The integral $\int_{0}^{1} \frac{ \log (1-x)}{1+x^2}dx$

Question

Recently a very interesting result

$\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$

has been proved in a more than elegant way. See Show that $\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$

Here we find the value of the integral

$I=\int_{0}^1 \frac{\log(1-x)}{1+x^2} dx$

Let $x=\tan t$, then

$I=\int_{0}^{\pi/4} \log(1-\tan t) dt.$

By IVth property, we get

$I=\int_0^{\pi/4} \log (1-\tan(\pi/4-t)) dt=\int_{0}^{\pi/4} \log \left( 1-\frac{1-\tan x}{1+\tan x}\right) dt=\int_{0}^{\pi/4} \log (2 \tan x)~dx-\int_{0}^{\pi/4}\log(1+\tan x)dx=\frac{\pi}{4} \log 2+J-K.$

$J=\int_{0}^{\pi/4} \log \tan x ~dx=-C,$ see Definite integral $\int_0^{\pi/4}\log\left(\tan{x}\right)\ dx$

Let us work for $K=\int_{0}^{\pi/4} \log(1+\tan x) dx$, by IV property, again we get

$K=\int_0^{\pi/4} \log (1+\tan(\pi/4-t)) dt=\int_{0}^{\pi/4} \log \left( 1+\frac{1-\tan x}{1+\tan x}\right) dt=\int_{0}^{\pi/4} \log 2~ dx-K \implies K=\frac{\pi}{8} \log 2.$

Finally, we get $I=\frac{\pi}{8}\log 2-C,$ where $C$ is the Catalan constant.

What could be other interesting ways of finding $I$?

Answer 1

Perhaps a cleaner way is to consider the integral $$ \int_1^\infty \frac{\ln(u-1)}{1+u^2}du. $$ First, perform the substitution $u\rightarrow (u+1)/(u-1) = 1 + 2/(u-1)$ to get $$ \int_1^\infty \frac{\ln(u-1)}{1+u^2}du = \int_1^\infty \frac{\ln\left(\frac{2}{u-1}\right)}{1+u^2}du = \frac{\pi}{4}\ln2 -\int_1^\infty \frac{\ln(u-1)}{1+u^2}\Longrightarrow \int_0^1\frac{\ln(u-1)}{1+u^2} = \frac{\pi}{8}\ln 2. $$ Now use the substitution $u\rightarrow x^{-1}$ to get $$ \int_1^\infty \frac{\ln(u-1)}{1+u^2}du = \int_0^1\frac{\ln(x^{-1}-1)}{1+x^2}dx = \int_0^1\frac{\ln(1-x)}{1+x^2}dx -\int_0^1\frac{\ln x}{1+x^2}dx = \frac{\pi}{8}\ln 2. $$ Rearranging gives us an expression for the original integral: $$ \int_0^1\frac{\ln(1-x)}{1+x^2}dx = \frac{\pi}{8}\ln 2 + \int_0^1\frac{\ln x}{1+x^2}dx = \frac{\pi}{8}\ln 2 - C, $$ where $\int_0^1\ln(x)/(1+x^2)dx = -C$ is apparently a well-known integral.

Answer 2

$$\int_0^1\frac{\arctan x}{x(1+x)}dx=\int_0^1\frac{\arctan x}{x}dx-\int_0^1\frac{\arctan x}{1+x}dx$$ where the first integral is $G$ and by $x\to \frac{1-x}{1+x}$, we have

$$\int_0^1\frac{\arctan x}{1+x}dx=\int_0^1\frac{\pi/4-\arctan x}{1+x}dx$$

$$\Longrightarrow \int_0^1\frac{\arctan x}{1+x}dx=\frac{\pi}{8}\ln(2).$$

Answer 3

Split the integrand\begin{align} \int_{0}^1 \frac{\ln(1-x)}{1+x^2} dx =& \int_{0}^1 \underset{= \frac\pi8 \ln2} {\frac{\ln \sqrt2}{1+x^2} }dx + \int_{0}^1 \underset{=-G}{\frac{\ln \frac{1-x}{1+x}}{1+x^2}}\overset{x\to\frac{1-x}{1+x}}{dx} + \int_{0}^1 \underset{K=- K=0}{\frac{\ln \frac{1+x} {\sqrt2 }}{1+x^2} }\overset{x\to\frac{1-x}{1+x}}{dx} = \frac\pi8 \ln2-G \end{align}

Answer 4

Let $x=\tan\theta$, then $$ \begin{aligned} I & =\int_0^{\frac{\pi}{4}} \ln (1-\tan \theta) d \theta \\ & =\int_0^{\frac{\pi}{4}}\ln (\cos \theta-\sin \theta)d\theta-\int_0^{\frac{\pi}{4}}\ln (\cos \theta)d \theta \end{aligned} $$ For the first one,$$ \begin{aligned}\int_0^{\frac{\pi}{4}}[\ln (\cos \theta-\sin \theta)= & \frac{1}{2} \int_0^{\frac{\pi}{4}} \ln (1-\sin 2 \theta) d \theta\\ = & \frac{1}{4} \int_0^{\frac{\pi}{2}} \ln (1-\sin \theta) d \theta\\ = & \frac{1}{4}\int_0^{\frac{\pi}{2}} \ln (1-\cos \theta) d \theta \\ = & \frac{1}{4}\int_0^{\frac{\pi}{2}} \ln \left(2 \sin ^2 \frac{\theta}{2}\right) d \theta \\ = & \frac{\pi}{8} \ln 2+\frac12 \int_0^{\frac{\pi}{2}} \ln \left(\sin \frac{\theta}{2}\right) d \theta \\ = & \frac{\pi}{8} \ln 2+\int_0^{\frac{\pi}{4}} \ln (\sin \theta) d \theta \end{aligned} $$ Now we can conclude that $$ I =\frac{\pi}{8} \ln 2+\int_0^{\frac{\pi}{4}} \ln (\tan \theta) d \theta \\ =\frac{\pi}{8} \ln 2-G $$ where the last answer comes from the post.