Two hard integrals: $\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx$ and $\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx$

Question

I found two integrals that seem hard to evaluate: $$I_1=\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx$$ $$I_2=\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx$$ I am just a beginner in logarithmic integral. So, I searched to find substitution like $x=\frac{1}{1+x}$, $x=\frac{1}{1-x}$, or $x=\frac{1-x}{1+x}$, but they didn't work. Can I ask some ideas from every one? Thank you.

EDIT: After using Mathematica, with MZIntegrate paclet gives closed-form:

$$\begin{align}I_1&=\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx\\&=G^2+\frac{5 \text{Li}_4\left(\frac{1}{2}\right)}{4}+\frac{35}{32} \zeta (3) \log (2)-\frac{119 \pi ^4}{5760}+\frac{5 \log ^4(2)}{96}-\frac{5}{96} \pi ^2 \log ^2(2)\\I_2&=\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx\\&=2 G^2+\frac{35}{16} \zeta (3) \log (2)-\frac{199 \pi ^4}{5760}\end{align}$$ where $G$ is Catalan's constant.

Answer 1

The routine integrals $\int_0^1 \frac{\ln^2t}{1-t}dt = 2\zeta(3)$, $\int_0^1 \frac{\ln^2t}{1+t}dt = \frac32\zeta(3)$, $\int_0^1 \frac{\ln^3t}{1-t}dt = -\frac{\pi^4}{15}$, $\int_0^1 \frac{\ln^2t}{1+t^2}dt =\frac{\pi^3}{16}$, $\int_0^1 \frac{\ln t}{1+t^2}dt =-G$, and $\int_0^1 \frac{\ln t}{1+t}dt =-\frac{\pi^2}{12}$ are used below without elaboration.

\begin{align} I_2=&\int_0^1\frac{\ln^2x\ln(1+x^2)}{1-x}dx\\ =& \int_0^1\ln(1+x^2)\ d\left( \int_0^x \frac{\ln^2t}{1-t}dt\right)\\ =& \ \ln2 \int_0^1 \frac{\ln^2t}{1-t}dt -\int_0^1\frac{2x}{1+x^2}\int_0^x\frac{\ln^2t}{1-t} \overset{t=xy}{dt}\\ =&\ 2\ln2\zeta(3) +2\int_0^1\int_0^1\frac{\ln^2xy}{1+y^2}\left(\frac{1+xy}{1+x^2}-\frac1{1-xy}\right)dxdy\tag1 \end{align} Note that, with $\ln^2xy =\ln^2x +2\ln x\ln y +\ln^2 y$ \begin{align} &\int_0^1\int_0^1\frac{(1+xy)\ln^2xy}{(1+x^2)(1+y^2)}dxdy\\ =& \int_0^1 \int_0^1\frac{\ln^2x +2\ln x\ln y +\ln^2 y}{(1+x^2)(1+y^2)}+\frac1{16}\frac{\ln^2x +2\ln x\ln y +\ln^2 y}{(1+x)(1+y)}\ dxdy\\ =&\ 2G^2 +\frac3{16}\ln2\zeta(3)+\frac{37\pi^4}{1152}\\ \\ & \int_0^1\int_0^1\frac{\ln^2xy}{(1+y^2)(1-xy)}\overset{x=t/y}{dx}dy\\ =& \ \frac12\int_0^1 d\left(\ln\frac{y^2}{1+y^2}\right)\int_0^y \frac{\ln^2t}{1-t}dt \overset{ibp}=-\ln2\zeta(3)+\frac{\pi^4}{15}+\frac12I_2 \end{align} Plug into (1) to obtain \begin{align} I_2 = &\ 2\ln2\zeta(3)+2 \left(2G^2 +\frac3{16}\ln2\zeta(3)+\frac{37\pi^4}{1152} \right)+2\ln2\zeta(3)-\frac{2\pi^4}{15}-I_2\\ =&\ 2G^2 +\frac{35}{16}\ln2\zeta(3)-\frac{199\pi^4}{5760} \end{align}

Answer 2

Just some remarks from a higher perspective. Integrals: $$\tag{1} \int_0^1 f(x) dx$$ when $f(x)$ is one of $$\small \frac{\log^n x \log(1+x^2)}{1\pm x},\frac{\log^{2n} x \log(1+x^2)}{1+x^2},\frac{x\log^{2n} x \log(1+x^2)}{1+x^2},\frac{\log^{2n} x \arctan x}{1- x},\frac{\log^{n} x \arctan x}{1+ x},\frac{x\log^{2n} x \arctan x}{1 + x^2},\frac{\log^{2n+1} x \arctan x}{1 + x^2},\frac{\log^{2n} x\log(1\pm x)}{1 + x^2},\frac{x\log^{2n+1}x \log(1\pm x)}{1 + x^2},\frac{\log^{2n}x \log(1\pm x) \log(1+x^2)}{x},\frac{\log^{2n+1}x \arctan x \log(1+x^2)}{x},\frac{\log^{2n+1}x \arctan x \log(1\pm x)}{x},\frac{\log^{2n+1} x \text{Li}_{2m}(x)}{1+x^2}, \frac{\log^{2n} x \text{Li}_{2m+1}(x)}{1+x^2}, \frac{x \log^{2n+1} x \text{Li}_{2m+1}(x)}{1+x^2} \text{ etc...}$$ all have results of form $$\tag{2}\int_0^1 f(x) dx = \sum_{i=1}^{w} a_i L(i)L(w-i)$$ here $a_i \in \mathbb{Q}$, $L(s) =\zeta(s)$ or $\beta(s)$ are Riemann zeta or Dirichlet beta, $w$ is the weight of integrand, namely degree of numerator plus one.

---

OP's $I_2$ belongs to above cases; $I_1$ also, had the power of $\log$ were even; the odd power case is more difficult.

For $I_1$, the fact that MZIntegrate can do it already implies a mechanical proof exists. If you want magical/elegant solutions, perhaps Conrad Valean or Ali Shadhar can shed light on this. A highly non-trivial extension is $\int_{0}^{1}\frac{\log^3{(x)}\log{(1\pm x)}\log{(1+x^2)}}{x}dx$, you can put it into MZIntegrate to see what it looks like, I think Conrad and Ali have yet to found a solution to this one.

For example: _{$$\int_0^1 \frac{\log ^5(x) \log \left(x^2+1\right) \arctan x}{x} dx = 720 \beta(8)-\frac{25 \pi ^5 \zeta (3)}{32}-\frac{15 \pi ^3 \zeta (5)}{2}-\frac{123825 \pi \zeta (7)}{2048}$$ $$\int_0^1 \frac{\log^4 x \log(1+x)}{1+x^2} dx = \frac{7 \pi ^4 G}{30}+2 \pi ^2 \beta(4)-48 \beta(6)+\frac{5}{128} \pi ^5 \log (2)$$ $$\int_0^1 \frac{x \log^5 x \log(1-x)}{1+x^2} dx = -\frac{\pi ^4 \zeta (3)}{8}-\frac{75 \pi ^2 \zeta (5)}{128}+\frac{129495 \zeta (7)}{2048}-\frac{33}{512}\pi ^6 \log (2)$$ $$\int_0^1 \frac{\log^4 x \log(1-x) \log(1+x^2)}{x} dx = -\frac{5 \pi ^5 G}{32}-\frac{1}{2} 3 \pi ^3 \beta(4)-12 \pi \beta(6)+\frac{121 \pi ^4 \zeta (3)}{2560}+\frac{105 \pi ^2 \zeta (5)}{512}+\frac{121557 \zeta (7)}{1024}$$}

when weight $w\geq 7$, MZIntegrate is not able to evaluate them directly. I will add this functionality on the next update.

If one replaces $1+x^2$ occurring in $f(x)$ by $1+x+x^2$ or $1-x+x^2$ or $1+x^4$ etc, one obtain similar expressions, except now $L$ has to be other Dirichlet L-functions.

---

Proof sketch: for all cases $f(x)$ above, the integral $\int_0^1 f(x) dx$ can be converted to colored multiple zeta values (CMZVs) of depth 2 and level 4 and weight $w$. There is a result saying that whenever depth and weight of a CMZV satisfies certain parity condition, its real/imaginary part reduces to lower depths CMZVs. In our case, we have reduction into depth 1 CMZVs, which when level = 4 is simply $\zeta(s)$ or $\beta(s)$, giving $(2)$. QED.

Answer 3

Express the first integral via integration by parts in terms of the second and a third integral $$ I_1=\int_{0}^{1}\frac{\ln x\ln(1-x)\ln (1+x^2)}{x}dx=\frac12I_2 -I_3\tag1$$ where \begin{align} I_2=&\int_{0}^{1}\frac{\ln^2x\ln (1+x^2)}{1-x}dx = 2G^2 +\frac{35}{16}\ln2\zeta(3)-\frac{199\pi^4}{5760}\\ I_3=& \int_{0}^{1}\frac{x\ln^2 x\ln(1-x)}{1+x^2}dx =-\frac54\text{Li}_4\left(\frac{1}{2}\right)+\frac{13 \pi ^4}{3840}+\frac{5\pi^2}{96} \ln ^22-\frac{5 }{96}\ln ^42 \end{align}

Similarly to $I_2$, $I_3$ is relatively manageable and is evaluated in the Appendix. Substitute above into (1) to obtain $$I_1=\frac54\text{Li}_4\left(\frac{1}{2}\right)+ G^2+\frac{35}{32} \ln2\zeta (3) -\frac{119 \pi ^4}{5760}-\frac{5\pi^2}{96} \ln ^22+\frac{5 }{96}\ln ^42 $$

---

Appendix: (to be completed) \begin{align} I_3=\frac18 \int_{0}^{1}\frac{\ln^2 x\ln (1-x)}{1+x}\ \overset{x\to x^2}{dx} - \int_{0}^{1}\frac{x\ln^2x\ln(1+x)}{1+x^2}dx\\ \end{align} \begin{align} \int_{0}^{1}\frac{\ln^2x\ln(1-x)}{1+x}dx=&\ -4\text{Li}_4\left(\frac{1}{2}\right) +\frac{\pi^4}{90}+\frac{\pi^2}6\ln^22-\frac16\ln^42\\ \int_{0}^{1}\frac{x\ln^2x\ln(1+x)}{1+x^2}dx=&\ \frac34 \text{Li}_4\left(\frac{1}{2}\right)-\frac{23\pi^4}{11520}-\frac{\pi^2}{32}\ln^22+\frac1{32}\ln^42 \end{align}

Answer 4

Solutions by Cornel Ioan Valean

It's straightforward to show that $\displaystyle \int_0^1 x^{n-1}\log(x)\log(1-x)\textrm{d}x=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\zeta(2)\frac{1}{n}$, and this is easily extracted by differentiating once with respect to $n$ the well-known result {with a Beta function (involving derivates) structure}, $\displaystyle\int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-\frac{\psi(n+1)+\gamma}{n}$. Observe the latter result, when $n$ is a positive integer, may be put in the form with harmonic numbers, that is $\displaystyle\int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-\frac{H_n}{n}$. The last integral form can be found elementary calculated in (Almost) Impossible Integrals, Sums, and Series, page $59$.

Returning to the first main integral and using the previous results, we get $$\int_0^1\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}\textrm{d}x=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}\int_0^1x^{2n-1}\log(x)\log(1-x)\textrm{d}x $$ $$=-\frac{1}{2}\zeta(2)\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{n^2}+\frac{1}{4}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{n^3}+\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}^{(2)}}{n^2}$$ $$=G^2-\frac{5}{16}\log^2(2)\zeta(2)+\frac{35}{32}\log(2)\zeta(3)-\frac{119}{64}\zeta(4)+\frac{5}{96}\log^4(2)+\frac{5}{4}\operatorname{Li}_4\left(\frac{1}{2}\right),$$ and the last equality follows by using the fact that the last two series are computed here, and they are some of the most difficult to calculate harmonic series (of low weight class, $\le7$) in the mathematical literature.

For the second main integral, we have the more general result,

\begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x^2}{2}\right)}{1-x}\textrm{d}x \end{equation*} \begin{equation*} =m\zeta (m+1)- 2^{-m} \left(1-2^{-m+1}\right) \log(2 ) \zeta (m) -\sum_{k=0}^{m-1}\beta(k+1)\beta(m-k) \end{equation*} \begin{equation*} -\sum_{k=1}^{m-2}2^{- m-1}(1-2^{-k})(1-2^{-m+k+1}) \zeta (k+1)\zeta (m-k), \end{equation*} where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$ represents the $n$th generalized harmonic number of order $m$,
$\zeta$ denotes the Riemann zeta function and $\beta$ designates the Dirichlet beta function.

A very simple solution may be found in this answer that exploits the symmetry in double integrals.

End of story