How to evaluate $\int_0^1 \ln ^3(1+x) \ln (1-x) d x$?

Question

QUESTION:How to evaluate $$\int_0^1 \ln ^3(1+x) \ln (1-x) d x$$?

I'm not sure of the closed form of the integral, as I haven't evaluated it yet. However, after evaluating the integral $$\int_0^1 \ln (1+x) \ln (1-x) \, dx,$$ I thought about integrating $$\int_0^1 \ln^3(1+x) \ln (1-x) \, dx.$$

I am also interested in evaluation of $$\int_0^1 \ln^n(1+x) \ln (1-x) \, dx.$$ If possible

Here is my attempt

\begin{aligned} & I=\int_0^1 \ln ^3(1+x) \ln (1-x) d x \\ & \text { Let }: a=\ln (1-x) \wedge b=\ln (1+x) \\ & \text { and } \quad b^3 a=\frac{1}{8}(a+b)^4-\frac{1}{8}(a-b)^4-a^3 b \\ & \Rightarrow I=\frac{1}{8} \underbrace{\int_0^1 \ln ^4\left(1-x^2\right) d x}_A-\frac{1}{8} \underbrace{\int_0^1 \ln ^4\left(\frac{1-x}{1+x}\right) d x}_B \\ & -\underbrace{\int_0^1 \ln ^3(1-x) \ln (1+x) d x}_C \\ & A=\frac{1}{2} \int_0^1 x^{-\frac{1}{2}} \ln ^4(1-x) d x=\left.\frac{1}{2} \mathrm{~B}^{(4)}\left(\frac{1}{2}, a\right)\right|_{a=1} \\ & =\frac{1}{2} \mathrm{~B}\left(\frac{1}{2}, a\right)\left\{\left[\psi^{(0)}(a)-\psi^{(0)}\left(a+\frac{1}{2}\right)\right]^4\right. \\ & +6\left[\psi^{(1)}(a)-\psi^{(1)}\left(a+\frac{1}{2}\right)\right]\left[\psi^{(0)}(a)-\psi^{(0)}\left(a+\frac{1}{2}\right)\right]^2 \\ & +4\left[\psi^{(2)}(a)-\psi^{(2)}\left(a+\frac{1}{2}\right)\right]\left[\psi^{(0)}(a)-\psi^{(0)}\left(a+\frac{1}{2}\right)\right] \\ & \left.+3\left[\psi^{(1)}(a)-\psi^{(1)}\left(a+\frac{1}{2}\right)\right]^2+\psi^{(3)}(a)-\psi^{(3)}\left(a+\frac{1}{2}\right)\right\}\left.\right|_{a=1} \\ & \text { But }: \psi^{(0)}=-\gamma \wedge \psi^{(0)}\left(\frac{3}{2}\right)=2-2 \ln (2)-\gamma \\ \text{and to $n \in \mathbb{N}^{+}$:}\\ & \star \psi^{(n)}(1)-\psi^{(n)}\left(\frac{3}{2}\right)=(-1)^{n+1} n!\left[2^{n+1}+\left(2-2^{n+1}\right) \zeta(n+1)\right] \\ & \Rightarrow A=(2 \ln (2)-2)^4+6(4-2 \zeta(2))(2 \ln (2)-2)^2 \\ & +4(12 \zeta(3)-16)(2 \ln (2)-2)+3(4-2 \zeta(2))^2+96-84 \zeta(4) \\ & \Rightarrow A=-48 \ln ^2(2) \zeta(2)+96 \ln (2)(\zeta(3)+\zeta(2))+16 \ln ^4(2) \\ & -64 \ln ^3(2)+192 \ln ^2(2)-54 \zeta(4)-96 \zeta(3)-96 \zeta(2) \\ & -384 \ln (2)+384 \ldots(\alpha) \\ & \end{aligned}

As you can see, I was able to solve the integral $A$, but I don't have any ideas for integrals $B$ and $C.$

Answer 1

Solution for integral B

Letting $t=\frac{1-x}{1+x}$ transforms the integral into $$ \begin{aligned} I & =\int_0^1 \ln ^4 t \frac{2 d t}{(1+t)^2} \\ & =2 \int_0^1 \frac{\ln ^4 t}{(1+t)^2} d t\\&= \left. I^{(4)}(a)\right|_{a=0} \end{aligned} $$ where $$ \begin{aligned} I(a) & =\int_0^1 \frac{t^a}{(1+t)^2} d t \\ & =\sum_{n=1}^{\infty}(-1)^{n-1} n \int_0^1 t^{a+n-1} d t \\ & =\sum_{n=1}^{\infty} \frac{(-1)^{n-1} n}{a+n} \end{aligned} $$ Differentiating $I(a)$ w.r.t.$a$ by four times at $a=0$ yields $$ \begin{aligned} I^{(4)}(0) & =\left.\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x(-1)^4 4!}{(a+n)^5}\right|_{a=0} \\ & =24 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^4}\\ & =24\left[\sum_{n=1}^{\infty} \frac{1}{n^4}-2 \sum_{n=1}^{\infty} \frac{1}{(2 n)^4}\right] \\ & =21 \zeta(4) \\ & =\frac{21 \pi^4}{90} \\ & =\frac{7 \pi^4}{30} \end{aligned} $$ Hence we get $$ \boxed{\int_0^1 \ln ^4\left(\frac{1-x}{1+x}\right) d x=2 \cdot \frac{7 \pi^4}{30}=\frac{7 \pi^4}{15}} $$ In general,

$$\int_0^1 \ln ^n\left(\frac{1-x}{1+x}\right) d x=(-1)^n 2 n!\left(1-\frac{1}{2^{n-1}}\right) \zeta(n)$$

Answer 2

$\color{brown}{\textbf{Integral B.}}$

$$I_B=\int\limits_0^1 \ln^4\left(\dfrac{1+x}{1-x}\right)\,\text dx =16\int\limits_0^1 \operatorname{arctanh}^4 x\,\text dx =\dfrac7{15}\pi^4,$$ $$I_B\approx 45.45757\ 58158\ 67804\ 04367\ 21552\ 54729.$$ Really, $$I_B=16\int\limits_0^1 \operatorname{arctanh}^4 x\,\text dx =16\int\limits_0^\infty \dfrac{t^4\,\text dt}{\cosh^2t} =64\int\limits_0^\infty \dfrac{t^4\,e^{-2t}\,\text dt}{\left(1+e^{-2t}\right)^2}$$ $$=64\sum\limits_{k=1}^\infty (-1)^{k+1} k\int\limits_0^\infty t^4 e^{-2kt}\,\text dt =64\sum\limits_{k=1}^\infty (-1)^{k+1} k\cdot \dfrac3{4k^5}$$ $$=48\sum\limits_{k=1}^\infty \dfrac{(-1)^{k+1}}{k^4} =\dfrac7{15}\pi^4.$$

$\color{brown}{\textbf{Main integral.}}$

Substitution $$x=e^t-1,\quad \text dx=e^t\text dt,\quad \ln(1+x)=t,$$ $$I=\int\limits_0^1 \ln^3(1+x)\ln(1-x)\,\text dx =\int\limits_0^{\ln2} t^3 e^t \ln\left(2-e^{t}\right)\,\text dt$$ $$=\int\limits_0^{\ln2} t^3 e^t \left(\ln2+\ln\left(1-\frac12e^t\right)\right)\,\text dt=I_0+I_1,$$ where $$I_0=\ln 2\int\limits_0^{\ln2}t^3 e^t\,\text dt,\quad I_1=\int\limits_0^{\ln2} t^3 e^t \ln\left(1-\frac12e^t\right)\,\text dt.\tag1$$ Then $$I_0=\ln2\cdot e^t\left(t^3-3t^2+6t-6\right)\bigg|_0^{\ln2} =2(\ln2)^4-6(\ln2)^3+12(\ln2)^2-12\ln2+6\ln2,$$ $$I_0=2\ln^4(2)-6\ln^3(2)+12\ln^2(2)-6\ln2.\tag2$$ Also, $$I_1=\int\limits_0^{\ln2} t^3 e^t \ln\left(1-\frac12e^t\right)\,\text dt =-\int\limits_0^{\ln2} t^3 e^t \sum\limits_{k=1}^{\infty}\dfrac1k\dfrac{e^{kt}}{2^k} \,\text dt =-\sum\limits_{k=1}^{\infty}\dfrac{2^{-k}}{k}\,I_{1,k+1},$$ where $$I_{1,k}=\int\limits_0^{\ln2} t^3 e^{kt}\text dt =\dfrac1{k^4}\int\limits_0^{\ln2} (kt)^3 e^{kt}\text d(kt) =\dfrac1{k^4}\int\limits_0^{k\ln2} t^3 e^{t}\text dt$$ $$=\dfrac1{k^4}e^t\left(t^3-3t^2+6t-6\right)\bigg|_0^{k\ln2}$$ $$=\dfrac6{k^4}+2^k\left(\dfrac{\ln^3(2)}{k}-\dfrac{3\ln^2(2)}{k^2}+\dfrac{6\ln2}{k^3}-\dfrac{6}{k^4}\right),$$ Let $$J_0=-\sum\limits_{k=1}^{\infty}\dfrac{2^{-k}}{k}\,\dfrac6{(k+1)^4} =-24+π^2(1-\ln(2))-6\ln^2(2)$$ $$+2\ln^3(2)+\ln(64) +12\operatorname{Li_4}\left(\dfrac12\right)+\dfrac{21}2 ζ(3),$$ $$\begin{align} &J_1=-\sum\limits_{k=1}^{\infty}\dfrac{2\ln^3(2)}{k(k+1)}=-2\ln^3(2),\\[4pt] &J_2=\sum\limits_{k=1}^{\infty}\dfrac{6\ln^2(2)}{k(k+1)^2}=(12-\pi^2)\ln^2(2),\\[4pt] &J_3=-\sum\limits_{k=1}^{\infty}\dfrac{12\ln(2)}{k(k+1)^3}=2\ln(2)(π^2-18+6 ζ(3)),\\[4pt] &J_4=\sum\limits_{k=1}^{\infty}\dfrac{12}{k(k+1)^4}=48-2\pi^2-\dfrac2{15}π^4- 12ζ(3). \end{align}$$ Then $$\begin{align} &I_1=J_0+J_1+J_2+J_3+J_4=24-\pi^2-\dfrac2{15}\pi^4-\dfrac32\zeta(3) \\[4pt] &+\big(\pi^2-30+12\zeta(3)\big)\ln(2)+(6-\pi^2)\ln^2(2) +12\operatorname{Li_4}\left(\dfrac12\right), \end{align}\tag3$$ $$\color{green}{\mathbf{\begin{align}&I=24-\pi^2-\dfrac2{15}\pi^4-\dfrac32\zeta(3)+\big(\pi^2-36+12\zeta(3)\big)\ln(2)+(18-\pi^2)\ln^2(2)\\[4pt] &-6\ln^3(2)+2\ln^4(2)+12\operatorname{Li_4}\left(\dfrac12\right)\approx -0.19480\ 62619\ 27738\ 36514\ 85561\ 20791.\end{align}\tag4}}$$ Obtained result corresponds with WA calculations.

Answer 3

Too long for a comment: For a first solution, use a similar strategy to the one here: https://math.stackexchange.com/q/4917671.

For a second solution, based on reducing the calculations to harmonic series, combine $$\log^3(1+x) = \sum (-1)^n \frac{{x}^{n + 1}}{n + 1} (H_n^2 - H_n^{(2)})$$ together with the integral identity $$\int_0^1 x^{n-1}\log(1-x) dx = - \frac{H_n}{n},$$ both presented in (Almost) Impossible Integrals, Sums, and Series (2019). If interested, similar and powerful identities also involving skew-harmonic numbers can be found in the sequel, More (Almost) Impossible Integrals, Sums, and Series (2023).