Is there a simple way to evaluate $\int_0^1 \arctan x \ln \left(\frac{1-x}{1+x}\right) d x?$

Question

After trying many methods but fail, I try the following substitution.

Letting $t=\frac{1-x}{1+x}$ preserves the interval and transforms the integral into $$I=\int_0^1 \arctan x \ln \left(\frac{1-x}{1+x}\right) d x = 2 \int_0^1 \arctan \left(\frac{1-t}{1+t}\right) \frac{\ln t}{(1+t)^2} d t$$ Noting that $$\tan \left(\frac{\pi}{4}-\arctan t\right)=\frac{1-t}{1+t} \Leftrightarrow \frac{\pi}{4}-\arctan t=\arctan \left(\frac{1-t}{1+t}\right) $$ $$ I=\frac{\pi}{2} \underbrace{\int_0^1 \frac{\ln t}{(1+t)^2} d t }_K-2 \underbrace{\int_0^1\frac{\arctan t}{(1+t)^2} \ln t d t}_L $$

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Using integration by parts, we have $$ \begin{aligned} \int \frac{\ln t}{(1+t)^2} d t & =-\int \ln t d\left(\frac{1}{1+t}\right)=-\frac{\ln t}{1+t}+\int \frac{d t}{t(1+t)} \\ & =-\frac{\ln t}{1+t}+\ln t-\ln (1+t)=\frac{t \ln t}{1+t}-\ln (1+t) \cdots (*) \\\Rightarrow \quad K&=-\ln 2 \end{aligned} $$

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Using (*), we can tackle the integral $L$ by integration by parts again, $$ \begin{aligned} L= & \underbrace{ {\left[\arctan t\left(\frac{t \ln t}{1+t}-\ln (1+t)\right)\right]_0^1 }}_{=-\frac{\pi}{4}\ln 2 } - \underbrace{ \int_0^1 \frac{t \ln t}{\left(1+t^2\right)(1+t)} d t}_{M} + \underbrace{ \int_0^1 \frac{\ln (1+t)}{1+t^2} d t}_{N} \end{aligned} $$

Using power series and partial fractions, $$ \begin{aligned} M&=-\frac{1}{2} \int_0^1 \frac{\ln t}{1+t} d t+\frac{1}{2} \int_0^1 \frac{t \ln t}{1+t^2} d t+\frac{1}{2} \int_0^1 \frac{\ln t}{1+t^2} dt \\&=-\frac{1}{2} \sum_{k=0}^{\infty}(-1)^k \int_0^1 t^k \ln t d t+\frac{1}{2} \sum_{k=0}^{\infty}(-1)^k \int_0^1 t^{2k +1} \ln tdt +\frac{1}{2} \sum_{k=0}^{\infty}(-1)^k \int_0^1 t^{2 k} \ln t d t\\&= -\frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2}+\frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+2)^2}+\frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)^2}\\&= -\frac{3}{8}\left(1-\frac{1}{2}\right) \zeta(2)+\frac{1}{2} G\\&=-\frac{\pi^2}{32}+\frac{G}{2}\end{aligned} $$

Letting $t=\tan \theta$ transforms $$ N=\int_0^{\frac{\pi}{4}} \ln (1+\tan \theta) d \theta \stackrel{x\mapsto\frac{\pi}{4}-x}{=} \int_0^{\frac{\pi}{4}} \ln \left(\frac{2}{1+\tan \theta}\right) d \theta= \frac{\pi}{8} \ln 2 $$

Grouping them together yields $$I=-\frac{\pi}{4} \ln 2+\frac{\pi^2}{16}-G$$

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Is there a simple way to evaluate $\int_0^1 \arctan x \ln \left(\frac{1-x}{1+x}\right) d x?$

Your comments and alternatives are highly appreciated.

Answer 1

$$\arctan\left(x\right)=-\frac{i}{2} \log\left(\frac{1+ix}{1-ix}\right), \quad\tanh^{-1}\left(x\right)=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$$

$$\log\left(\frac{1}{a}\right) = -\log\left(a\right)$$

$$\log\left(\frac{1-x}{1+x}\right)=-2\tanh^{-1}\left(x\right)$$

$$-2\int_{0}^{1}\arctan\left(x\right)\tanh^{-1}\left(x\right)dx$$

$$\arctan\left(x\right)=-\frac{i}{2}\left[\log\left(1+ix\right)-\log\left(1-ix\right)\right], \quad -2\tanh^{-1}\left(x\right)=\log\left(1-x\right)-\log\left(1+x\right)$$

$$\arctan\left(x\right) \cdot -2\tanh^{-1}\left(x\right) = -\frac{i}{2} \left[-\log\left(1-x\right)\log\left(1-ix\right)+\log\left(1-x\right) \log\left(1+ix\right)+\log\left(1+x\right)\log\left(1-ix\right)-\log\left(1+x\right)\log\left(1+ix\right)\right]$$

$$ -\frac{i}{2} \left[-\int_{0}^{1}\log\left(1-x\right)\log\left(1-ix\right)dx+\int_{0}^{1}\log\left(1-x\right)\log\left(1+ix\right)dx+\int_{0}^{1}\log\left(1+x\right)\log\left(1-ix\right)dx-\int_{0}^{1}\log\left(1+x\right)\log\left(1+ix\right)dx\right] $$

$$-\frac{i}{2}\left[-I_1 + I_2 + I_3 - I_4\right]$$

$$I_1 = \int_{0}^{1} \log\left(1-x\right)\log\left(1-ix\right)dx$$

$$\log\left(1-x\right) = -\sum_{n=1}^{\infty} \frac{x^n}{n}$$

$$\log\left(1-ix\right) = -\sum_{m=1}^{\infty} \frac{x^m e^{i\frac{\pi}{2}m}}{m}$$

$$\prod = \sum_{\substack{n=1 \\ m=1}}^{\infty}\frac{x^{n+m}e^{i \frac{\pi}{2}m}}{nm}$$

$$\int_{0}^{1} x^{n+m}dx = \frac{1}{n+m+1}$$

$$\sum_{\substack{n=1 \\ m=1}}^{\infty}\frac{e^{i \frac{\pi}{2}m}}{nm\left(n+m+1\right)}$$ $H$ denotes the harmonic numbers, and $\mathbf{C}$ denotes Catalan's constant $$\sum_{n=1}^{\infty}\frac{1}{n^2+mn+n}=\frac{H^{m+1}}{m+1}$$ $$ \sum_{m=1}^{\infty} \frac{e^{i \frac{\pi}{2}m} H_{m+1}}{m^2+m}= \frac{\pi^2}{16}-\frac{\log^2\left(2\right)}{8} - \mathbf{C} + i \frac{\pi \log\left(2\right)}{8}$$

$$\boxed{I_1 = \frac{\pi^2}{16}-\frac{\log^2\left(2\right)}{8} - \mathbf{C} + i \frac{\pi \log\left(2\right)}{8}}$$

$$I_2 = \int_{0}^{1}\log\left(1-x\right)\log\left(1+ix\right)dx$$

$$\boxed{I_2 = -I_1}$$

$$I_3 = \int_{0}^{1}\log\left(1+x\right)\log\left(1-ix\right)dx$$

$$\boxed{I_3=-I_1}$$

$$I_4=\int_{0}^{1}\log\left(1+x\right)\log\left(1+ix\right)dx$$

$$\boxed{I_4=I_1}$$

$$-2\int_{0}^{1}\arctan\left(x\right)\tanh^{-1}\left(x\right)dx = -\frac{i}{2}\left[-4I_1\right]$$

$$\boxed{\int_{0}^{1} \arctan\left(x\right) \log\left(\frac{1-x}{1+x}\right)dx = -\frac{\pi\log\left(2\right)}{4} + \frac{\pi^2}{8} - \frac{\log^2\left(2\right)}{4} - 2\mathbf{C}}$$