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While reviewing an old book the integral \begin{align} J_{3} = \int_{0}^{1} \left( \frac{\sin^{-1}(x)}{x} \right)^{3} \ dx = \frac{\pi}{2}\left( 3 \ln 2 - \frac{\pi^{2}}{8} \right). \end{align} was asked to be shown as an exercise. Is it possible to obtain a complete derivation of this result? As an extension of this first integral what are the values of the integrals \begin{align} J_{1} = \int_{0}^{1} \frac{\sin^{-1}(x)}{x} \ dx \end{align} and \begin{align} J_{2} = \int_{0}^{1} \left( \frac{\sin^{-1}(x)}{x} \right)^{2} \ dx \end{align}

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I have tried on several occations to find a general formula for this integral, everytime it confounds me! $J_4, J_5,\ldots$ anyone? –  gar Jun 4 at 17:33

3 Answers 3

First make the substitution $t = \arcsin x$.

Then

$$ \int_{0}^{1} \frac{\arcsin^{3} x}{x^{3}} \ dx = \int_{0}^{\pi /2} \frac{t^{3}}{\sin^{3} t} \ \cos t \ dt$$

Now integrate by parts by letting $u = t^{3}$ and $ \displaystyle dv = \frac{\cos t}{\sin^{3}t} \ dt$.

$$ = t^{3} \left(\frac{-1}{2 \sin^{2}t} \right)\Big|^{\pi /2}_{0} + \frac{3}{2} \int_{0}^{\pi /2} \frac{t^{2}}{\sin^{2} t} \ dt$$

$$ =- \frac{\pi^{3}}{16} + \frac{3}{2} \int_{0}^{\pi /2} \frac{t^{2}}{\sin^{2} t} \ dt $$

Integrate by parts again by letting $u = t^{2}$ and $\displaystyle dv = \frac{1}{\sin^{2} t} \ dt$.

$$ = - \frac{\pi^{3}}{16} + \frac{3}{2} \left(-t^{2} \cot t \Big|^{\pi/2}_{0} + 2 \int_{0}^{\pi/2} t \cot t \ dt \right)$$

$$ =- \frac{\pi^{3}}{16} + 3 \int_{0}^{\pi/2} t \cot t \ dt $$

Integrate by parts one more time.

$$ = - \frac{\pi^{3}}{16} + 3 \left( t \log(\sin t) \Big|^{\pi /2}_{0} - \int_{0}^{\pi /2} \log(\sin t) \ dt\right)$$

$$= - \frac{\pi^{3}}{16} - 3 \int_{0}^{\pi/2} \log(\sin t) \ dt $$

$$ = - \frac{\pi^{3}}{16} - 3 \left(- \frac{\pi \ln 2}{2} \right)$$

http://math.stackexchange.com/a/378785/16033

$$= - \frac{\pi^{3}}{16} + \frac{3 \pi}{2} \ln 2 $$

$$ = \frac{\pi}{2} \left(3 \ln 2 - \frac{\pi^{2}}{8} \right)$$

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+1 Cool, now I can upvote ;-p –  achille hui Jun 4 at 15:48
    
@achillehui Thanks for pointing out the mistake. And thanks for the upvote. –  Random Variable Jun 4 at 15:52
    
You are welcome. I think it will be useful to include some links to one of those questions that evaluate the integral $\log(\sin(x))$ like this and that –  achille hui Jun 4 at 16:02
    
@achillehui Usually I take too much time before I post my answer and someone posts the answer before me. This time I didn't take enough time and made a bunch of typos. A little embarrassing. I knew immediately how to approach the evaluation because I had come across this family of integrals years ago. –  Random Variable Jun 4 at 16:29
    
I have the same problem of taking too long to write an answer and making tons of typos. You at least beat me in answering this question this time! –  achille hui Jun 4 at 16:32

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{J_{3}\equiv\int_{0}^{1}\bracks{\arcsin\pars{x} \over x}^{3}\,\dd x ={\pi \over 2}\bracks{3\ln\pars{2} - {\pi^{2} \over 8}}:\ {\large ?}}$

Set $\ds{\quad x\equiv \sin\pars{t}\quad\imp\quad t=\arcsin\pars{x}}$: \begin{align} J_{3}&\equiv\int_{0}^{\pi/2}\,{t^{3} \over\sin^{3}\pars{t}}\, \bracks{\cos\pars{t}\,\dd t} =\left. -\,\half\,{t^{3} \over \sin^{2}\pars{t}}\right\vert_{0}^{\pi/2} +\int_{0}^{\pi/2}{1 \over 2\sin^{2}\pars{t}}\,3t^{2}\,\dd t \\[3mm]&=-\,\half\pars{\pi \over 2}^{3} +3\color{#c00000}{\int_{0}^{\pi/2}{t^{2}\,\dd t \over 1 - \cos\pars{2t}}} \tag{1} \end{align}

\begin{align}&\color{#c00000}{% \int_{0}^{\pi/2}{t^{2}\,\dd t \over 1 - \cos\pars{2t}}} ={1 \over 8}\int_{0}^{\pi}{t^{2}\,\dd t \over 1 - \cos\pars{t}} ={1 \over 16}\int_{-\pi}^{\pi}{t^{2}\,\dd t \over 1 - \cos\pars{t}} \\[3mm]&={1 \over 16} \int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ < \pi}} {-\ln^{2}\pars{z} \over 1 - \pars{z^{2} + 1}/\pars{2z}}\,{\dd z \over \ic z} =-\,{1 \over 8}\,\ic \int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ < \pi}} {\ln^{2}\pars{z}\,\dd z \over \pars{z - 1}^{2}} \\[3mm]&={1 \over 8}\,\ic\int_{-1}^{0} {\ln^{2}\pars{-x} + 2\pi\ic\ln\pars{-x} - \pi^{2} \over \pars{x - 1}^{2}}\,\dd x + {1 \over 8}\,\ic\int_{0}^{-1} {\ln^{2}\pars{-x} - 2\pi\ic\ln\pars{-x} - \pi^{2} \over \pars{x - 1}^{2}}\,\dd x \\[3mm]&=-\,{\pi \over 2}\int_{0}^{1} {\ln\pars{x}\,\dd x \over \pars{x + 1}^{2}} =-\,{\pi \over 2}\sum_{n = 1}^{\infty}\pars{-1}^{n}n \int_{0}^{1}\ln\pars{x}x^{n - 1}\,\dd x \\[3mm]&=-\,{\pi \over 2}\sum_{n = 1}^{\infty}\pars{-1}^{n}n \lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}x^{\mu + n - 1}\,\dd x ={\pi \over 2}\ \underbrace{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n}}_{\ds{=\ \ln\pars{2}}}\ =\ {\pi \over 2}\,\ln\pars{2} \end{align}

By replacing this result in $\pars{1}$: $$\color{#44f}{\large% J_{3}\equiv\int_{0}^{1}\bracks{\arcsin\pars{x} \over x}^{3}\,\dd x ={\pi \over 2}\bracks{3\ln\pars{2} - {\pi^{2} \over 8}}} $$

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Very nice answer. +1 I like how you changed the contour. –  Random Variable Jun 4 at 17:00
    
@RandomVariable Thanks. I like contour integrals a lot. –  Felix Marin Jun 4 at 17:14

This is for the extension of the question you asked:

\begin{align*} J_1 &= \int_{0}^{1} \, \frac{\arcsin{x}}{x} \, dx \\ &= \int_{0}^{\pi/2} \, \frac{t}{\sin{t}}\, \cos{t}\, dt \\ &= t\, \log{\sin{t}}\Big|_0^{\pi/2}-\int_{0}^{\pi/2} \, \log{\sin{t}}\, dt \\ &= 0 + \frac{\pi}{2}\log{2} \end{align*}

\begin{align*} J_2 &= \int_{0}^{1} \, \left(\frac{\arcsin{x}}{x}\right)^2 \, dx \\ &= \int_{0}^{\pi/2} \, \left(\frac{t}{\sin{t}}\right)^2\, \cos{t} \, dt \\ &= -\frac{t^2}{\sin{t}}\Big|_0^{\pi/2}+\int_{0}^{\pi/2} \, \frac{2\, t}{\sin{t}}\, dt \\ &= -\frac{\pi^2}{4}+4\, G \approx 1.19646127643654 \end{align*}

where $G$ is the catalan's constant

Update:

Using the results in generalized integral $I_n$,

\begin{align*} J_4 &= \frac{1}{16} \, \pi^{4} - \frac{1}{2} \, \pi^{2} + G {\left(\pi^{2} + 8\right)} - \frac{1}{96} \, \psi^{(3)}\left( \frac{1}{4}\right) \approx 1.49222813527376\\ J_5 &= -\frac{1}{128} \, \pi^{5} - \frac{5}{48} \, \pi^{3} + \frac{5}{12} \, {\left(6 \, \pi + \pi^{3}\right)} \log\left(2\right) - \frac{15}{8} \, \pi \zeta(3) \approx 1.69763017912507\\ J_7 &= -\frac{\pi^{7}}{768} - \frac{77 \, \pi^{5}}{1920} - \frac{7 \, \pi^{3}}{48} + \frac{7 \, \pi}{8} \log\left(16\right) + \frac{7}{60} \, {\left(\pi^{5} + 25 \, \pi^{3}\right)} \log\left(2\right) + \frac{105 \, \pi}{8}\, \zeta(5) - \frac{7}{8} \, {\left(15 \, \pi + 2 \, \pi^{3}\right)} \zeta(3) \approx 2.29253050578831 \end{align*}

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Nice work on $J_{4} - J_{6}$. –  Leucippus Jun 5 at 14:08
    
Thanks! But note that it's $J_7$, no known closed form yet for $J_6$. I could not find a general formula in any table of integrals. –  gar Jun 5 at 14:33

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