8

I want to evaluate $$\int_0 ^ 1 \frac{\ln (x)(1+\ln(x) +\ln(1-x))}{x^2+1}dx$$.

This question is from RMM.

I tried to separate the integral into:

$$\begin{align*} & \int_0 ^ 1 \frac{\ln (x)(1+\ln(x) +\ln(1-x))}{x^2+1}dx \\ &= \int_0 ^ 1 \frac{\ln (x)}{x^2+1}dx+\int_0 ^ 1 \frac{\ln^2 (x)}{x^2+1}dx + \int_0 ^ 1 \frac{\ln (x)\ln(1-x)}{x^2+1}dx \\ &= \int_0^{\frac{\pi}{4}} \ln(\tan(x))dx+\int_0^{\frac{\pi}{4}} \ln^2(\tan(x))dx+\int_0^{\frac{\pi}{4}}\ln(\tan(x))\ln(1-\tan(x)) dx \end{align*}$$

I have tried all the integration methods that I know, but I couldn't reach any useful result.

user170231
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Mathematics enjoyer
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2 Answers2

5

Utilize $ 2\ln x\ln(1-x) =\ln^2x+\ln^2(1-x)-\ln^2\frac x{1-x}$ to break up the integral as follows

\begin{align} I=&\int_0 ^ 1 \frac{\ln x\ [1+\ln x+\ln(1-x)]}{x^2+1}dx\\ =& \int_0 ^ 1 \frac{\ln x}{x^2+1}dx + \frac32\int_0 ^ 1 \frac{\ln^2x}{x^2+1}dx\\ &\>\>\>\>\> + \frac12 \int_0 ^ 1 \frac{\ln^2(1-x)}{x^2+1}dx -\frac12 \int_0 ^ 1 \frac{\ln^2\frac x{1-x}}{x^2+1}dx \end{align} where the first three integrals are familiar $\int_0 ^ 1 \frac{\ln x}{x^2+1}dx=-G$ $$\int_0 ^ 1 \frac{\ln^2x}{x^2+1}dx=\frac{\pi^3}{16}, \>\>\> \int_0 ^ 1 \frac{\ln^2(1-x)}{x^2+1}dx=2\Im{\text{Li}_3(\frac{1+i}2)}$$ and the last integral can be further broken up as \begin{align} & \int_0 ^ 1 \frac{\ln^2\frac x{1-x}}{x^2+1}\overset{t= \frac x{1-x} }{dx} =\int_0^\infty \frac{\ln^2t}{t^2+2t +2}dt\\ =& \int_0^\infty \frac{t^2\ln^2t}{t^4+4}\overset{t\to\frac{\sqrt2}t}{dt} -2 \int_0^\infty \frac{t\ln^2t}{t^4+4}\overset{t^2\to 2t}{dt} +2 \int_0^\infty \frac{\ln^2t}{t^4+4}\overset{t\to\sqrt2 t}{dt} \\ =& -\frac{\ln^22}8\int_0^\infty\frac{1}{t^2+1}dt -\frac18 \int_0^\infty\frac{\ln^2t}{t^2+1}dt \\ &\>\>\>\>\> -\frac{\ln^22}{2\sqrt2}\int_0^\infty\frac{1}{t^4+1}dt+ \sqrt2 \int_0^\infty\frac{\ln^2t}{t^4+1}dt\\ =& \ \frac{5\pi^3}{64}+\frac\pi{16}\ln^2 2 \end{align} Substitute above results into $I$ to obtain $$ I =-G +\frac{7\pi^3}{128}-\frac{\pi}{32}\ln^22 + \Im{\text{Li}_3(\frac{1+i}2)} $$

Quanto
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4

Splitting the integrand into three and evaluating them one by one. $$ \int_0^1 \frac{\ln x(1+\ln x+\ln (1-x))}{x^2+1} d x= \int_0^1 \frac{\ln x}{x^2+1} d x+\int_0^1 \frac{\ln ^2 x}{x^2+1} d x+\int_0^1 \frac{\ln x \ln (1-x)}{x^2+1} d x $$

As we know that $$\int_0^1 \frac{\ln x}{x^2+1} d x=-G$$

Considering \begin{aligned} J(a) & =\int_0^1 \frac{x^a}{x^2+1} d x =\frac{1}{4}\left[\psi\left(\frac{a+3}{4}\right)-\psi\left(\frac{a+1}{4}\right)\right] \end{aligned} Then $$ \int_0^1 \frac{\ln^2x}{x^2+1} =J^{\prime \prime}(0)=\frac{1}{64}\left[\psi^{\prime \prime}\left(\frac{3}{4}\right)-\psi^{\prime \prime}\left(\frac{1}{4}\right)\right]=\frac{\pi^3}{16} $$

Using the result in post, $$\ \displaystyle\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx=\frac{3\pi^3}{64}+\frac{\pi}{16}\ln^22-\Im\operatorname{Li}_3(1+i),$$ we can conclude that $$ \boxed{\int_0^1 \frac{\ln x(1+\ln x+\ln (1-x))}{x^2+1} d x =-G+\frac{7 \pi^3}{64}+\frac{\pi}{16} \ln ^2 2-\Im \operatorname{Li_3}(1+i)} $$

Lai
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