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

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

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 $y=\mathrm{e}^{-i2u}$ $$\int \ln\left(1 - \mathrm{e}^{-i2u} \right) du = \frac{i}{2} \int \frac{\ln(1-y)}{y} dy = -\frac{i}{2} \operatorname{Li}_{2}(y) = -\frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2u} \tag{3} \label{eq:20161030-3}$$

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.

• When you type \mathrm{ln}2 instead of \ln2, you don't get proper spacing and you see $\mathrm{ln}2$ instead of $\ln2$. Notice this difference: $$a\ln b \qquad a\ln(b)$$ These are a\ln b and a\ln(b). You see that the spacing to the right of \ln depends on context. Similarly on the left, although this example doesn't show that. MathJax is based ultimately on the way TeX handles math notation, and TeX was created by Donald Knuth, not by illiterate cavemen who don't think about things like this when they write software. I edited accordingly. $\qquad$ – Michael Hardy Oct 31 '16 at 2:25
• math.stackexchange.com/questions/37829 – Aforest Oct 31 '16 at 2:29

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

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

• Wow, I have no idea how you pull beautiful solutions like this out so quickly! – user223391 Oct 31 '16 at 2:22
• @ZacharySelk: no, it should be correct. $I$ is the opposite of the integral appearing in the LHS of $(3)$ ($\log\sin\theta$ is clearly negative over the integration range). – Jack D'Aurizio Oct 31 '16 at 2:24
• Oops, I see that. Well done! – user223391 Oct 31 '16 at 2:25

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,

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

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

• There should be a negative in the begging. – Ahmed S. Attaalla Feb 12 '17 at 17:38