_0^1 \frac{\ln^2(1+x)\ln^2 x}{1+x}dx\overset{?}=\end{align}

To be continued...

Second addendum.

$m\geq 0, n\geq 0$, integers \begin{align}W_{m,n}&=\int_0^{\frac{1}{2}} \frac{\ln^m(1-x)\ln^nx}{1-x}dx\end{align} Since, \begin{align}W_{2,2}&\overset{\text{IBP}}=-\frac{1}{3}\Big[\ln^3(1-x)\ln^2 x\Big]_0^{\frac{1}{2}}+\frac{2}{3}\underbrace{\int_0^{\frac{1}{2}} \frac{\ln^3(1-x)\ln x}{x}dx}_{u=1-x}\\ &=\frac{\ln^5 2}{3}+\frac{2}{3}\underbrace{\int_0^1\frac{\ln(1-u)\ln^3 u}{1-u}du}_{=J_6}-\frac{2}{3}W_{1,3}\\ W_{3,1}&\overset{\text{IBP}}=-\frac{1}{4}\Big[\ln^4(1-x)\ln x\Big]_0^{\frac{1}{2}}+\frac{1}{4}\underbrace{\int_0^{\frac{1}{2}} \frac{\ln^4(1-x)}{x}dx}_{u=1-x}\\ &=\frac{\ln^5 2}{4}+\frac{1}{4}\int_0^1 \frac{\ln^4 u}{1-u}du-\frac{1}{4}A_4=\frac{\ln^5 2}{4}+6\zeta(5)-\frac{1}{4}A_4\\ \int_0^{\frac{1}{2}}\frac{\ln^4\left(\frac{x}{1-x}\right)}{1-x}dx&=A_4-4W_{1,3}+6W_{2,2}-4W_{3,1}+\underbrace{W_{4,0}}_{=\frac{\ln^5 2}{5}}\\ &=2A_4+\frac{6\ln^5 2}{5}-24\zeta(5)-8W_{1,3}+4J_6\\ J_4&\overset{u=\frac{x}{1+x}}=\int_0^{\frac{1}{2}}\frac{\ln^2(1-u)\ln^2\left(\frac{u}{1-u}\right)}{1-u}du=W_{2,2}-2W_{3,1}+W_{4,0}\\ &=\frac{A_4}{2}+\frac{\ln^5 2}{30}-12\zeta(5)-\frac{2W_{1,3}}{3}+\frac{2J_6}{3} \end{align} then, \begin{align}-\frac{1}{8}\int_0^{\frac{1}{2}}\frac{\ln^4\left(\frac{x}{1-x}\right)}{1-x}dx+\frac{3}{2}J_4&=\frac{A_4}{2}-\frac{\ln^5 2}{10}-15\zeta(5)+\frac{J_6}{2}\\ J_4&=\frac{A_4}{3}-\frac{\ln^5 2}{15}-10\zeta(5)+\frac{1}{12}\underbrace{\int_0^{\frac{1}{2}}\frac{\ln^4\left(\frac{x}{1-x}\right)}{1-x}dx}_{u=\frac{x}{1-x}}+\frac{J_6}{3}\\ &=\frac{A_4}{3}-\frac{\ln^5 2}{15}-\frac{65\zeta(5)}{8}+\frac{J_6}{3}\\ \end{align} Since, \begin{align}A_4=\ln^5 2-\frac{2\pi^2\ln^3 2}{3}+\frac{21\zeta(3)\ln^2 2}{2}+24\text{Li}_4\left(\frac{1}{2}\right)\ln 2+24\text{Li}_5\left(\frac{1}{2}\right)\end{align} then, \begin{align}J_4&=\frac{4\ln^5 2}{15}-\frac{2\pi^2\ln^3 2}{9}+\frac{7\zeta(3)\ln^2 2}{2}+8\text{Li}_4\left(\frac{1}{2}\right)\ln 2-\frac{65\zeta(5)}{8}+8\text{Li}_5\left(\frac{1}{2}\right)+\frac{J_6}{3}\end{align} To terminate the computation, i have to compute $\displaystyle J_6=\int_0^1 \frac{\ln(1-x)\ln^3 x}{1-x}dx$.

The use of Beta function, harmonic sums are possible ways, but there is an alternative.

To be continued...

Epilogue.

\begin{align}J_6&\overset{\text{IBP}}=\left[\ln(1-x)\left(\int_0^x\frac{\ln^3 t}{1-t}dt-\int_0^1\frac{\ln^3 t}{1-t}dt\right)\right]_0^1+\\&\int_0^1\int_0^1\left(\frac{x\ln^3(tx)}{(1-x)(1-tx)}- \frac{\ln^3 t}{(1-x)(1-t)}\right)dtdx\\ &=\int_0^1\int_0^1\left(\frac{\ln^3 x+3\ln^2x\ln t+3\ln x\ln^2 t}{(1-x)(1-t)}-\frac{\ln^3(tx)}{(1-t)(1-tx)}\right)dtdx\\ &\overset{\text{Fubini}}=-2\pi^2\zeta(3)+\int_0^1\int_0^1\left(\frac{\ln^3 x}{(1-x)(1-t)}-\frac{\ln^3(tx)}{(1-t)(1-tx)}\right)dtdx\\ &=-2\pi^2\zeta(3)+\int_0^1\left(\frac{1}{1-t}\left(\int_0^1 \frac{\ln^3 x}{1-x}dx\right)-\frac{1}{t(1-t)}\left(\int_0^t\frac{\ln^3 u}{1-u}du\right)\right)dt\\ &=-2\pi^2\zeta(3)+\int_0^1\left(\frac{1}{1-t}\left(\int_t^1 \frac{\ln^3 x}{1-x}dx\right)-\frac{1}{t}\left(\int_0^t\frac{\ln^3 u}{1-u}du\right)\right)dt\\ &\overset{\text{IBP}}=-2\pi^2\zeta(3)-J_6+\underbrace{\int_0^1\frac{\ln^4 t}{1-t}dt}_{=24\zeta(5)}\\ J_6&=\boxed{\int_0^1 \frac{\ln(1-t)\ln^3 t}{1-t}dt=12\zeta(5)-\pi^2\zeta(3)}\\ J_4&=\\&\boxed{\int_0^1 \frac{\ln^2(1+t)\ln^2 t}{1+t}dt=\frac{4\ln^5 2}{15}-\frac{2\pi^2\ln^3 2}{9}+\frac{7\zeta(3)\ln^2 2}{2}+8\text{Li}_4\left(\frac{1}{2}\right)\ln 2-\frac{\pi^2\zeta(3)}{3}-\frac{33\zeta(5)}{8}+8\text{Li}_5\left(\frac{1}{2}\right)}\\ J_2&=\\&\boxed{\int_0^1\!\frac{\ln^2(1-t)\!\ln^2(1+t)}{1+t}\!\!dt\!=\frac{7\ln^5 2}{30}\!-\!\frac{2\pi^2\ln^3 2}{9}+4\zeta(3)\ln^2 2-\frac{\pi^4\ln 2}{20}-\frac{\pi^2\zeta(3)}{3}+\frac{63\zeta(5)}{8}-4\text{Li}_5\left(\frac{1}{2}\right)}\\ J&=\\&\boxed{\int_0^1 \frac{\ln^4(1-t^2)}{1+t}dt=\frac{16\ln^5 2}{5}-\frac{8\pi^2\ln^3 2}{3}+48\zeta(3)\ln^2 2-\frac{3\pi^4\ln 2}{5}-4\pi^2\zeta(3)+72\zeta(5)} \end{align} Proof completed.

Bonus. \begin{align}\int_0^1 \frac{\ln(1+x)\ln^3 x}{1+x}dx&=\frac{1}{4}\int_0^1 \frac{\ln^4(1+x)}{1+x}dx-\underbrace{\frac{1}{4}\int_0^1 \frac{\ln^4\left(\frac{x}{1+x}\right)}{1+x}dx}_{u=\frac{x}{1+x}}+\frac{3}{2}J_4+\\&\frac{1}{4}\int_0^1\frac{\ln^4 x}{1+x}dx-\underbrace{\int_0^1 \frac{\ln^3(1+x)\ln x}{1+x}dx}_{\text{IBP}}\\ &=\frac{\ln^5 2}{20}-\frac{A_4}{4}+\frac{3}{2}J_4+\frac{45\zeta(5)}{8}+\frac{K_4}{4}=\boxed{\frac{87\zeta(5)}{16}-\frac{\pi^2\zeta(3)}{2}} \end{align}

Answer 3

Hopefully an Answer in progress If I can work on it but just consider it as a long comment right now.

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$$I=\int_{0}^{1}\frac{\ln^4\left(1-x^{2}\right)}{1+x}dx$$

Rewrite: $$\ln\left(1-x^{2}\right)=\ln\left(\frac{1-x}{1+x}\right)+2\ln\left(1+x\right)$$ and expand.

Consider : $$I(m,n)=\int_{0}^{1}\frac{\ln\left(\frac{1-x}{1+x}\right)^{m}\ln\left(1+x\right)^{n}}{1+x}dx$$

Then, $$I=I\left(4,0\right)+16I\left(0,4\right)+24I\left(2,2\right)+32I\left(1,3\right)+8I\left(3,1\right)$$

By Wolfram: $$16I(0,4)=\frac{16}{5}\ln^5\left(2\right)$$

Which is also a term directly present in the answer given in OP.
This makes me think that this could be the correct approach.

Also by Wolfram,

$$I(4,0)=\frac{45}{2}\zeta(5)$$