How to evaluate Integral:$\int^1_0\ln\left({\frac{1+x}{1-x}}\right)\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)dx$

Question

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

Let $x=\sin(y)$.

$$\begin{align}I&=\int^{\frac{\pi}{2}}_0 \cos(y)\ln\left({\frac{1+\sin(y)}{1-\sin(y)}}\right)\ln\left({\frac{1+\cos(y)}{1-\cos(y)}}\right)dy\\&=4\int^{\frac{\pi}{2}}_0 \cos(y)\ln\left({\frac{1+\sin(y)}{\cos(y)}}\right)\ln\left({\frac{1+\cos(y)}{\sin(y)}}\right)dy\end{align}$$

I'm stuck on the solution here. Any help on how to evaluate this integral?

This integration is from the book Special Functions.

The answer is given in the book as equal to:$$I=8G-\frac{\pi^2}{2}$$where $G=\int_0^1 \frac{\sinh ^{-1} x}{\sqrt{1-x^2}}dx$ is Catalan’s constant.

Answer 1

Continue with the substitution $x=\sin t$ \begin{align} &\int^1_0\ln{\frac{1+x}{1-x}}\ln{\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}{dx}\\ =& \ 4\int_0^{\pi/2}\tanh^{-1}(\sin t) \tanh^{-1}(\cos t)\ d(\sin t)\\ \overset{ibp}=&\ 4 \int_0^{\pi/2}\tanh^{-1}(\sin t)dt -4 \int_0^{\pi/2}\tan t\tanh^{-1}(\cos t)dt\\ =&\ 4\cdot (2G)- 4\cdot \frac{\pi^2}8=8G -\frac{\pi^2}2 \end{align} where $$\int_0^{\pi/2}\tanh^{-1}(\sin t)dt\overset{\sin t=\frac{1- s^2}{1+s^2}}=-2\int_0^1\frac {\ln s}{s^2+1}ds =2G $$ $$\int_0^{\pi/2}\tan t\tanh^{-1}(\cos t)dt\overset{s^2=\tan\frac t2}=\int_0^1\frac {\ln s}{s^2-1}ds=\frac{\pi^2}8 $$

Answer 2

\begin{align}J&=\int^1_0\ln\left({\frac{1+x}{1-x}}\right)\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)dx\\ &\overset{u=\sqrt{\frac{1-x}{1+x}}}=16\int_0^1\frac{u\ln\left(\frac{1-u}{1+u}\right)\ln u}{(1+u^2)^2}du\\ &\overset{\text{IBP}}=-8\underbrace{\left[\frac{\ln\left(\frac{1-u}{1+u}\right)\ln u}{1+u^2}\right]_0^1}_{=0}+8\underbrace{\int_0^1\frac{\ln\left(\frac{1-u}{1+u}\right)}{u(1+u^2)}du}_{=J_1}-16\underbrace{\int_0^1\frac{\ln u}{1-u^4}du}_{=J_2}\\ J_1&=\underbrace{\int_0^1\frac{\ln\left(\frac{1-u}{1+u}\right)}{u}du}_{=J_{1,1}}-\underbrace{\int_0^1\frac{u\ln\left(\frac{1-u}{1+u}\right)}{1+u^2}du}_{=J_{1,2}}\\ J_{1,1}&\overset{z=\frac{1-u}{1+u}}=2\int_0^1\frac{\ln z}{1-z^2}dz=2\int_0^1\frac{\ln z}{1-z}dz-\underbrace{\int_0^1\frac{2z\ln z}{1-z^2}dz}_{w=z^2}\\ &=-2\zeta(2)+\frac12\zeta(2)=\boxed{-\frac32\zeta(2)}\\ J_{1,2}&\overset{z=\frac{1-u}{1+u}}=\int_0^1\frac{(1-z)\ln z}{(1+z)(1+z^2)}dz=\int_0^1\frac{\ln z}{1+z}dz-\underbrace{\int_0^1\frac{z\ln z}{1+z}dz}_{w=z^2}\\ &=\int_0^1\frac{\ln z}{1+z}dz-\frac14\int_0^1\frac{\ln w}{1+w}dw=\frac34\int_0^1\frac{\ln z}{1+z}dz\\ &=\frac34\int_0^1\frac{\ln z}{1-z}dz-\frac34\underbrace{\int_0^1\frac{2z\ln z}{1-z^2}dz}_{w=z^2}=-\frac34\zeta(2)+\frac38\zeta(2)=\boxed{-\frac38\zeta(2)}\\ &\boxed{J_1=J_{1,1}-J_{1,2}=-\frac{9}{8}\zeta(2)}\\ J_2&=\frac12\int_0^1\frac{\ln u}{1-u}du-\frac14\underbrace{\int_0^1\frac{2u\ln u}{1-u^2}du}_{z=u^2}+\frac12\int_0^1\frac{\ln u}{1+u^2}du\\ &=-\frac12\zeta(2)+\frac18\zeta(2)-\frac{\text{G}}2=\boxed{-\frac38\zeta(2)-\frac{\text{G}}2}\\ &\boxed{J=8J_1-16J_2=-3\zeta(2)+8\text{G}=8\text{G}-\frac{\pi^2}{2}}\\ \end{align} NB: I assume that,\begin{align}\int_0^1\frac{\ln x}{1-x}dx&=-\zeta(2)\\ \zeta(2)=\frac{\pi^2}{6} \end{align}

Answer 3

\begin{align} &\int^1_0\ln{\frac{1+x}{1-x}}\ln{\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}{dx}\\ \overset{IBP}=& \int^1_0\ln{\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\;\mathrm d\bigg((1+x)\ln(1+x)+(1-x)\ln(1-x)\bigg)\\ =&2\int_0^1\bigg((1+x)\ln(1+x)+(1-x)\ln(1-x)\bigg)\frac{1}{x\sqrt{1-x^2}}\;\mathrm dx\\ \overset{x\to\sin x}=&2\int_0^{\pi/2}\bigg(\frac{2\ln\cos(x)}{\sin x}+\ln(1+\sin x))-\ln(1-\sin x)\bigg)\;\mathrm dx\\ =&4\int_0^{\pi/2}\frac{\ln\cos(x)}{\sin x}\;\mathrm dx+2\int_0^{\pi/2}\ln(\frac{1+\sin x}{1-\sin x})\;\mathrm dx\\ =&-\frac{\pi^2}2-4\int_0^{\pi/2}\ln\tan(\frac x2)dx\\ =&-\frac{\pi^2}2+8G. \end{align}

Answer 4

Using the same strategy as before:

$$\begin{align*} I &= 4 \int_0^1 \operatorname{artanh}x \operatorname{arsech}x \, dx \\ &= 4 \int_0^1 \int_x^1 \frac{\operatorname{artanh} x}{y\sqrt{1-y^2}} \, dx\, dy = 4 \int_0^1 \int_0^y \cdots \, dx\,dy \\ &= 4 \int_0^1 \frac{y \operatorname{artanh}y + \frac12 \log\left(1-y^2\right)}{y\sqrt{1-y^2}} \, dy & \text{by parts} \\ &= 4 \underbrace{\int_0^1 \frac{\operatorname{artanh}y}{\sqrt{1-y^2}} \, dy}_{2G} + 2 \int_0^1 \frac{\log\left(1-y^2\right)}{y\sqrt{1-y^2}} \, dy \\ &= 8G + 4 \underbrace{\int_0^1 \frac{\log y}{1-y^2}}_{-\frac{\pi^2}8} & y\to\sqrt{1-y^2} \\ &= \boxed{8G - \frac{\pi^2}2} \end{align*}$$

The first labeled integral is listed here. The second is well-known and may be evaluated by exploiting Taylor series, for instance.

Answer 5

Proceeding similar to this,

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

Using the below Taylor series,

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

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

Integrating by parts, we get

If interested in all the steps/working of the integration by parts, see here

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

$$\color{red}{\int_0^1 \frac{x^{2n+1}}{\sqrt{1-x^2}}\,dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma(n+1)}{\Gamma\left(n+\frac{3}{2}\right)}=\frac{4^n}{(2n+1)\binom{2n}{n}}}$$

Found the above result here

Using the above result,

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

Now comes the hard part,

These are known sums in disguise, but only after partial fraction (complete partial fraction gives diverging series),

$$\color{red}{\frac{1}{(2n+1)^2(2n+2)}=\frac{1}{(2n+1)^2}-\frac{1}{(2n+1)(2n+2)}}$$

So,

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

After a long search, I found this expansion,

Starting from,

$$\color{red}{\arcsin(x)=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(2n+1)}x^{2n+1}}$$

$$\implies \sum_{n=0}^{\infty} \frac{4^n}{\displaystyle(2 n+1)^2\binom{2n}{n}} = \int_0^1 \frac{\arcsin(t)}{t \sqrt{1-t^2}}\,dt$$

Now to evaluate this integral, $$\int_0^1 \frac{\arcsin(t)}{t \sqrt{1-t^2}}\,dt\underset{t\to\sin t}\implies \int_0^\frac{\pi}{2}\frac{t}{\sin t}\,dt$$

Now this is a popular integral (whose proof is here)

$$\therefore \color{red}{\sum_{n=0}^{\infty} \frac{4^n}{\displaystyle(2 n+1)^2\binom{2n}{n}} = 2G}\tag1$$

Now using the below Taylor series which I found here,

$$\color{red}{(\arcsin x)^2=\sum_{n \ge 0}\frac{4^n}{(2n+1)(n+1)\binom{2n}{n}}x^{2n+2}}$$

Set $x=1$ and dividing both sides by $2$,

$$\therefore \color{blue}{\sum_{n \ge 0}\frac{4^n}{(2n+1)(2n+2)\binom{2n}{n}}=\frac{\pi^2}{8}}\tag2$$

Our integral is $4\times (1)-4\times (2)$

$$\therefore{\color{green}{\int^1_0\ln\left({\frac{1+x}{1-x}}\right)\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)dx=8G-\frac{\pi^2}{2}}}$$

Answer 6

$$\begin{align} I&=4\int_0^1\color{green}{\tanh^{-1}x}\color{purple}{\tanh^{-1}\sqrt{1-x^2}}\color{green}{dx}\\ &\stackrel{IBP}{=}0-4\int_0^1\left(x\tanh^{-1}x+\frac12\ln(1-x^2)\right)\left(-\frac1{x\sqrt{1-x^2}}\right)dx\\ &\stackrel{\color{blue}{x=\tanh t},\,\color{red}{t=\sqrt{1-x^2}}}{=}\color{blue}{4\int_0^1\frac{\tanh^{-1}x}{\sqrt{1-x^2}}dx}+\color{red}{2\int_0^1\frac{\ln(1-x^2)}{x\sqrt{1-x^2}}dx}\\ &=\color{blue}{4\int_0^\infty\frac t{\cosh t}dt}+\color{red}{4\int_0^1\frac{\ln t}{1-t^2}dt}\\ &=\color{blue}{8\int_0^\infty\frac t{e^t+e^{-t}}dt}+2\int_0^1\frac{\ln t}{1-t}dt+2\int_0^1\frac{\ln t}{1+t}dt\,\,\,\text{(Make the sub $t=e^{-u}$)}\\ &=\color{blue}{8\beta(2)}-2\int_0^\infty\frac{u}{e^u-1}dt-2\int_0^1\frac{u}{e^u+1}dt\\ &=8\beta(2)-2\zeta(2)-2\eta(2)\\ &=8\beta(2)-2(\tfrac{\pi^2}6)-2(\tfrac{\pi^2}{12})\\ &=8G-\frac{\pi^2}2 \end{align}$$

$\zeta(s)$: Riemann zeta function

$\eta(s)$: Dirichlet eta function

$\beta(s)$: Dirichlet beta function

$IBP$: Integration by parts