Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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}

share|cite|improve this question
    
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 '14 at 17:33

EDIT: I changed the second half of my answer.

$$ \begin{align} \int_{0}^{1} \frac{\arcsin^{3} (x)}{x^{3}} \, dx &= \int_{0}^{\pi /2} \frac{t^{3}}{\sin^{3} (t)} \, \cos (t) \, dt \\ &= t^{3} \left(-\frac{1}{2 \sin^{2}(t)} \right)\Bigg|^{\pi /2}_{0} + \frac{3}{2} \int_{0}^{\pi /2} \frac{t^{2}}{\sin^{2} (t)} \, dt \\ &=- \frac{\pi^{3}}{16} + \frac{3}{2} \int_{0}^{\pi /2} \frac{t^{2}}{\sin^{2} (t)} \, dt \\ &= - \frac{\pi^{3}}{16} + \frac{3 \pi}{2} \int_{0}^{\infty} \frac{t}{\cosh^{2}(t)} \, dt \tag{1}\\&= - \frac{\pi^{3}}{16} + \frac{3\pi}{2} \lim_{b \to \infty} \left(t \tanh (t) \Big|_{0}^{b} - \int_{0}^{b} \tanh (t) \, dt \right) \\&= - \frac{\pi^{3}}{16} + \frac{3 \pi}{2} \lim_{b \to \infty} \left(b \tanh (b) - \ln (\cosh b) \right) \\&= - \frac{\pi^{3}}{16} + \frac{3 \pi}{2} \lim_{b \to \infty} \ln \left(\frac{2 \, e^{b \tanh (b)}}{e^{b} + e^{-b}} \right) \\ &= - \frac{\pi^{3}}{16} + \frac{3 \pi}{2} \, \ln (2) \tag{2} \\ &= \frac{\pi}{2} \left(3 \ln (2) - \frac{\pi^{2}}{8} \right)\end{align}$$


$(1)$ Integrate the function $ \frac{z^{2}}{\sin^{2}(z)}$ around a rectangular contour with vertices at $0$, $\pi/2$, $\pi/2 + iR$, and $iR$. The value of the integral is purely imaginary on the left side of the rectangle. And as $R \to \infty$, the integral vanishes along the top of the rectangle since $\left|\frac{1}{\sin^{2}(z)} \right|$ decays exponentially as $\text{Im}(z) \to \infty$.

$(2)$ $\tanh (b) \to 1$ as $b \to \infty$

share|cite|improve this answer
    
+1 Cool, now I can upvote ;-p – achille hui Jun 4 '14 at 15:48
    
@achillehui Thanks for pointing out the mistake. And thanks for the upvote. – Random Variable Jun 4 '14 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 '14 at 16:02
    
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 '14 at 16:32
    
Syntax error in your notation - unmatched single pipe $|$ – Tomas Jun 5 '14 at 10:20

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

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

share|cite|improve this answer
    
Very nice answer. +1 I like how you changed the contour. – Random Variable Jun 4 '14 at 17:00
    
@RandomVariable Thanks. I like contour integrals a lot. – Felix Marin Jun 4 '14 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*}

share|cite|improve this answer
    
Nice work on $J_{4} - J_{6}$. – Leucippus Jun 5 '14 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 '14 at 14:33

$$ \begin{align} &\int_0^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{1}\\ &=2\int_0^{\pi/4}\big(\log(2)+\log(\sin(t))+\log(\cos(t))\big)\,\mathrm{d}t\tag{2}\\ &=\frac\pi2\log(2)+2\int_0^{\pi/4}\log(\sin(t))\,\mathrm{d}t+2\int_{\pi/4}^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{3}\\ &=\frac\pi2\log(2)+2\int_0^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{4}\\ &=-\frac\pi2\log(2)\tag{5} \end{align} $$ Explanation:
$(2)$: substitute $t\mapsto2t$ and $\sin(2t)=2\sin(t)\cos(t)$
$(3)$: substitute $t\mapsto\pi/2-t$ in the integral of $\log(\cos(t))$
$(4)$: combine domains of integration
$(5)$: subtract $(4)$ from $2$ times $(1)$ $$ \begin{align} \int_0^1\left(\frac{\sin^{-1}(x)}{x}\right)^3\,\mathrm{d}x &=\int_0^{\pi/2}\left(\frac{t}{\sin(t)}\right)^3\,\mathrm{d}\sin(t)\tag{6}\\ &=-\frac12\int_0^{\pi/2}t^3\,\mathrm{d}\frac1{\sin^2(t)}\tag{7}\\ &=-\frac12\frac{\pi^3}8+\frac32\int_0^{\pi/2}\left(\frac{t}{\sin(t)}\right)^2\,\mathrm{d}t\tag{8}\\ &=-\frac12\frac{\pi^3}8-\frac32\int_0^{\pi/2}t^2\,\mathrm{d}\cot(t)\tag{9}\\ &=-\frac12\frac{\pi^3}8+3\int_0^{\pi/2}t\cot(t)\,\mathrm{d}t\tag{10}\\ &=-\frac12\frac{\pi^3}8+3\int_0^{\pi/2}t\,\mathrm{d}\log(\sin(t))\tag{11}\\ &=-\frac12\frac{\pi^3}8-3\int_0^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{12}\\ &=\bbox[5px,border:2px solid #C0A000]{-\frac12\frac{\pi^3}8+\frac{3\pi}2\log(2)}\tag{13} \end{align} $$ Explanation:
$\phantom{1}(6)$: $x=\sin(t)$
$\phantom{1}(7)$: prepare for integration by parts
$\phantom{1}(8)$: integrate by parts
$\phantom{1}(9)$: prepare for integration by parts
$(10)$: integrate by parts
$(11)$: prepare for integration by parts
$(12)$: integrate by parts
$(13)$: apply $(5)$ $$ \begin{align} \int_0^1\left(\frac{\sin^{-1}(x)}{x}\right)^2\,\mathrm{d}x &=\int_0^{\pi/2}\left(\frac{t}{\sin(t)}\right)^2\,\mathrm{d}\sin(t)\tag{14}\\ &=-\int_0^{\pi/2}t^2\,\mathrm{d}\frac1{\sin(t)}\tag{15}\\ &=-\frac{\pi^2}4+2\int_0^{\pi/2}\frac{t}{\sin(t)}\,\mathrm{d}t\tag{16}\\ &=-\frac{\pi^2}4-2\int_0^{\pi/2}t\,\mathrm{d}\log(\csc(x)+\cot(x))\tag{17}\\ &=-\frac{\pi^2}4+2\int_0^{\pi/2}\log(\csc(x)+\cot(x))\,\mathrm{d}t\tag{18}\\ &=-\frac{\pi^2}4+2\int_0^{\pi/2}\big(\log(1+\cos(x))-\log(\sin(x))\big)\,\mathrm{d}t\tag{19}\\ &=-\frac{\pi^2}4+2\left(2\mathrm{G}-\frac\pi2\log(2)+\frac\pi2\log(2)\right)\tag{20}\\ &=\bbox[5px,border:2px solid #C0A000]{-\frac{\pi^2}4+4\mathrm{G}}\tag{21} \end{align} $$ Explanation:
$(14)$: $x=\sin(t)$
$(15)$: prepare for integration by parts
$(16)$: integrate by parts
$(17)$: prepare for integration by parts
$(18)$: integrate by parts
$(19)$: $\csc(x)+\cot(x)=\frac{1+\cos(x)}{\sin(x)}$
$(20)$: apply $(8)$ from this answer and $(5)$ from above
$(21)$: simplify

where $\mathrm{G}$ is Catalan's constant. $$ \begin{align} \int_0^1\left(\frac{\sin^{-1}(x)}{x}\right)\mathrm{d}x &=\int_0^{\pi/2}\left(\frac{t}{\sin(t)}\right)\mathrm{d}\sin(t)\tag{22}\\ &=\int_0^{\pi/2}t\,\mathrm{d}\log(\sin(t))\tag{23}\\ &=-\int_0^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{24}\\ &=\bbox[5px,border:2px solid #C0A000]{\frac\pi2\log(2)}\tag{25} \end{align} $$ Explanation:
$(22)$: $x=\sin(t)$
$(23)$: prepare for integration by parts
$(24)$: integrate by parts
$(25)$: apply $(5)$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.