Compute $\int_0^1 \frac{\arcsin(x)}{x}dx$

$$\int_0^1 \frac{\arcsin(x)}{x}dx$$

This is a proposed for a Calculus II exam, and I have absolutely no idea how to solve it. Tried using Frullani or Lobachevsky integrals, or beta and gamma functions, but I can't even find a way to start it. Wolfram Alpha gives a kilometric solution, but I know that cannot be the only answer. Any help appreciated!

Solution 1:

Let $y=\arcsin x\;\Rightarrow\;\sin y =x\;\Rightarrow\;\cos y\ dy=dx$, then $$ \int_0^1 \frac{\arcsin(x)}{x}dx=\int_0^{\Large\frac\pi2}y\cot y\ dy. $$ Now use IBP by taking $u=y$ and $dv=\cot y\ dy\;\Rightarrow\;v=\ln(\sin x)$, then \begin{align} \int_0^{\Large\frac\pi2}y\ \cot y\ dy&=\left.y\ln(\sin y)\right|_0^{\Large\frac\pi2}-\int_0^{\Large\frac\pi2}\ln(\sin y)\ dy\\ &=-\int_0^{\Large\frac\pi2}\ln(\sin y)\ dy. \end{align} The last integral can be evaluated by using property $$ \int_a^b f(x)\ dx=\int_a^b f(a+b-x)\ dx. $$ We obtain $$ \int_0^{\Large\frac\pi2}\ln(\sin y)\ dy=-\frac\pi2\ln2, $$ where $$ \int_0^{\Large\frac\pi2}\ln(\sin y)\ dy=\int_0^{\Large\frac\pi2}\ln(\cos y)\ dy\quad\Rightarrow\quad\text{by symmetry}. $$ Thus $$ \int_0^1 \frac{\arcsin(x)}{x}dx=\large\color{blue}{\frac\pi2\ln2}. $$

Solution 2:

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\arcsin\pars{x} \over x}\,\dd x:\ {\large ?}}$

\begin{align}&\color{#66f}{\large\int_{0}^{1}{\arcsin\pars{x} \over x}\,\dd x} =-\int_{0}^{1}{\ln\pars{x} \over \root{1 - x^{2}}}\,\dd x =-{1 \over 4}\int_{0}^{1}x^{-1/2}\ln\pars{x}\pars{1 - x}^{-1/2}\,\dd x \\[3mm]&=-{1 \over 4}\lim_{\mu \to -1/2}\partiald{}{\mu} \int_{0}^{1}x^{\mu}\pars{1 - x}^{-1/2}\,\dd x =-{1 \over 4}\lim_{\mu \to -1/2}\partiald{}{\mu} \bracks{\Gamma\pars{\mu + 1}\Gamma\pars{1/2} \over \Gamma\pars{\mu + 3/2}} \\[3mm]&=-{1 \over 4}\,\Gamma\pars{\half}\braces{% {\Gamma\pars{1/2} \over \Gamma\pars{1}}\,\bracks{% \overbrace{\Psi\pars{\half}}^{\ds{-\gamma - 2\ln\pars{2}}}\ -\ \overbrace{\Psi\pars{1}}^{\ds{-\gamma}}}}\ =\ \half\,\Gamma^{2}\pars{\half}\ln\pars{2} \\[3mm]&=\color{#66f}{\large\half\,\pi\ln\pars{2}} \quad\mbox{with}\quad\Gamma\pars{\half} = \root{\pi}\quad\mbox{and}\quad\Gamma\pars{1} = 1. \end{align}