Calculus – Integral of Combination Log and Inverse Trig Function

Question

Does the following integral have a closed-form ?:

\begin{equation}
\int_{0}^{1}{\ln\left(\,x\,\right) \over 1 + x}\,\arccos\left(\,x\,\right)
\,{\rm d}x
\end{equation}

This integral has been posted in Integral and Series a week ago but it remains unsolved. So, I decide to post it here.

Could anyone here please help me to find the closed-form preferably with elementary ways ( high school methods ) ?.

Any help would be greatly appreciated. Thank you.

Answer 1

First, we set $x=\cos\theta\ \color{red}{\Rightarrow}\ dx=-\sin\theta\ d\theta$, then \begin{align} \mathcal{I}=\int_0^1\frac{\ln x}{1+x}\arccos(x)\ dx=-\int_0^{\Large\frac{\pi}2}\frac{\sin\theta}{1+\cos\theta}\cdot\theta\ \ln (\cos\theta)\ d\theta.\tag1 \end{align} Using the fact that \begin{align} \tan\left(\frac\theta2\right)=\frac{\sin\theta}{1+\cos\theta} \end{align} and setting $y=\frac\theta2$, then $(1)$ turns out to be \begin{align} \mathcal{I}=-\int_0^{\Large\frac{\pi}2}\theta\tan\left(\frac\theta2\right)\ln (\cos\theta)\ d\theta=-4\int_0^{\Large\frac{\pi}4}y\tan\left(y\right)\ln (\cos2y)\ dy.\tag2 \end{align} Now, setting $t=\tan y\ \color{red}{\Rightarrow}\ dy=\dfrac{dt}{1+t^2}$ yields \begin{align} \mathcal{I}&=-4\int_0^1 t\arctan(t)\ln \left(\frac{1-t^2}{1+t^2}\right)\ \frac{dt}{1+t^2}\\ &=4\int_0^1\left[ \frac{t\arctan(t)}{1+t^2}\ln \left(1+t^2\right)- \frac{t\arctan(t)}{1+t^2}\ln \left(1-t^2\right)\right]\ dt.\tag3 \end{align} Actually, the second integral in $(3)$ has been evaluated by Kirill and it is equal to \begin{align} \int_0^1 \frac{t\arctan(t)}{1+t^2}\ln \left(1-t^2\right)\ dt=-\frac{\pi^3}{48}-\frac{\pi}{8}\log^2 2 +\text{G}\log 2,\tag4 \end{align} where $\text{G}$ is is the Catalan's Constant. We may also refer to the following technique to evaluate \begin{align} \int_0^1 \frac{\ln \left(1+at\right)\ln \left(1+bt\right)}{1+ct}\ dt. \end{align} The first integral in $(3)$ can be evaluated by setting $t=\tan x$, we have \begin{align} \int_0^1 \frac{t\arctan(t)}{1+t^2}\ln \left(1+t^2\right)\ dt=-2\int_0^{\Large\frac{\pi}4}x\tan x\ln (\cos x)\ dx. \end{align} Now, applying IBP by setting $u=x$ and $$ v=\int\tan x\ln (\cos x)\ dx=-\ln^2(\cos x)\quad \color{red}{\Rightarrow}\quad \text{set}\ z= \cos x. $$ Hence \begin{align} \int_0^1 \frac{t\arctan(t)}{1+t^2}\ln \left(1+t^2\right)\ dt&=2x\ln^2(\cos x)\bigg|_{x=0}^{\Large\frac{\pi}4}-2\int_0^{\Large\frac{\pi}4}\ln^2 (\cos x)\ dx\\ &=\frac{\pi}8\ln^22-2\int_0^{\Large\frac{\pi}4}\ln^2 (\cos x)\ dx.\tag5 \end{align} Random Variable has successfully proven that $$ \int_0^{\Large\frac{\pi}4}\ln^2 (\sin x)\ dx=\frac{\pi^{3}}{192} + \frac{\text{G}}{2}\ln2 + \frac{3 \pi}{16} \ln^{2}2 - \Im \ \text{Li}_{3}(1-i)\tag6 $$ and Lucian has shown that $$ \int_0^{\Large\frac{\pi}4}\ln^2 (\sin x)\ dx+\int_0^{\Large\frac{\pi}4}\ln^2 (\cos x)\ dx=\frac{\pi^3}{24}+\dfrac\pi2\ln^22.\tag7 $$ Then by using $(3)$, $(4)$, $(5)$, $(6)$, and $(7)$, we finally obtain

\begin{align} \mathcal{I}=\int_0^1\frac{\ln x}{1+x}\arccos(x)\ dx=\color{purple}{2\text{G}\ln2+\frac\pi2\ln^22+\frac{\pi^3}{16}+4\,\Im\bigg[\operatorname{Li}_3(1-i)\bigg]}. \end{align}

The above result is exactly similar to Cleo's answer. It can be proven by using the following trilogarithm identity

$\displaystyle \operatorname{Li}_3(z)+\operatorname{Li}_3(1-z)+\operatorname{Li}_3\left(1-\frac1z\right)=\zeta(3)+\frac16\ln^3z+\zeta(2)\ln z-\frac12\ln^2z\ln(1-z). $$\tag8$

Setting $z=i$ to $(8)$, Wolfram Alpha shows that $\displaystyle\Im\bigg[\operatorname{Li}_3(1-i)\bigg]=-\Im\bigg[\operatorname{Li}_3(1+i)\bigg]$ since $$ \Im\bigg[\zeta(3)+\frac16\ln^3(i)+\zeta(2)\ln(i)-\frac12\ln^2(i)\ln(1-i)-\operatorname{Li}_3(i)\bigg]=0. $$ Thus, the another closed-form of $\mathcal{I}$ is

\begin{align} \mathcal{I}=\int_0^1\frac{\ln x}{1+x}\arccos(x)\ dx=\color{purple}{2\text{G}\ln2+\frac\pi2\ln^22+\frac{\pi^3}{16}-4\,\Im\bigg[\operatorname{Li}_3(1+i)\bigg]}. \end{align}