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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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  • $\begingroup$ I have tried on several occations to find a general formula for this integral, everytime it confounds me! $J_4, J_5,\ldots$ anyone? $\endgroup$
    – gar
    Jun 4 '14 at 17:33
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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 $b \to \infty$

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  • $\begingroup$ @achillehui Thanks for pointing out the mistake. And thanks for the upvote. $\endgroup$ Jun 4 '14 at 15:52
  • $\begingroup$ 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 $\endgroup$ Jun 4 '14 at 16:02
  • $\begingroup$ 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! $\endgroup$ Jun 4 '14 at 16:32
  • $\begingroup$ Syntax error in your notation - unmatched single pipe $|$ $\endgroup$
    – Tomas
    Jun 5 '14 at 10:20
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{J_{3}\equiv\int_{0}^{1}\bracks{\arcsin\pars{x} \over x}^{3}\,\dd x ={\pi \over 2}\bracks{3\ln\pars{2} - {\pi^{2} \over 8}}:\ {\large ?}}$

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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  • $\begingroup$ Very nice answer. +1 I like how you changed the contour. $\endgroup$ Jun 4 '14 at 17:00
  • $\begingroup$ @RandomVariable Thanks. I like contour integrals a lot. $\endgroup$ Jun 4 '14 at 17:14
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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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  • $\begingroup$ Nice work on $J_{4} - J_{6}$. $\endgroup$
    – Leucippus
    Jun 5 '14 at 14:08
  • $\begingroup$ 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. $\endgroup$
    – gar
    Jun 5 '14 at 14:33
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$$ \begin{align} &\int_0^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{1}\\ &=2\int_0^{\pi/4}\big(\log(2)+\log(\sin(t))+\log(\cos(t))\big)\,\mathrm{d}t\tag{2}\\ &=\frac\pi2\log(2)+2\int_0^{\pi/4}\log(\sin(t))\,\mathrm{d}t+2\int_{\pi/4}^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{3}\\ &=\frac\pi2\log(2)+2\int_0^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{4}\\ &=-\frac\pi2\log(2)\tag{5} \end{align} $$ 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 $(5)$

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