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

Question

I stumbled upon the interesting definite integral \begin{equation} \int\limits_0^1 \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2 \end{equation}

Here is my proof of this result.

Let $u=\sin^{-1}(x)$ then integrate by parts, \begin{align} \int \frac{\sin^{-1}(x)}{x} dx &= \int u \cot(u) du \\ &= u \ln\sin(u) - \int \ln\sin(u) du \tag{1} \label{eq:20161030-1} \end{align}

\begin{align} \int \ln\sin(u) du &= \int \ln\left(\frac{\mathrm{e}^{iu} - \mathrm{e}^{-iu}}{i2} \right) du \\ &= \int \ln\left(\mathrm{e}^{iu} - \mathrm{e}^{-iu} \right) du \,- \int \ln(i2) du \\ &= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \int \ln\mathrm{e}^{iu} du \,-\, u\ln(i2) \\ &= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \frac{i}{2}u^{2} -u\ln2 \,-\, ui\frac{\pi}{2} \tag{2} \label{eq:20161030-2} \end{align}

To evaluate the integral above, let $y=\mathrm{e}^{-i2u}$ \begin{equation} \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du = \frac{i}{2} \int \frac{\ln(1-y)}{y} dy = -\frac{i}{2} \operatorname{Li}_{2}(y) = -\frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2u} \tag{3} \label{eq:20161030-3} \end{equation}

Now we substitute equation \eqref{eq:20161030-3} into equation \eqref{eq:20161030-2}, then substitute that result into equation \eqref{eq:20161030-1}, switch variables back to (x), and apply limits, \begin{align} \int\limits_{0}^{1} \frac{\sin^{-1}(x)}{x} dx &= \sin^{-1}(x)\ln(x) + \sin^{-1}(x)\left(\ln2 + i\frac{\pi}{2}\right) \\ &- \frac{i}{2}[\sin^{-1}(x)]^{2} + \frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2\sin^{-1}(x)} \Big|_0^1 \\ &= \frac{\pi}{2}\ln2 \end{align}

I would be interested in seeing other solutions.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Answer 2

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 $$ \int_{0}^{\pi/2}\log\sin(\theta)\,d\theta = \frac{\pi}{2}\lim_{n\to +\infty}\frac{1}{n}\log\left(\frac{2n}{2^n}\right) = -\frac{\pi}{2}\log(2).\tag{3}$$

Answer 3

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,

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

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

Answer 4

\begin{array}{r} \displaystyle \int_{0}^{1} \frac{\arcsin x}{x} d x=& \int_{0}^{1} \arcsin x d(\ln x) \stackrel{IBP}{=} \displaystyle -\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^{2}}} d x \end{array} Let $\displaystyle I(a)=\int_{0}^{1} \frac{x^{a}}{\sqrt{1-x^{2}}} d x$ and $x=\sin \theta$, then our integral becomes $$ \begin{aligned} I(a) &=\int_{0}^{\frac{\pi}{2}} \sin ^{2\left(\frac{a+1}{2}\right)-1} \theta \cos ^{2\left(\frac{1}{2}\right)-1} \theta d \theta \\ &=\frac{1}{2} B\left(\frac{a+1}{2}, \frac{1}{2}\right) \\ &=\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{a+1}{2}\right)}{\Gamma\left(\frac{a}{2}+1\right)} \end{aligned} $$ Using logarithmic differentiation yields $$ \frac{I^{\prime}(a)}{I(a)}=\frac{1}{2}\left[\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a}{2}+1\right)\right] $$ Putting $a=0$ gives $$ \begin{aligned} I^{\prime}(0) &=\frac{I(0)}{2} \cdot\left(\psi\left(\frac{1}{2}\right)-\psi(1)\right)=\frac{\pi}{4}(-\ln 4)=-\frac{\pi}{2} \ln 2 \end{aligned} $$ Now we can conclude that $$\displaystyle \int_{0}^{1} \frac{\arcsin x}{x} dx = -\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^{2}}} dx =-I^{\prime}(0)= \frac{\pi}{2} \ln 2$$

Answer 5

Transform our integral by letting $\theta=\arcsin x$, then $$ \begin{aligned} \int_{0}^{1} \frac{\arcsin x}{x} d x &=\int_{0}^{\frac{\pi}{2}} \frac{\theta \cos \theta d \theta}{\sin \theta} \\ &=\int_{0}^{\frac{\pi}{2}} \theta d(\ln (\sin \theta)) \\ &=-\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d \theta \end{aligned} $$

By my post, $\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d \theta=-\frac{\pi}{2}\ln 2$, we can now conclude that $$\boxed{\int_{0}^{1} \frac{\arcsin x}{x} d x= \frac{\pi}{2}\ln 2}$$

Answer 6

$$\int_{0}^{1} \frac{\sin^{-1}x}{x} dx = \int_{0}^{1} \int_{0}^{1} \frac{\sqrt{1-x^2}}{1-x^2+x^2y^2} dy\ dx \overset{x=\sin t} =\int_0^1 \frac{\pi}{2(1+y)}dy=\frac{\pi}{2}\ln2 $$