I need help to evaluate integral:$I=\int^1_0\frac{1}{x}\ln\left({\frac{1+x}{1-x}}\right)\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)dx$

Question

I need help to evaluate integral:$$I=\int^1_0\frac{1}{x}\ln\left({\frac{1+x}{1-x}}\right)\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)dx$$

let $x=\sin(y)$ therfore:

\begin{align*}I&=\int^{\frac{\pi}{2}}_0\frac{\cos(y)}{\sin(y)}\ln\left({\frac{1+\sin(y)}{1-\sin(y)}}\right)\ln\left({\frac{1+\cos(y)}{1-\cos(y)}}\right)dy \\ &= 4\int^{\frac{\pi}{2}}_0\frac{\cos(y)}{\sin(y)}\ln\left({\frac{1+\sin(y)}{\cos(y)}}\right)\ln\left({\frac{1+\cos(y)}{\sin(y)}}\right)dy \\ &= ^{Ibp}\int^{\frac{\pi}{2}}_0 \frac{\ln(\sin(y))}{\sin(y)}\ln\left({\frac{1+\sin(y)}{\cos(y)}}\right)dy - \int^{\frac{\pi}{2}}_0 \frac{\ln(\sin(y))}{\cos(y)}\ln\left({\frac{1+\cos(y)}{\sin(y)}}\right)dy \\&= -\int^{\frac{\pi}{2}}_0 \frac{\ln(\tan(y))}{\cos(y)}\ln\left({\frac{1+\cos(y)}{\sin(y)}}\right)dy \end{align*} This integral is from the book Special Functions by Yury. The result of the integral is given as equal to:$$I=\pi^2\ln(2)$$

I need help proving this.

Answer 1

With $x=\sin t$ \begin{align} &\int^1_0\frac{1}{x}\ln{\frac{1+x}{1-x}}\ln{\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\ dx\\ = &\ 4\int^{\pi/2}_0\frac{\tanh^{-1}(\sin t) \ \tanh^{-1} (\cos t)}{\tan t}dt\\ = &\ 4 \int_ 0^{\pi/2}\int_0^{\pi/2} \frac{\cos s\cos t\ \tanh^{-1} (\cos t)}{1 -\sin^2 s\ \sin^2 t}\ \overset{ibp} {dt}\ ds \\ = &\ 4 \int_ 0^{\pi/2}\int_ 0^{\pi/2} \frac{\tanh^{-1} (\sin s \sin t)}{\tan s \sin t}{dt}\ ds \\ =& \ 2 \pi \int_ 0^{\pi/2} \frac{s}{\tan s} ds = 2\pi \cdot \frac\pi2\ln2=\pi^2 \ln2 \end{align}

where \begin{align} & \int_ 0^{\pi/2} \frac{\tanh^{-1} (\sin s\sin t)}{\sin t}{dt}\\ =& \ \frac12\int_0^{\pi/2}\int_{-s}^s\frac{\cos u}{1+\sin u\cos t}du \ dt= \frac12\int_{-s}^s(\frac\pi2-u)du=\frac{\pi s}2 \end{align}

Answer 2

$$I=\int^1_0\frac{1}{x}\ln\left({\frac{1+x}{1-x}}\right)\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)dx$$

As the bounds are from $0$ to $1$ , needless to say that this series stands for $|x|<1$,

$$\ln\left(\frac{1+x}{1-x}\right)=2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$$

A simple transformation taken from here

$$I=2\int^1_0\frac{1}{x}\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)dx $$

$$I=2\sum_{n=0}^{\infty}\frac{1}{2n+1}\int_0^1x^{2n}\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)\,dx$$

Integrating by parts, we get

$$I=4\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}}\,dx$$

Solving that integral is done here

Basically substitute $\sin(\theta)=x$ and then using the Wallis product

$$2\int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}}\,dx=\frac{\pi}{4^n} \binom{2 n}{n}$$

Using this,

Basically the sum is solved using a well known identity of $\binom{2 n}{n}$ and then using this awesome technique; derivative of the beta function

$$\color{blue}{\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\frac{\binom{2 n}{n}}{4^n}=\frac{1}{2}\pi\ln(2)}$$

Refer this and this for better understanding

$$I=2\pi\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\frac{\binom{2 n}{n}}{4^n} =\pi^2\ln(2)$$

$$\boxed{\color{red}{\int^1_0\frac{1}{x}\ln\left({\frac{1+x}{1-x}}\right)\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)dx=\pi^2\ln(2)}}$$

Quick run of the "Integrating by parts" claimed above,

$$J=\int x^{2n}\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)\,dx$$

$$J=\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)\int x^{2n}\,dx-\int \frac{d}{dx}\left(\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)\right)\int x^{2n}\,dx\,dx$$

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

$$J=\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)\frac{x^{2n+1}}{2n+1}+\frac{2}{2n+1}\int \frac{x^{2n+1}}{x\sqrt{1-x^2}}\,dx$$

$$J=\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)\frac{x^{2n+1}}{2n+1}+\frac{2}{2n+1}\int \frac{x^{2n}}{\sqrt{1-x^2}}\,dx$$

Applying the bounds $[0,1]$,

$$\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)\frac{x^{2n+1}}{2n+1}|_0^1=0$$

$$J=\frac{2}{2n+1}\int \frac{x^{2n}}{\sqrt{1-x^2}}\,dx$$

Answer 3

$$\newcommand{artanh}{\operatorname{artanh}} \newcommand{arsech}{\operatorname{arsech}} \begin{align*} I &= \int_0^1 \frac4x \artanh x \arsech x \, dx \\ &= \int_0^1 \frac4x \left(\int_0^1 \frac x{1-x^2y^2} \, dy\right) \left(\int_1^\tfrac1x \frac{dz}{z\sqrt{1-x^2z^2}}\right) \, dx \\ &= \int_{x=0}^1 \int_{y=0}^1 \int_{z=1}^\tfrac1x \cdots \, dz\, dy \, dx \\ &= \int_{z=1}^\infty \int_{y=0}^1 \int_{x=0}^\tfrac1z \frac4{z\left(1-x^2y^2\right)\sqrt{1-x^2z^2}} \, dx \, dy \, dz \\ &= \int_{z=1}^\infty \int_{y=0}^1 \int_{X=0}^\infty \frac4{z\left(1+\left(z^2-y^2\right)X^2\right)} \, dX \, dy \, dz & X=\frac x{\sqrt{1-x^2z^2}} \\ &= \int_{z=1}^\infty \int_{y=0}^1 \frac{2\pi}{z\sqrt{z^2-y^2}} \, dy \, dz \\ &= \int_{z=0}^1 \int_{y=0}^z \frac{2\pi}{z\sqrt{1-y^2}} \, dy \, dz & y\to\frac yz \\ &= 2\pi \underbrace{\int_0^1 \frac{\arcsin z}z\,dz}_{=\tfrac\pi2\log2} = \boxed{\pi^2\log2} \end{align*}$$

The last integral and equivalent forms are evaluated/further linked here.