Is the closed form of $\int_0^1 \frac{x\ln^a(1+x)}{1+x^2}dx$ known in the literature?

Question

We know how hard these integrals

$$\int_0^1 \frac{x\ln(1+x)}{1+x^2}dx; \int_0^1 \frac{x\ln^2(1+x)}{1+x^2}dx; \int_0^1 \frac{x\ln^3(1+x)}{1+x^2}dx; ...$$

can be. So I decided to come up with a generalization and I succeeded:

With $x=1/(1+y)$, we have

$$\int_0^1 \frac{y\ln^a(1+y)}{1+y^2}dy=(-1)^a\left[\int_{1/2}^1\frac{\ln^a(x)}{x}dx+\int_{1/2}^1\frac{1-2x}{x^2+(1-x)^2}\ln^a(x)dx\right]$$

$$=(-1)^a\left[\frac{(-1)^a}{a+1}\ln^{a+1}(2)+\Re\int_{1/2}^1\frac{1+i}{1-(1+i)x}\ln^a(x)dx\right].$$

For the remaining integral, we need to find $\int_{1/2}^1\frac{z}{1-zx}\ln^a(x)dx:$

$$\int_{1/2}^1\frac{z}{1-zx}\ln^a(x)dx=\int_{0}^1\frac{z}{1-zx}\ln^a(x)dx-\underbrace{\int_0^{1/2}\frac{z}{1-zx}\ln^a(x)dx}_{x=y/2}$$

$$=(-a)^aa!\,\text{Li}_{a+1}(z)-\int_0^{1}\frac{\frac{z}2}{1-\frac{z}2y}\ln^a(y/2)dy$$

$$=(-a)^aa!\,\text{Li}_{a+1}(z)-\sum_{k=0}^a{a\choose k}(-\ln(2))^{a-k}\int_0^1 \frac{\frac{z}2}{1-\frac{z}2y}\ln^k(y)dy$$

$$=(-a)^aa!\,\text{Li}_{a+1}(z)-\sum_{k=0}^a{a\choose k}(-\ln(2))^{a-k}(-1)^kk!\,\text{Li}_{k+1}\left(\frac{z}2\right)$$

$$=(-1)^aa!\,\left[\text{Li}_{a+1}(z)-\sum_{k=0}^a\frac{\ln^{a-k}(2)}{(a-k)!}\text{Li}_{k+1}\left(\frac{z}2\right)\right].\quad\quad (*)$$

Using this integral, we get

$$\int_0^1\frac{x\ln^a(1+x)}{1+x^2}dx=\frac{\ln^{a+1}(2)}{a+1}+a!\,\Re\left[\text{Li}_{a+1}(1+i)-\sum_{k=0}^a\frac{\ln^{a-k}(2)}{(a-k)!}\text{Li}_{k+1}\left(\frac{1+i}2\right)\right].$$

Question: Is this generalization known in the mathematical literature? and if so, may I get the reference?

Thanks,

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Bonus: Following the same approach above, using $\frac{1}{x^2+(1-x)^2}=\Im\,\frac{1+i}{1-(1+i)x}$, also gives $$\int_0^1\frac{\ln^a(1+x)}{1+x^2}dx=a!\,\Im\left[\text{Li}_{a+1}(1+i)-\sum_{k=0}^a\frac{\ln^{a-k}(2)}{(a-k)!}\text{Li}_{k+1}\left(\frac{1+i}2\right)\right].$$

Also if we divide the identity in $(*)$ by $z$ then integrate $\int_0^1$ w.r.t $z$, we obtain

$$\int_{1/2}^1\frac{\ln(1-x)\ln^a(x)}{x}dx=(-1)^{a-1}a!\,\left[\zeta(a+2)-\sum_{k=0}^a\frac{\ln^{a-k}(2)}{(a-k)!}\text{Li}_{k+2}\left(\frac{1}2\right)\right].$$

Answer 1

Yes, such generalizations are in at least one upcoming book preparing to enter the mathematical literature :-) (probably new entries)

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A solution by Cornel Ioan Valean

In the following, I'll present a more generalized result of the one in the main post. First, we note that $$\int_0^1 \frac{\log^n(1+x)}{a+ x }\textrm{d}x$$ $$=\int_0^1 (\log((a+x)/(a-1)))'\log^n(1+x)\textrm{d}x$$ $$=\log^n(2)\log\left(\frac{a+1}{a-1}\right)-n\int_0^1\frac{\log((a+x)/(a-1))}{1+x}\log^{n-1}(1+x)\textrm{d}x$$ $$\text{$\biggr\{$observe that $\int \frac{\log(1+(1+x)/(a-1))}{(1+x)/(a-1)}((1+x)/(a-1))'\textrm{d}x=-\operatorname{Li}_2\left(\frac{1+x}{1-a}\right)$$\biggr\}$}$$ $$=\log^n(2)\log\left(\frac{a+1}{a-1}\right)-n\int_0^1\left(-\operatorname{Li}_2\left(\frac{1+x}{1-a}\right)\right)'\log^{n-1}(1+x)\textrm{d}x$$ $$=\log^n(2)\log\left(\frac{a+1}{a-1}\right)+n\log^{n-1}(2)\operatorname{Li}_2\left(\frac{2}{1-a}\right)$$ $$-n(n-1)\int_0^1\operatorname{Li}_2\left(\frac{1+x}{1-a}\right)\frac{\log^{n-2}(1+x)}{1+x}\textrm{d}x$$ $$\text{$\biggr\{$use that $\int \frac{1}{1+x}\operatorname{Li}_n\left(\frac{1+x}{1-a}\right)\textrm{d}x=\operatorname{Li}_{n+1}\left(\frac{1+x}{1-a}\right)$ and integrate by parts$\biggr\}$}$$ $$=\log^n(2)\log\left(\frac{a+1}{a-1}\right)+n\log^{n-1}(2)\operatorname{Li}_2\left(\frac{2}{1-a}\right)$$ $$-n(n-1)\int_0^1 \left(\operatorname{Li}_3\left(\frac{1+x}{1-a}\right)\right)' \log^{n-2}(1+x)\textrm{d}x$$ $$=\log^n(2)\log\left(\frac{a+1}{a-1}\right)+n\log^{n-1}(2)\operatorname{Li}_2\left(\frac{2}{1-a}\right)-n(n-1)\log^{n-2} (2)\operatorname{Li}_3\left(\frac{2}{1-a}\right)$$ $$+n(n-1)(n-2)\int_0^1 \left(\operatorname{Li}_3\left(\frac{1+x}{1-a}\right)\right)' \log^{n-3}(1+x)\textrm{d}x$$ $$\text{\{and then continue to integrate by parts for another $n-3$ times\}}$$ $$=\log^n(2)\log\left(\frac{a+1}{a-1}\right)+\sum_{k=1}^n(-1)^{k-1}\frac{n!}{(n-k)!}\log^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac{2}{1-a}\right)$$ $$+(-1)^n n! \operatorname{Li}_{n+1}\left(\frac{1}{1-a}\right),\tag1$$ where for $a \in (-1,0)$ we understand the result as a Cauchy principal value.

If we replace $a$ by $i a$ in $(1)$ and then take the real part of both sides, we obtain $$\int_0^1 \frac{x\log^n(1+x)}{a^2+ x^2}\textrm{d}x$$ $$=\sum_{k=1}^n(-1)^{k-1}\frac{n!}{(n-k)!}\log^{n-k}(2) \Re\biggr\{\operatorname{Li}_{k+1}\left(\frac{2}{1-ia}\right)\biggr\}+(-1)^n n! \Re\biggr\{\operatorname{Li}_{n+1}\left(\frac{1}{1-ia}\right)\biggr\},$$ which brings us to the desired generalization. At last, we let $a \mapsto 1$.

Q.E.D.

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More powerful generalizations by Cornel Ioan Valean

\begin{equation*} i) \ \int_0^1 \frac{x\operatorname{arctanh}^n(x)}{1+a^2 x^2}\textrm{d}x \end{equation*} \begin{equation*} =\frac{n!}{a^2 2^n}\left((1-2^{-n})\zeta(n+1)+ \Re\biggr\{\operatorname{Li}_{n+1}\left(\frac{a^2-1}{a^2+1}+i\frac{2a}{1+a^2}\right)\biggr\}\right); \end{equation*} \begin{equation*} ii) \ \int_0^1 \frac{x\operatorname{arctanh}^n(x)}{1-a^2 x^2}\textrm{d}x \end{equation*} \begin{equation}\label{ih2} =\frac{n!}{a^2 }\left((1-2^n)4^{-n}\zeta(n+1)-2^{-n-1}\operatorname{Li}_{n+1}\left(\frac{a-1}{a+1}\right)-2^{-n-1}\operatorname{Li}_{n+1}\left(\frac{a+1}{a-1}\right)\right); \end{equation} $$iii) \ \int_0^1 \frac{\operatorname{arctanh}^n(x)}{1+a^2 x^2}\textrm{d}x=\frac{n!}{a2^n}\Im\biggr\{\operatorname{Li}_{n+1}\left(\frac{a^2-1}{a^2+1}+i\frac{2a}{1+a^2}\right)\biggr\};$$ $$iv) \ \int_0^1 \frac{\operatorname{arctanh}^n(x)}{1-a^2 x^2}\textrm{d}x=\frac{n!}{a 2^{n+1} }\left(\operatorname{Li}_{n+1}\left(\frac{a-1}{a+1}\right)-\operatorname{Li}_{n+1}\left(\frac{a+1}{a-1}\right)\right).$$

Solution: largely speaking, all results are almost straightforward if we let the variable change $\displaystyle x\mapsto \frac{1-x}{1+x}$ and combine it with the use of $\displaystyle \int_0^1 \frac{z\log^n(t)}{1-z t}\textrm{d}t=(-1)^n n!\operatorname{Li}_{n+1}(z), \ z\in \mathbb{C}\setminus (1, \infty)$. As regards the results from the points $i)$ and $iii)$, after the variable change, the partial fraction decomposition and splitting the integral, we need to consider the denominator of the integrand of the harder resulting integral to be expressed by using complex numbers before using the auxiliary integral result above.

Fascinating results that also use the previous results $$\int_0^1 \frac{x\log(1-x)\log(1+x)}{a^2+x^2}\textrm{d}x$$ $$ =-\frac{3}{8}\zeta(3)+\frac{1}{4}\operatorname{Li}_3\left(\frac{1}{1+a^2}\right)-\frac{1}{2}\Re\biggr\{\operatorname{Li}_3\left(\frac{1-a^2}{1+a^2}+i\frac{2a}{1+a^2}\right)\biggr\}$$ $$=-\frac{3}{8}\zeta(3)+\frac{1}{4}\operatorname{Li}_3(\cos^2\theta)-\frac{1}{2}\sum_{n=1}^{\infty} \frac{\cos(2n\theta)}{n^3}$$

and $$ \int_0^1 \frac{x\log(1-x)\log^2(1+x)}{a^2+x^2}\textrm{d}x$$ $$ =\frac{1}{3}\arctan^4(a)-\frac{2}{3}\pi \arctan^3(|a|)+2\zeta(2)\arctan^2(a)-\frac{15}{8}\zeta(4)$$ $$ -\frac{1}{2}\operatorname{Li}_4\left(\frac{1}{1+a^2}\right) +2\sum _{n=1}^{\infty } \frac{1}{n^4}(1+a^2)^{-n/2}\cos(n\arctan(a))$$ $$ =\frac{1}{3}\theta^4-\frac{2}{3}\pi |\theta|^3+2\zeta(2)\theta^2-\frac{15}{8}\zeta(4)-\frac{1}{2}\operatorname{Li}_4(\cos^2(\theta)) +2\sum _{n=1}^{\infty } \frac{\cos(n\theta)\cos^n(\theta)}{n^4},$$ where $a=\tan(\theta)$. I also exploited that $\displaystyle \int_0^1 \frac{x\log^n(1-x)}{a^2+x^2}\textrm{d}x=(-1)^n n! \Re\biggr\{\operatorname{Li}_{n+1}\left(\frac{1}{1\pm i a}\right)\biggr\}$.

A note: Such results and many others alike will be found in the sequel of (Almost) Impossible Integrals, Sums, and Series.