Prove $\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2$

A real-analytic solution. Through the substitution $x=\sin\theta$ and integration by parts, our integral becomes

$$ I = \int_{0}^{\pi/2}\theta\cot(\theta)\,d\theta = -\int_{0}^{\pi/2}\log\sin(\theta)\,d\theta \tag{1}$$ and since $$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right) = \frac{2n}{2^n}\tag{2} $$ is a well-known identity, by Riemann sums $$ \int_{0}^{\pi/2}\log\sin(\theta)\,d\theta = \frac{\pi}{2}\lim_{n\to +\infty}\frac{1}{n}\log\left(\frac{2n}{2^n}\right) = -\frac{\pi}{2}\log(2).\tag{3}$$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{\arcsin\pars{x} \over x}\,\dd x & \,\,\,\stackrel{\mbox{i.b.p.}}{=}\,\,\, -\int_{0}^{1}{\ln\pars{x} \over \root{1 - x^{2}}}\,\dd x \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, -\,{1 \over 4}\int_{0}^{1}{x^{-1/2}\ln\pars{x} \over \root{1 - x}}\,\dd x \\[5mm] & = \left.-\,{1 \over 4}\,\partiald{}{\mu}\int_{0}^{1}x^{\mu}\pars{1 - x}^{-1/2}\,\dd x\,\right\vert_{\ \mu\ =\ -1/2} \\[5mm] & = \left.-\,{1 \over 4}\,\partiald{}{\mu}\bracks{\Gamma\pars{\mu + 1}\Gamma\pars{1/2} \over \Gamma\pars{\mu + 3/2}}\,\right\vert_{\ \mu\ =\ -1/2} \\[5mm] & = -\,{1 \over 4}\,\root{\pi}\, {\Gamma'\pars{1/2}\Gamma\pars{1} - \Gamma'\pars{1}\Gamma\pars{1/2} \over \Gamma^{2}\pars{1}} \\[5mm] & = -\,{1 \over 4}\,\root{\pi}\,\bracks{% \Gamma\pars{1 \over 2}\Psi\pars{1 \over 2} - \Gamma\pars{1}\Psi\pars{1}\Gamma\pars{1 \over 2}} \\[5mm] & = -\,{1 \over 4}\,\pi\bracks{\Psi\pars{1 \over 2} + \gamma} = -\,{1 \over 4}\,\pi\bracks{-2\ln\pars{2}} = \bbox[#ffe,10px,border:1px dotted navy]{\ds{{1 \over 2}\,\pi\ln\pars{2}}} \end{align}

Integration by parts reduces the integral to,

$$\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^2}} dx$$

And the substitution $x=\sin u$ reduces the integral to,

$$I=\int_{0}^{\frac{\pi}{2}} \ln (\sin u) du$$

And the substitution $v=\frac{\pi}{2}-x$ reduces the integral to,

$$I=\int_{0}^{\frac{\pi}{2}} \ln (\cos v) dv$$

$$I=\int_{0}^{\frac{\pi}{2}} \ln (\cos u) du$$

Now adding the integrals and noting properties of logarithms we have,

$$2I=\int_{0}^{\frac{\pi}{2}} \left( \ln (2 \sin x \cos x)-\ln 2\right) dx$$

Double angle,

$$2I=\int_{0}^{\frac{\pi}{2}} \ln (\sin 2x) dx -\frac{\pi}{2} \ln 2$$

The substitution $s=2x$ gives $$2I=\frac{1}{2}\int_{0}^{\pi} \ln (\sin s) ds -\frac{\pi}{2} \ln 2$$

But

$$\int_{0}^{\pi} \ln (\sin s) ds=2I$$

Follows from the substitution $w=\frac{\pi}{2}-s$ and the evenness of the function $f(w)=\ln (\cos w)$:

$$\int_{0}^{\pi} \ln (\sin s) ds$$ $$=-\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \ln (\cos w) dw$$

$$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln (\cos w)dw $$ $$=2 \int_{0}^{\frac{\pi}{2}} \ln (\cos w) dw=2I$$

So,

$$2I=I-\frac{\pi}{2}\ln 2$$

$$I=-\frac{\pi}{2}\ln 2$$