Generalizing $\int_0^1 \frac{\log(1-x)\log^n(1+x)}{1+x} \, dx$

Question

I have been looking for the generalized version of the below integral,

$$I(n)=\int_0^1 \frac{\log(1-x)\log^n(1+x)}{1+x} \, dx$$

So far I have derived the results only for $n=1$,

$$I(1)=\int_0^1 \frac{\log(1-x)\log(1+x)}{1+x} \, dx = \frac{1}{3}\log^3(2)-\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$

Now trying for $n=2,3,4$ using all variations of identities led to integrals which do not seem to have any nice closed forms

I am aware of a similar generalization performed here

$$ \int_0^1 \frac{\log^n (1-x) \log^{n-1} (1+x)}{1+x} dx=\frac{1}{2} \left(\sum _{m=1}^n \sum _{k=1}^m +\sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n\right) \frac{\binom{n}{k}\binom{n-1}{m-k} \log ^{-m+2 n-1}(2) }{-k+m+1}\left(k \underset{a\to 0}{\text{lim}}\underset{b\to 1}{\text{lim}}\frac{\partial ^{m}B(a,b)}{\partial a^{k-1}\, \partial b^{-k+m+1}}+(-1)^m \log ^{m+1}(2)\right)+\frac{\log ^{2 n}(2)}{n}$$

Is there any alternate method to compute for $n>1$, and build a generalization ?

Answer 1

\begin{align} I(n)=&\int_0^1 \frac{\ln(1-x)\ln^n(1+x)}{1+x} \, dx\\ \overset{ibp}=&\ \frac1{n+1} \int_0^1 \frac{\ln^{n+1}(1+x)-\ln^{n+1}2}{1-x} \, \overset{x\to 1-x}{dx}\\ =&\ \frac1{n+1} \sum_{k=1}^{n+1}\binom {n+1}k (\ln2)^{n+1-k}\int_0^1\frac{\ln^k(1-\frac x2)}x dx\\ \end{align} Note that \begin{align} J(k)=&\int_0^1\frac{\ln^k(1-\frac x2)}x \overset{t= 1-\frac x2 }{dx} =\int_0^{1}\frac{\ln^k t}{1-t} dt -\int_0^{1/2}\frac{\ln^k t}{1-t}\overset{t\to \frac12t}{ dt}\\ =& \ (-1)^k k!\ \text{Li}_{k+1}(1) - \sum_{j=0}^k \binom kj (\ln2)^{k-j}(-1)^j j! \ \text{Li}_{j+1}(\frac12) \end{align} Thus \begin{align} I(n) = &\ \frac1{n+1} \sum_{k=1}^{n+1}\binom {n+1}k (\ln2)^{n+1-k}J(k)\\ =&\ \sum_{k=1}^{n+1}\frac{n!\ln^{n+1}2}{(n+1-k)!} \bigg[\ln^k\frac12\ \text{Li}_{k+1}(1)- \sum_{j=0}^k \frac{\ln^j\frac12\ \text{Li}_{j+1}(\frac12)}{(k-j)!}\bigg] \end{align}

Answer 2

Change of variable:

$$y=1+x$$

$$I(n)=\int_1^2 \frac{\ln(2-y)\ln^ny}{y} \, dy$$

Now consider

$$I(a)=\int_1^2y^a\ln(2-y)\, dy$$

We get back to the original integral by differentiating $I(a)$ with respect to $a$ $n$ times and then taking $a=-1$

Using expansion

$$\ln\left (1-\frac{y}{2} \right )=-\sum_{k=1}^{\infty}\frac{y^k}{k2^k}$$

we get after a simple integration:

$$I(a)=\ln 2\frac{2^{a+1}-1}{a+1}+\sum_{k=1}^{\infty}\frac{1}{k2^k(k+a+1)}-\sum_{k=1}^{\infty}\frac{2^{k+a+1}}{k2^k(k+a+1)}$$

Now, by differentiating $I(a)$ with respect to $a$ $n$ times and then taking $a=-1$ we reach the desired result:

$$I(n)=\frac{(\ln 2)^{n+2}}{n+1}+(-1)^nn!\left [\text{Li}_{n+2} \left ( \frac{1}{2} \right ) -\zeta(n+2)-\sum_{i=1}^{n}(-1)^{i}\frac{(\ln 2)^{i}}{i!}\zeta(n+2-i)\right ]$$

An addition

An alternative form that does not use the polylogarithm and zeta function:

$$I(n)=\frac{(\ln 2)^{n+2}}{n+1}-(-1)^nn!\sum_{k=1}^{\infty} \frac{1- \frac{1}{2^k}+f(k,n)}{k^{n+2}}$$

where $$f(k,n)=\sum_{i=1}^{n}(-1)^{i}\frac{(k\ln 2)^{i}}{i!}$$ The previous result follows directly from this last form

Answer 3

First, assume $u = 1 + x$. Then, solve it with integration by parts, as $v=log(2-u)$ and $w=\frac{log^n(u)}{u}$. The result of indefinite integral is $\frac{1}{n+1}\left(log(2-u).log^{n+1}(u)+\int{\frac{log^{n+1}(u)}{2-u}du} \right)$. You can apply the definite version of it to get the result of generalized $I(n)$.