Evaluate $\int_0^1 \frac{\ln x \sin^{-1}x}{x(1-x^2)}dx $

Question

Question: how to evaluate the following integral $$\int_0^1 \frac{\ln x \sin^{-1}x}{x(1-x^2)}dx $$ I tried partial fractions as follows $$\int_0^1 \frac{\ln x \sin^{-1}x}{x(1-x^2)}dx = \int_0^1 \bigg(\frac{\ln x \sin^{-1}x}{x}+\frac{x\ln x \sin^{-1}x}{1-x^2}\bigg)dx $$ I was able to handle the first integral by integration-by-parts $$\int_0^1 \frac{\ln x \sin^{-1}x}{x}dx=-\frac12 \int_0^1 \frac{\ln^2 x }{\sqrt{1-x^2}}dx $$ This integral is known to be $\frac{\pi^3}{24}+\frac\pi2\ln^2(2)$. But, the second integral seems to be hard to do and I was getting nowhere.

Answer 1

Consider \begin{align} I(a)=&\int_0^{\pi/2} \frac{\tan^{-1}(a\tan t)\ln \sin t }{\sin t\cos t}dt \\ I’(a)=& \int_0^{\pi/2} \frac{\sec^2t\ln \sin t}{1+a^2\tan^2t}dt \\ =&-\int_0^1 \int_0^{\pi/2}\frac{y \sec^2t}{(1+a^2 \tan^2t)(y^2+\tan^2t)}dt\ dy\\ =& -\frac\pi2\int_0^1 \frac1{1+ay}dy= -\frac{\pi\ln(1+a)}{2a} \end{align} Then \begin{align} \int_0^1 &\frac{\ln x \sin^{-1}x}{x(1-x^2)}dx \overset{x=\sin t}= \int_0^{\pi/2} \frac{t\ln \sin t }{\sin t\cos t}dt \\ &=I(1)=\int_0^1 I’(a)da =-\frac\pi2 \int_0^1\frac{\ln(1+a)}{a}da=-\frac{\pi^3}{24} \end{align}

Answer 2

For the second integral, I made the substitution $$x=\sin u$$ After simplifying, I got the integral down to $$\int_0^{\pi/2}u\tan u \ln(\sin u) \ du$$.

I then used King’s Rule. $$u\rightarrow \frac{\pi}{2}-u$$ This gave $$\int_0^{\pi/2}\left(\frac{\pi}{2}-u\right)\cot u \ln(\cos u) \ du$$

Multiplying this out, $$\frac{\pi}{2}\int_0^{\pi/2}\frac{\ln(\cos u)}{\tan u}-\int_0^{\pi/2}u\cot u \ln(\cos u) \ du$$

The first of the two integrals evaluates to $-\frac{\pi^2}{24}$. The second one evaluates to $-\frac{\pi}{4}\ln^2(2)$. Thus, combining the results, you get $$\int_0^1 \frac{x\ln x\sin^{-1}x}{1-x^2} \ dx = -\frac{\pi^3}{48}+ \frac{\pi}{4}\ln^2(2)$$

See the two links below as sources

How to integrate $\int_{0}^{\frac{\pi}{2}}\frac{x\ln{\cos(x)}}{\tan x}dx$?

How do you calculate $\int_0^{\pi/2}\tan(x)\ln(\sin(x))\,dx$?

Note: For this one, after applying King’s Rule, you find that the integral in this thread is equal to the one mentioned above.

Answer 3

$$I=\int_0^1 \frac{\ln x \sin^{-1}x}{x(1-x^2)}dx\overset{x=\sin t}{=}\int_0^{\pi/2}\frac{t\ln\sin t}{\sin t\cos t}dt\overset{t=\frac \pi2-x}{=}\int_0^{\pi/2}\frac{(\frac\pi2-x)\ln\cos x}{\sin x\cos x}dx$$ $$\Rightarrow 2I=\int_0^{\pi/2}\left(\frac\pi2\ln\cos x+x\ln\tan x\right)\frac{dx}{\sin x\cos x}$$ $$=\int_0^{\pi/2}\left(\frac\pi2\frac{\ln\cos x}{\ln\tan x}+x\right)\ln\tan x\,d\big(\ln\tan x\big)$$ $$\overset{IBP}{=}-\frac12\int_0^{\pi/2}\left(\ln^2(\tan x)-\frac\pi2\tan x\ln\tan x-\frac\pi2\frac{\ln\cos x}{\cos x\sin x}\right)dx$$ $$\overset{\tan x=t}{=}-\frac12\int_0^\infty\frac{\ln^2t}{1+t^2}dt+\frac\pi4\int_0^\infty\left(\frac{t\ln t}{1+t^2}-\frac{\ln(1+t^2)}{2t}\right)dt$$ $$=-\frac{\pi^3}{16}+\frac\pi{16}\int_0^\infty\left(\frac{\ln x}{1+x}-\frac{\ln(1+x)}{x}\right)dx$$ $$=-\frac{\pi^3}{16}+\frac\pi{16}\int_0^1\left(\frac{\ln x}{1+x}-\frac{\ln(1+x)}{x}\right)dx+\frac\pi{16}\int_1^\infty\left(\frac{\ln x}{1+x}-\frac{\ln(1+x)}{x}\right)dx$$ Making the substitution $t=\frac1x$ in the last integral, $$2I=-\frac{\pi^3}{16}-\frac\pi4\int_0^1\frac{\ln(1+t)}tdt=-\frac{\pi^3}{16}-\frac{\pi^3}{48}=-\frac{\pi^3}{12}$$ $$I=-\frac{\pi^3}{24}$$

Answer 4

Let $x=\sin \theta$ followed by integration by parts, then $$ \begin{aligned} I & =\int_0^{\frac{\pi}{2}} \frac{\theta \ln (\sin \theta)}{\sin \theta \cos \theta} d \theta \\ & =\int_0^{\frac{\pi}{2}} \theta \ln (\sin \theta) d(\ln (\tan \theta)) \\ & =- \underbrace{ \int_0^{\frac{\pi}{2}} \frac{\theta}{\tan \theta} \ln (\tan \theta) d \theta}_{J} - \underbrace{ \int_0^{\frac{\pi}{2}} \ln (\sin \theta) \ln (\tan \theta) d \theta}_{K} \end{aligned} $$ For $J$, let $t=\tan \theta$, then $$J=\int_0^{\infty} \frac{\tan ^{-1} t}{t\left(1+t^2\right)} \ln t d t= J’(1)$$ where $$J(a)=\int_0^{\infty} \frac{\tan ^{-1}(a t) \ln t}{t\left(1+t^2\right)} d t $$ whose derivative is

$$ \begin{aligned} J^{\prime}(a) & =\int_0^{\infty} \frac{\ln t}{\left(1+a^2 t^2\right)\left(1+t^2\right)} d t \\ & =\frac{1}{1-a^2} \int_0^{\infty}\left(\frac{\ln t}{1+t^2}-\frac{a^2 \ln t}{1+a^2 t^2}\right) d t \\ & =-\frac{a^2}{1-a^2} \int_0^{\infty} \frac{\ln t}{1+a^2 t^2} d t \\ & =\frac{a}{1-a^2} \int_0^{\infty} \frac{\ln a+\ln t}{1+t^2} d t \quad (\textrm{ via }t\to\frac 1{at})\\ & =\frac{a \pi \ln a}{2\left(1-a^2\right)} \end{aligned} $$ Hence $$ \begin{aligned} J& =\frac{\pi}{2} \int_0^1 \frac{a \ln a}{1-a^2} d a \\&=\frac{\pi}{2} \sum_{n=0}^{\infty} \int_0^1 a^{2 n+1} \ln a d a \\ & = -\frac{\pi^2}{2} \sum_{n=0}^{\infty} \frac{1}{4(n+1)^2} \\&=-\frac{\pi^3}{48} \end{aligned} $$ For $K$, we refer to the post 1

$$\int_0^{\pi/2}\ln^2(\sin\theta)\,\mathrm d\theta = \frac{1}{24} \left(\pi^3 + 3\pi \ln^2(4)\right)$$

and post 2 $$\int_0^{\frac{\pi}{2}} \ln (\sin \theta) \ln (\cos \theta) d \theta=\frac{\pi}{2}\log^2 2-\frac{\pi^3}{48}$$ Grouping them yields $$\int_0^{\frac{\pi}{2}} \ln (\sin \theta) \ln (\tan \theta) d \theta= \int_0^{\frac{\pi}{2}} \ln^2 (\sin \theta)-\int_0^{\frac{\pi}{2}} \ln (\sin \theta)\ln (\cos \theta) =\frac{\pi^3}{16} $$ We now conclude that $$ \boxed{\int_0^1 \frac{\ln x \sin ^{-1} x}{x\left(1-x^2\right)} d x=-\frac{\pi^3}{24}} $$

Answer 5

Using the Taylor expansions of $\arctan y$ and $\log\left(1+y^2\right)$, together with some special cases of polylogarithms:

$$\begin{align*} I &= \int_0^1 \frac{\log x \arcsin x}{x\left(1-x^2\right)} \, dx \\ &= \int_0^\infty \frac{\log\frac y{\sqrt{1+y^2}} \arctan y}y \, dy & y=\frac x{\sqrt{1-x^2}} \\ &= \int_0^1 \frac{\arctan y \log y}y \, dy - \frac\pi4 \int_0^1 \frac{\log\left(1+y^2\right)}y \, dy & y\to\frac1y \text{ for }y>1 \\ &= \sum_{n\ge0} \frac{(-1)^{n+1}}{(2n+1)^3} - \frac\pi8 \sum_{n\ge1} \frac{(-1)^{n+1}}{n^2}\\ &= \frac\pi8 \sum_{n\ge1} \frac{(-1)^n}{n^2} - \Im \sum_{n\ge1} \frac{i^n}{n^3}\\ &= \frac\pi8 \operatorname{Li}_2(-1) - \Im \operatorname{Li}_3(i) \\ &= \frac\pi8 \left(-\frac{\pi^2}{12}\right) + \Im \frac{3\zeta(3)-i\pi^3}{32} = \boxed{-\frac{\pi^3}{24}} \end{align*}$$