4

Evaluate the integral in a closed-form :

$$I=\int_0^{1}\frac{\ln x\ln (1-x^{2})}{1+x^{2}}\mathrm dx$$

My attempt :

After put $x=\tan y$ we obtain:

$$I=\int_0^{\frac{π}{4}}\ln (\tan x)\ln (1-\tan^{2} x)dx$$

$$1-\tan^{2} x=(1+\tan x)(1-\tan x)$$ $$\ln (1+\tan x)=\ln(\sin x+\cos x)-\ln(\cos x)=\ln (\cos (\frac{π}{4}-x))-\ln (\cos x)$$

Also:

$$\ln (1-\tan x)=\ln(\cos x-\sin x)-\ln(\cos x)=\ln (\cos (\frac{π}{4}+x))-\ln (\cos x)$$

So:

$$\ln (\tan x)\ln (1-\tan^{2} x)$$

$$=(\ln (\sin x)-\ln(\cos x)(\ln(\cos (\frac{π}{4}-x))-\ln (\cos x))(\ln(\cos (\frac{π}{4}+x))-\ln(\cos x))$$

Now I have many integrals. How can I evaluate them? Let me know if anyone has other ideas.

Quanto
  • 97,352
Ellen Ellen
  • 2,319
  • 3
    Split the logarithm in two, and use the values given here: https://math.stackexchange.com/a/2972249/515527 It won't look nice. – Zacky Jul 15 '19 at 08:58
  • Is there anything wrong with current answers given which discourages you from accepting one of them, to show future users that this question has been answered? If so, please say what could be improved. – mrtaurho Jul 23 '19 at 11:45

3 Answers3

4

As pointed out by Zacky the solution may be found here given by pisco. Adopting the notation they use there we see that

\begin{align*} I&=\int_0^1\frac{\log(x)\log(1-x^2)}{1+x^2}\mathrm dx\\ &=\int_0^1\frac{\log(x)\log(1-x)}{1+x^2}\mathrm dx+\int_0^1\frac{\log(x)\log(1+x)}{1+x^2}\mathrm dx\\ &=I_{ab}+I_{ac} \end{align*}

In the end of their post they get that

\begin{align*} I_{ab}&=\Im\left(\operatorname{Li}_3\left(\frac{1+i}2\right)\right)-\frac{\pi^3}{128}-\frac\pi{32}\log^2(2)\\ I_{ac}&=-3\Im\left(\operatorname{Li}_3\left(\frac{1+i}2\right)\right)+\frac{11\pi^3}{128}+\frac{3\pi}{32}\log^2(2)-2G\log(2) \end{align*}

Thus,

$$\small\therefore~I~=~\int_0^1\frac{\log(x)\log(1-x^2)}{1+x^2}\mathrm dx~=~-2\Im\left(\operatorname{Li}_3\left(\frac{1+i}2\right)\right)+\frac{5\pi^3}{64}+\frac\pi{16}\log^2(2)-2G\log(2)$$

clathratus
  • 17,161
mrtaurho
  • 16,103
2

\begin{align} \int_0^1\frac{\ln x\ln(1-x^2)}{1+x^2}\ dx=\int_0^1\frac{\ln x\ln(1-x^4)}{1+x^2}\ dx-\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx \end{align} The first integral was nicely done here by Bennett Gardiner. $$\int_0^1\frac{\ln x\ln(1-x^4)}{1+x^2}\ dx=\frac{\pi^3}{16}-3G\ln2$$ and I managed to calculate the second integral here. $$\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx=\frac{3\pi^3}{32}+\frac{\pi}8\ln^22-\ln2~G-2\text{Im}\operatorname{Li_3}(1+i)$$ Thus $$\int_0^1\frac{\ln x\ln(1-x^2)}{1+x^2}\ dx=2\text{Im}\operatorname{Li_3}(1+i)-\frac{\pi^3}{32}-\frac{\pi}8\ln^22-2\ln2~G$$

Ali Shadhar
  • 25,498
  • 1
    Interesting. Combining our two results we can conclude that $$\Im\left[\operatorname{Li}_3(1+i)+\operatorname{Li}_3\left(\frac{1+i}2\right)\right]~=~\frac{3\pi}{32}\log^2(2)+\frac{7\pi^3}{128}$$ Which seems to work out numerically. – mrtaurho Jul 16 '19 at 21:07
  • 1
    @mrtaurho we can conclude this relation without comparing the two result, we can just use the tri-logarithmic identity: $$\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)+\operatorname{Li}_3\left(1-\frac1x\right)=\frac16\ln^3x-\frac12\ln^2x\ln(1-x)+\zeta(2)\ln x+\zeta(3)$$ – Ali Shadhar Jul 16 '19 at 21:12
  • Then let's call it a validation of this particular Trilogarithmic identity. But which argument do you chose to obtain the given equality? I could not figure it out. – mrtaurho Jul 16 '19 at 21:15
  • Sorry mrtaurho I mean let $x=1+i$ – Ali Shadhar Jul 16 '19 at 21:38
  • That explains a lot, though. It didn't work out with $x=i$ at all (I tried ^^). Thank you nevertheless! – mrtaurho Jul 16 '19 at 21:40
  • Yes sorry about that. I remember I did it before and thought it was i not 1+i. – Ali Shadhar Jul 16 '19 at 21:42
2

Presented below is a self-contained evaluation. Let $J=\int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx$

\begin{align} &\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx \overset{x\to\frac{1-x}{1+x}}= J -2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx-2G\ln2+\frac{\pi^3}{16}\\ &\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}dx \overset{(0,1)+\overset{x\to 1/x}{(1,\infty)}} = 2J -2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx+\frac{\pi^3}{16} \end{align}

Eliminate $J$ to get

$$\int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx =\frac12\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}dx-K-2G\ln2+\frac{\pi^3}{32} \tag1$$

where $K=\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx =2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right) $. Also

$$\int_0^1 \frac{\ln x\ln(1-x)}{1+x^2}dx =\frac12 K-\frac12\int_0^1\frac{\ln\frac x{1-x}}{1+x^2}dx +\frac12\int_0^1\frac{\ln^2x}{1+x^2}dx\tag2 $$

Then, (1) + (2)

$$ \int_0^1 \frac{\ln x\ln(1-x^2)}{1+x^2}dx =-2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right)-2G\ln2+\frac{\pi^3}{16} +\frac12 A \tag3$$

where \begin{align} A =& \int_0^1 \underset{t=1-x}{\frac{\ln^2(1-x)}{1+x^2}}dx- \int_0^1\underset{t=\frac x{1-x}}{\frac{\ln\frac x{1-x}}{1+x^2}}dx +\underset{t=1+x}{ \int_0^\infty \frac{\ln^2(1+x)}{1+x^2}}dx \\ =& \int_0^\infty {\frac{4t \ln^2 t}{4+t^4}}dt \overset{t^2=2u}= \frac14 \int_0^\infty \frac{\ln^2(2u)}{1+u^2}du = \frac{\pi}{8}\ln^22+\frac{\pi^3}{32} \end{align}

Plug into (3) to obtain $$ \int_0^1 \frac{\ln x\ln(1-x^2)}{1+x^2}dx =-2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right) -2G\ln2 +\frac{\pi}{16}\ln^22+\frac{5\pi^3}{64}$$

Quanto
  • 97,352