Nice pair of trilog integrals $\int_0^z \frac{\log ^2(x) \log (1\pm x)}{1\mp x} \, dx$

Question

Recently, in the wake of the solution of Compute $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$, I stumbled on this symmetric pair of integrals

$$i_{\pm}(z) = \int_0^z \frac{\log ^2(x) \log (1\pm x)}{1\mp x} \, dx$$

I tried several integrations by part and a series expansion but I could not solve them.

(a) Can you do better?

(b) A slightly easier version asks for the case $z=1$.

Here we have numerically $i_{+}(1) = 0.345691, i_{-}(1) = -0.235752$

Remark: the integrals would be easy if the denominator were identical with the argument of the logarithm.

Answer 1

I am going to solve $(b)$

Rgarding $i_+(1)$, use

$$\frac{\ln(1+x)}{1-x}=\sum_{n=1}^\infty \overline{H}_nx^n\tag1$$

multiply both sides by $\frac{\ln^2x}{x}$ then $\int_0^1$ and use the fact that $\int_0^1 x^{n-1}\ln^2x\ dx=\frac{2}{n^3}$ we get

$$2\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}=\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\ dx+\int_0^1\frac{\ln^2x\ln(1+x)}{x}\ dx$$

where

$$\int_0^1\frac{\ln^2x\ln(1+x)}{x}\ dx=-\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^1 x^{n-1}\ln^2(x)\ dx=-2\sum_{n=1}^\infty\frac{(-1)^n}{n^4}=\frac74\zeta(4)$$

and $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}$ can be found using the generalization

$$\sum_{k = 1}^\infty \frac{\overline H_k}{k^m} = \zeta (m) \log 2 - \frac{1}{2} m \zeta (m + 1) + \eta (m + 1) + \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1).$$ set $m=3$ $$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}=\frac74\ln2\zeta(3)-\frac5{16}\zeta(4)$$

Also I managed to calculate it here in a different way ( check the bonus).

Combine the two results we get

$$i_+(1)=\frac72\ln2\zeta(3)-\frac{19}{8}\zeta(4)$$

To find $i_-(1)$, just replace $x$ with $-x$ in $(1)$ and follow the same process coming across $\sum_{n=1}^\infty\frac{(-1)^n\overline{H}_n}{n^3}$ which is manageable using parity, but there is a nicer way

$$i_+(1)-i_-(1)=\int_0^1\ln^2(x)\left(\frac{\ln(1+x)}{1-x}-\frac{\ln(1-x)}{1+x}\right)\ dx$$

$$\overset{IBP}{=}\underbrace{-\ln^2(x)\ln(1-x)\ln(1+x)\bigg|_0^1}_{0}+2\int_0^1\frac{\ln x\ln(1-x)\ln(1+x)}{x}\ dx$$

the last integral can be done using the algebraic identity $4ab=(a+b)^2-(a-b)^2$ where $a=\ln(1-x)$ and $b=\ln(1+x)$ and you can find different solutions here

$$\int_0^1\frac{\ln x\ln(1-x)\ln(1+x)}{x}=2\operatorname{Li}_4\left(\frac12\right)-\frac12\ln^22\zeta(2)+\frac74\ln2\zeta(3)-\frac{27}{16}\zeta(4)+\frac1{12}\ln^42$$

$$\Longrightarrow i_-(1)=\zeta(4)+\ln^22\zeta(2)-4\operatorname{Li}_4\left(\frac12\right)-\frac1{6}\ln^42$$

Answer 2

Partial answer (I will examine just one of the cases and applied for $z = 1$).

I will analyse the following:

$$\int_0^1 \frac{\log^2(x)\log(1+x)}{1-x}\ \text{d}x$$

Let's start with $x \to e^z$ which transforms the integral into

$$\int_{-\infty}^0 \frac{z^2 e^z}{1-e^z}\log(1+e^z)\ \text{d}z$$

Use the Geometric series for

$$\frac{1}{1-e^z} = \sum_{k\in\mathbb{W}} e^{zk}$$

Where $\mathbb{W} = \mathbb{N} + \{0\}$.

Whence

$$\sum_{k\in\mathbb{W}}\int_{-\infty}^0 z^2 e^{z(k+1)}\log(1 + e^z)\ \text{d}z$$

We can now use the integration by parts with the following choice

$$f'(z) = z^2 e^{z(k+1)}$$ $$g(z) = \log(1 + e^z)$$

From which

$$f(z) = \frac{e^{(k+1) z} \left((k+1)^2 z^2-2 (k+1) z+2\right)}{(k+1)^3}$$ $$g'(z) = \frac{e^z}{1+e^z}$$

Hence

$$\sum_{k\in\mathbb{W}}\left[\left(\frac{e^{(k+1) z} \left((k+1)^2 z^2-2 (k+1) z+2\right)}{(k+1)^3}\right) \cdot \log(1 + e^z)\bigg|_{-\infty}^0 - \left(\frac{e^{(k+1) z} \left((k+1)^2 z^2-2 (k+1) z+2\right)}{(k+1)^3}\right)\cdot \frac{e^z}{1+e^z}\bigg|_{-\infty}^0 \right]$$

Computing the two limits is rather easy, and we ge

$$\sum_{k\in\mathbb{W}} \left(\frac{2\log(2)}{(1+k)^3} - \frac{1}{(1+k)^3}\right)$$

$$(\log(4) - 1)\sum_{k\in\mathbb{W}} \frac{1}{(1+k)^3}$$

The last sum is very well known, it's the Riemann Zeta function of three.

$$ \to (\log(4) - 1)\zeta(3)$$

Warning

The numerical value in this case is $\approx 0.464348(...)$ which is different from the true numerical one. I suspect I have made some error somewhere, hence I just wrote this down in order to read it clearly, and I will check it better later!

Answer 3

Partial self answer for $z=1$

Starting with the idea of @Mycroft I have transformed the remaining problem into the calculation of the following Euler sums

$$\sum_{k=0}^{\infty} \left\{\frac{H_{\frac{k-1}{2}}^{(2)}-H_{\frac{k}{2}}^{(2)}}{2 (k+1)^2},\frac{H_{\frac{k-1}{2}}^{(3)}-H_{\frac{k}{2}}^{(3)}}{4 (k+1)},\frac{H_{\frac{k-1}{2}}-H_{\frac{k}{2}}}{(k+1)^3}\right\}$$

Answer 4

Evaluated below are all four quadruplets $\int_0^1 \frac{\log ^2(x) \log (1\pm x)}{1\mp x} \, dx$.

\begin{align} I_-^- =&\int_0^1 \frac{\ln^2x \ln (1-x)}{1-x}dx =\int_0^1 \frac{\ln^2x}{1-x}\left(\int_0^1 \frac {-x}{1-x t}dt\right) dx\\ =&\int_0^1 \frac{1}{1-t}\left(\int_0^1 \frac {\ln^2x}{1-tx}dx - 2Li_3(1)\right) dt\\ =&\>2 \int_0^1\frac{Li_3(t)}{t}dt + 2 \int_0^1\frac{Li_3(t)-Li_3(1)}{1-t}\>\overset{ibp}{dt}\\ = &\>2Li_4(1) - Li_2^2(1)= -\frac{\pi^4}{180}\\ \\ I_-^+ =&\int_0^1 \frac{\ln^2x \ln (1+x)}{1-x}dx =\int_0^1 \frac{\ln^2x}{1-x}\left(\int_0^1 \frac {x}{1+x t}dt\right) dx\\ =&\int_0^1 \frac{1}{1+t}\left(2Li_3(1)-\int_0^1 \frac {\ln^2x}{1+tx}dx \right) dt\\ =&\>2 \int_0^1\frac{Li_3(-t)}{t}dt - 2 \int_0^1\frac{Li_3(-t)-Li_3(1)}{1+t}\>\overset{ibp}{dt}\\ = &\>2Li_4(-1)-2\ln2 Li_3(-1) - Li_2^2(-1)+2\ln2Li_3(1)\\= &\>\frac72\ln2\zeta(3)-\frac{19\pi^4}{720}\\ \\ I_+^+ =&\int_0^1 \frac{\ln^2x \ln (1+x)}{1+x}dx \>\>\>\>\>\>\>(x=\frac{t}{1-t})\\ =& \frac14 \ln^4 2 -\int_0^{1/2} \frac{\ln^2t\ln(1-t)}{1-t} dt +2 \int_0^{1/2} \frac{\ln t\ln^2(1-t)}{1-t}\>\overset{ibp}{dt} \\ =& \frac14 \ln^4 2 -\left(\frac12 I_-^{-} -\frac14\ln^42\right)+\frac23\left( \int_0^{1/2} \frac{\ln^3(1-t)}{t} \>\overset{t\to 1-t}{dt} -\ln^42 \right)\\ = & -\frac16\ln^42 -5Li_4(1) + \frac12Li_2^2(1) -\frac23\int_0^{1/2} \frac{\ln^3t}{1-t} \overset{t\to t/2}{dt} \\ &\hspace{5mm}-\frac23\left( -6Li_4(\frac12)-6\ln2Li_3(\frac12) -3\ln^2Li_2(\frac12)-\ln^42\right)\\ = & \>4Li_4(\frac12)+\frac72\ln2\zeta(3)-\frac{\pi^4}{24}-\frac{\pi^2}6\ln^22+\frac16\ln^42\\ \\ I_+^- = &\int_0^1 \frac{\ln^2x\ln(1-x)}{1+x}dx = \int_0^1 \ln(1-x)\> d\left( \int_1^x \frac{\ln^2 t}{1+t}dt\right) \\ =& \int_0^1 \frac1{1-x}\left(\int_0^x \frac{\ln^2t}{1+t} \overset{t=xy}{dt}-\int_0^1\frac{\ln^2t}{1+t}dt\right)dx\\ =& \int_0^1 \int_0^1 \frac{\ln^2x +2\ln x\ln y}{(1+y)(1-x)} dx dy-\int_0^1\int_0^1 \frac{\ln^2(xy)}{(1+y)(1+xy)} \overset{t=xy}{dx} dy\\ =& \>-2Li_2(1)Li_2(-1)+ 2\ln2 Li_3(1)-\int_0^1 \frac1{y(1+y)}\int_0^y\frac{\ln^2t}{1+t}dt \>\overset{ibp}{dy}\\ =& \>-2Li_2(1)Li_2(-1)+ 2\ln2 Li_3(1)+2\ln2Li_3(-1) +6Li_4(-1) -I_+^+ \\ =& \> -4Li(\frac12) + \frac{\pi^4}{90}+\frac{\pi^2}6\ln^22-\frac16\ln^42\\ \end{align}

Note that the special polylog values $Li_4(1)=\frac{\pi^4}{90},\>Li_4(-1)=-\frac{7\pi^4}{720}$, $Li_3(1)=\zeta(3)$, $ Li_3(-1)=-\frac34\zeta(3) $, $Li_3(\frac12)=\frac78\zeta(3)-\frac{\pi^2}{12}\ln^22+\frac16\ln^42$, $Li_2(1)=\frac{\pi^2}{6}$, $Li_2(-1)=-\frac{\pi^2}{12}$, $Li_2(\frac12)=\frac{\pi^2}{12}-\frac12\ln^22$ are used in above evaluations.