how to integrate $\int_0^1 \ln^4(1+x) \ln(1-x) \, dx$?

Question

I'm trying to evaluate the integral $$\int_0^1 \ln^4(1+x) \ln(1-x) \, dx,$$ and I'd like some help with my approach and figuring out the remaining steps.

or is it possible to evaluate $$\int_0^1 \ln^n(1+x) \ln(1-x) \, dx \; ?$$

My Attempt

First, I transformed the variable to simplify the integral: $$ \Omega = \int_0^1 \ln^4(1+x) \ln(1-x) \, dx \stackrel{1+x \rightarrow x}{=} \int_1^2 \ln^4(x) \ln(2-x) \, dx $$ This allowed me to split the integral into two parts: $$ \int_1^2 \ln^4(x) \left( \ln(2) + \ln\left(1-\frac{x}{2}\right) \right) \, dx = \ln(2) \underbrace{\int_1^2 \ln^4(x) \, dx}_{A} + \underbrace{\int_1^2 \ln^4(x) \ln\left(1-\frac{x}{2}\right) \, dx}_{B} $$ So we have: $$ \Omega = A \ln(2) + B $$

I was able to evaluate part $A$: $$ A = \int_1^2 \ln^4(x) \, dx = \left. \left( 24x + x \ln^4(x) - 4x \ln^3(x) + 12x \ln^2(x) - 24x \ln(x) \right) \right|_1^2 $$ Evaluating this, we get: $$ A = 24 + 2 \ln^4(2) - 8 \ln^3(2) + 24 \ln^2(2) - 48 \ln(2) $$

However, I'm stuck on how to handle part $B$. I think it involves expanding $\ln\left(1 - \frac{x}{2}\right)$ using a Taylor series, but I'm unsure how to proceed from here. Any suggestions are welcome

closed form: \begin{align*} &-6\left(8\text{Li}_{4}\left(\frac{1}{2}\right)+8\text{Li}_{5}\left(\frac{1}{2}\right)-\zeta(3)-8\zeta(5)+20\right)+\log(4)\left(6\zeta(3)(\log(4)-4)+96+(\log(2)-4)\log(2)(12+(\log(2)-1)\log(2))\right)\\ &-\frac{4}{3}\pi^{2}\left(-3+\log^{3}(2)+\log(8)-\log(2)\log(8)\right)-\frac{8}{15}\pi^{4}(\log(2)-1) \end{align*}

Thanks David G. Stork for the closed form!

Answer 1

$$\int\log(1-x)\log^4(x+1)dx$$

Integrating this by parts we get

$$(-x-\log(1-x)(1-x)+1)\log^4(x+1)-\int\frac{4(-x+\log(1-x)(x-1)+1)\log^3(x+1)}{x+1}dx$$

Factoring a $-4 $ out and focusing on the integral

$$\int\frac{4(-x+\log(1-x)(1-x)+1)\log^3(x+1)}{x+1}dx=\int\frac{(u+\log(2-u)(2-u)-2)\log^3(u)}{u}du\\=-\int\frac{\log(2-u)(u-2)\log^3(u)}{u}du-2\int\frac{\log^3(u)}{u}du+\int\log^3(u)du$$ Where $u=x+1$

Let's call $I_1=\int\frac{\log(2-u)(u-2)\log^3(u)}{u}du$ and $I_2=\int\frac{\log^3(u)}{u}du$ and $I_3=\int\log^3(u)du$

$$I_1=\int\frac{\log(2-u)(u-2)\log^3(u)}{u}du=\int\frac{\log(2-u)u\log^3(u)-2\log(2-u)\log^3(u)}{u}du\\=\int\log(2-u)\log^3(u)du-2\int\frac{\log(2-u)\log^3(u)}{u}du$$

We use integration by parts to the first integral and we get

$$(-u-\log(2-u)(2-u)+2)\log^3(u))-\int\frac{3(-u+\log(2-u)(u-2)+2)\log^2(u)}{u}du$$

Solving $$\int\frac{3(-u+\log(2-u)(u-2)+2)\log^2(u)}{u}du\\=-\int\frac{\log(2-u)(u-2)\log^2(u)}{u}du-2\int\frac{\log^2(u)}{u}du+\int\log^2(u)du$$

Solving $$\int\frac{\log(2-u)(u-2)\log^2(u)}{u}du\\=\int\log(2-u)\log^2(u)du-2\int\frac{\log(2-u)\log^2(u)}{u}du$$

Integrate by parts the first integral and get

$$\int\log(2-u)\log^2(u)du=(-u-\log(2-u)(2-u)+2)\log^2(u)-2\left (\int\frac{\log(2-u)(2-u)\log(u)}{u}du-2\int\frac{\log(u)}{u}du+\int\log(u) \right )\\=-2\left (\int\frac{\log(2-u)(2-u)\log(u)}{u}du-2(\frac{\log^2(u)}{2})+u\log(u)-u \right )$$

And to solve the other integral we write it as

$$\int\log(2-u)\log(u)du-2\int\frac{\log(2-u)\log(u)}{u}du\\$$

Integrating by parts the first one we get

$$\log(2-u)(u\log(u)-u)-\int\frac{u-u\log(u)}{2-u}du$$

We get

$$\int\frac{u-u\log(u)}{u-2}du=\int\frac{u\log(u)}{u-2}du-\int\frac{u}{u-2}du$$

To solve the first integral we can use the substitution $v=u-2$ and get

$$2\int\frac{\log(v+2)}{v}dv+=\int\frac{\log(\frac{v}{2}+1)}{v}dv+\log(2)\int\frac{1}{v}dv$$

this is just

$$-\operatorname{Li}_2(-\frac{v}{2})+\log(2)\log(v)$$

Working back we can see $$\int\log(v+2)dv=(v+2)\log(v+2)-v-2$$

That means $$2\int\frac{\log(v+2)}{v}dv+\int\log(v+2)=u\log(u)-u+2\log(2)\log(u-2)-2\operatorname{Li}_2(-\frac{u-2}{2})$$

And

$$\int\frac{u}{u-2}dv=u+2\log(u-2)-2$$

Therefore

$$\log(2-u)(u\log(u)-u)-\int\frac{u-u\log(u)}{2-u}dv\\=\log(2-u)(u\log(u)-u)-u\log(u)+2u-2\log(2)\log(u-2)+2\log(u-2)+2\operatorname{Li}_2(-\frac{u-2}{2})-2$$

Now solving

$$\int\frac{\log(2-u)\log(u)}{u}du$$

We substitute $v=\log(u)$

$$\int v\log(2-e^v)dv=\frac{v^2\log(2-e^v)}{2}-\int-\frac{v^2e^v}{2(2-e^v)}dv$$

To solve

$$\int-\frac{v^2e^v}{2(2-e^v)}dv\\=\log(-\frac{w}{2})\log^2(w+2)-2\int\frac{\log(-\frac{w}{2})\log(w+2)}{w+2}dw$$

We substitute $w=e^v-2$ , factor out constants and focus on the integral and get

$$\int\frac{\log^2(w+2)}{w}dw=\log(-\frac{w}{2})\log(w+2)-2\int\frac{\log(-\frac{w}{2})\log(w+2}{w+2}dw$$

To solve the integral we integrate by parts in order to solve using polylogarithms and get that

$$\int\log(2-u)\log(u)du-2\int\frac{\log(2-u)\log(u)}{u}du\\=\log(-\frac{u-2}{2})\log^2(u)-\log(2-u)\log^2(u)+\log(2-u)(u\log(u)-u)-u\log(u)+2\operatorname{Li}_2(\frac{u-2}{2}+1)\log(u)+2u-2$$

Also $$\int\frac{\log(u)}{u}du=\frac{\log^2(u)}{2}$$

Next we substitute $v=\log(u)$ and easily get that $$\int\frac{\log^2(u)}{u}du=\frac{\log^3(u)}{3}$$

Integrate by parts to get $$\int \log^2(u)du=u(\log^2(u)-2\log(u)+2)$$

By using the substitution $u=e^v$ , integrating by parts and using previous results we can find that $$\int\frac{\log(2-u)\log^3(u)}{u}du=$$

$$ \int \frac{\log(2 - x) \log^3(x)}{x} \, dx = 6 \, \text{Li}_5\left(\frac{x}{2}\right) - \text{Li}_2\left(\frac{x}{2}\right) \log^3(x) + 3 \, \text{Li}_3\left(\frac{x}{2}\right) \log^2(x) - 6 \, \text{Li}_4\left(\frac{x}{2}\right) \log(x) + \frac{1}{4} \log(2) \log^4(x) + \text{constant} $$

Pluging those results in we get that

$$I_1= \ln(2 - u) \ln^3(u) \, du - 2 \int \frac{\ln(2 - u) \ln^3(u)}{u} \, du = \ln\left(-\frac{u - 2}{2}\right) \frac{\ln^4(u)}{2} - \ln(2 - u) \frac{\ln^4(u)}{2} + \left(-u - \ln(2 - u)(2 - u) + 2\right) \ln^3(u) - 2 \ln\left(-\frac{u - 2}{2}\right) \ln^3(u) + 2 \ln(2 - u) \ln^3(u) + 2 \text{Li}_2\left(\frac{u - 2}{2} + 1\right) \ln^3(u) - 2 \ln^3(u) + 3 u \ln^2(u) - 3 \left(-u - \ln(2 - u)(2 - u) + 2\right) \ln^2(u) + 6 \ln\left(-\frac{u - 2}{2}\right) \ln^2(u) - 6 \ln(2 - u) \ln^2(u) + 6 \text{Li}_3\left(\frac{u - 2}{2} + 1\right) \ln^2(u) - 6 \text{Li}_2\left(\frac{u - 2}{2} + 1\right) \ln^2(u) + 6 \ln^2(u) + 6 \ln(2 - u) \left(u \ln(u) - u\right) - 18 u \ln(u) + 12 \text{Li}_4\left(\frac{u}{2}\right) \ln(u) + 12 \text{Li}_3\left(\frac{u}{2}\right) \ln(u) + 12 \text{Li}_2\left(\frac{u - 2}{2} + 1\right) \ln(u) + 24 u - 12 \text{Li}_5\left(\frac{u}{2}\right) - 12 \text{Li}_4\left(\frac{u}{2}\right) - 12 \ln(2) \ln(u - 2) + 12 \ln(u - 2) + 12 \text{Li}_2\left(-\frac{u - 2}{2}\right) - 12 \text{Li}_3\left(\frac{u - 2}{2} + 1\right) - 12$$

Now we can solve for $I_2$

$$I_2=\int\frac{\log^3(u)}{u}du=\frac{\log^4(u)}{4}$$

And $$I_3=\log^3(u)du=u\log^3(u)-3\int\log^2(u)du=x(\log^3(x)-3\log^2(x)+6\log(x)-6)$$

And finally we know that

$$I=-I_1-I_2+I_3$$

plugging them in and simplifying (with the use of various programs)

$$\int \ln(1-x)\ln^4(x+1)dx=\\\left(-2\ln^4\left(x+1\right)+8\ln^3\left(x+1\right)-24\ln^2\left(x+1\right)+48\ln\left(x+1\right)\right)\ln\left(1-\dfrac{x+1}{2}\right)+\left(\left(\ln\left(1-x\right)-1\right)x+\ln\left(1-x\right)-1\right)\ln^4\left(x+1\right)+\left(-8\operatorname{Li}_2\left(\dfrac{x+1}{2}\right)+\left(8-4\ln\left(1-x\right)\right)x-4\ln\left(1-x\right)+8\right)\ln^3\left(x+1\right)+\left(24\operatorname{Li}_3\left(\dfrac{x+1}{2}\right)+24\operatorname{Li}_2\left(\dfrac{x+1}{2}\right)+\left(12\ln\left(1-x\right)-36\right)x+12\ln\left(1-x\right)-36\right)\ln^2\left(x+1\right)+\left(-48\operatorname{Li}_4\left(\dfrac{x+1}{2}\right)-48\operatorname{Li}_3\left(\dfrac{x+1}{2}\right)-48\operatorname{Li}_2\left(\dfrac{x+1}{2}\right)+\left(96-24\ln\left(1-x\right)\right)x-24\ln\left(1-x\right)+96\right)\ln\left(x+1\right)+48\operatorname{Li}_5\left(\dfrac{x+1}{2}\right)+48\operatorname{Li}_4\left(\dfrac{x+1}{2}\right)+48\operatorname{Li}_3\left(\dfrac{x+1}{2}\right)+48\operatorname{Li}_2\left(\dfrac{x+1}{2}\right)+\left(24\ln\left(1-x\right)-120\right)x-24\ln\left(1-x\right)$$

Evaluating this from $0 \to 1$ we get that it is approximately $−0.1204561338771698$ Which is what Mathematica gives as well.