Integral $\int_0^1\frac{\arcsin^3 x}{x^2}\text{d}x=6\pi G-\frac{\pi^3}{8}-\frac{21}{2}\zeta(3)$

Question

Show that: $$\int_0^1\frac{\arcsin^3 x}{x^2}\text{d}x=6\pi G-\frac{\pi^3}{8}-\frac{21}{2}\zeta(3)$$ I evaluated this by some Fourier series. Is there any other method? Start with substitution of $$u=\arcsin x$$ Then we have to integrate $$\int_0^{\frac{\pi}{2}}\frac{u^3\cos u}{\sin^2 u}\text{d}u=-\int_0^{\frac{\pi}{2}}u^3\csc u\text{d}u$$ Since $$\int\csc u\text{d}u=\ln (\csc u-\cot u)=\ln \left(\frac{1-\cos x}{\sin x}\right)=\ln 2+2\ln \left(\sin \frac{x}{2}\right)-\ln \sin x$$ Thus $$\int_0^{\frac{\pi}{2}}u^2\csc u\text{d}u=\int_0^{\frac{\pi}{2}}u^2\text{d}\left(2\ln \frac{\sin u}{2}-\ln \sin u\right)$$ $$=-\frac{\pi^2}{4}\ln 2-2\int_0^{\frac{\pi}{2}}u\left(2\ln \sin \frac{u}{2}-\ln \sin u\right)$$ $$=-\frac{\pi^2}{4}\ln 2-4\int_0^{\frac{\pi}{2}}u\ln \sin \frac{u}{2}\text{d}u+2\int_0^{\frac{\pi}{2}}u\ln \sin u\text{d}u$$ $$=-\frac{\pi^2}{4}\ln 2+4\int_0^{\frac{\pi}{2}}u\left[\ln 2+\sum_{n=1}^{\infty}\frac{\cos nu}{n}\right]\text{d}u-\int_0^{\frac{\pi}{2}}u^2\cot u\text{d}u$$ $$=\frac{\pi^2}{4}\ln 2+4\sum_{n=1}^{\infty}\frac{1}{n}\int_0^{\frac{\pi}{2}}u\cos nu\text{d}u-\frac{\pi^2}{4}\ln 2+\frac78\zeta(3)$$ $$=\frac78\zeta(3)+2\sum_{n=1}^{\infty}\frac{-2+2\cos \frac{n\pi}{2}+n\pi \sin \frac{n\pi}{2}}{n^3}$$ $$=\frac78\zeta(3)-4\zeta(3)-\frac38\zeta(3)+2\pi G=2\pi G-\frac72\zeta(3)$$ Combine these gives $$\int_0^1\frac{\arcsin ^3x}{x^2}\text{d}x=6\pi G-\frac{\pi^3}{8}-\frac{21}{2}\zeta(3)$$

Answer 1

Integrating by parts twice, we get $$ \int_0^{\pi/2}x^2\,e^{ikx}\,\mathrm{d}x =i^{k-1}\frac{\pi^2}{4k}+i^k\frac\pi{k^2}+\frac2{k^3}\left(i^{k+1}-i\right) $$ Therefore, using $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$, $$ \begin{align} &\int_0^1\frac{\arcsin^3(x)}{x^2}\mathrm{d}x\\ &=\int_0^{\pi/2}\frac{x^3}{\sin^2(x)}\mathrm{d}\sin(x)\\ &=-x^3\csc(x)\Big]_0^{\pi/2}+3\int_0^{\pi/2}\csc(x)\,x^2\,\mathrm{d}x\\ &=-\frac{\pi^3}{8}+3\int_0^{\pi/2}x^2\,\frac{2ie^{-ix}\,\mathrm{d}x}{1-e^{-2ix}}\tag{$\lozenge$}\\ &=-\frac{\pi^3}{8}+6i\sum_{k=0}^\infty\int_0^{\pi/2}x^2\,e^{-(2k+1)ix}\,\mathrm{d}x\\ &=-\frac{\pi^3}{8}+6i\sum_{k=0}^\infty\left((-1)^k\frac{\pi^2}{8k+4}-i(-1)^k\frac\pi{(2k+1)^2}-\left((-1)^k-i\right)\frac2{(2k+1)^3}\right)\\ &=-\frac{\pi^3}{8}+6\sum_{k=0}^\infty\left(\frac{(-1)^k\pi}{(2k+1)^2}-\frac2{(2k+1)^3}\right)\tag{$\ast$}\\ &=-\frac{\pi^3}{8}-\frac{21}{2}\zeta(3)+6\pi\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\\ &=-\frac{\pi^3}{8}-\frac{21}{2}\zeta(3)+6\pi G \end{align} $$ where $G$ is Catalan's Constant. In $(\ast)$, we drop the imaginary part (which should be $0$).

There is a question about the convergence in $(\lozenge)$. To handle this, we can consider $$ x^2\frac{2ie^{-ix}}{1-e^{-2ix}}=\lim_{r\to1^-}x^2\frac{2ire^{-ix}}{1-r^2e^{-2ix}} $$ and the convergence in the sums is uniform as $r\to1^-$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\int_{0}^{1}{\arcsin^{3}\pars{x} \over x^{2}}\,\dd x} \\[2mm] & \sr{\arcsin\pars{x}\ \mapsto\ x}{=} \int_{0}^{\pi/2}{x^{3} \over \sin^{2}\pars{x}}\cos\pars{x}\dd x \\[5mm] \sr{\rm IBP}{=} & -\,{\pi^{3} \over 8} + 3\color{red}{\int_{0}^{\pi/2}{x^{2} \over \sin\pars{x}}\dd x}. \label{1}\tag{1} \\[5mm] & \mbox{Hereafter, I'll perform an integration along a quarter} \\ & \mbox{circle in the first quadrant. The}\ \color{red}{\tt \large red}\ \mbox{integral}\ \\ & \mbox{in}\ \pars{\ref{1}}\ \mbox{is the} \mbox{ "piece" along the arc. Namely,} \\[2mm] & \color{red}{\int_{0}^{\pi/2}{x^{2} \over \sin\pars{x}}\dd x} \\[5mm] = & \ \left.\Re\int_{x\ =\ 0}^{x\ =\ \pi/2} {\bracks{-\ic\ln\pars{z}}^{\,2} \over \pars{z^{2} - 1}/\pars{2\ic z}} {\dd z \over \ic z}\right\vert_{\,z\ =\ \exp\pars{\ic x}} \\[5mm] = & \ \left.2\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2} {\ln^{2}\pars{z}\over 1 - z^{2}}\,\dd z\right\vert_{\,z\ =\ \exp\pars{\ic x}} \\[5mm] = & \ -2\,\Re\int_{1}^{0} {\ln^{2}\pars{y} + \pi\ln\pars{y}\ic - \pi^{2}/4\, \over 1 + y^{2}}\,\ic\dd y \\[2mm] & -2\int_{0}^{1} {\ln^{2}\pars{x} \over 1 - x^{2}}\dd x \\[5mm] = & \ -2\pi\ \overbrace{\int_{0}^{1} {\ln\pars{y} \over 1 + y^{2}}\dd y}^{\ds{-G}}\ \\[2mm] & -\ 2\ \overbrace{\int_{0}^{1} {\ln^{2}\pars{x} \over 1 - x^{2}}\dd x}^{\ds{7\zeta\pars{3}/4}} = \color{red}{2\pi G - {7\zeta\pars{3} \over 2}}\label{2}\tag{2} \\[5mm] & \mbox{Therefore,}\ \pars{\ref{1}}\ \mbox{and}\ \pars{\ref{2}} \implies \\ & \color{#44f}{\int_{0}^{1}{\arcsin^{3}\pars{x} \over x^{2}}\,\dd x} \\[2mm] = & \ \bbx{\color{#44f}{6\pi G -{\pi^{3} \over 8} - {21\zeta\pars{3} \over 2}}} \approx 0.7682 \end{align}

Answer 3

Here's a suggestion that might be fruitful. It's not a full solution but it seems too messy to put into the comments. If somebody else completes the solution, feel free to add it to this.

You can use the substitution $\sin{t} = x$ to convert the integral to

$$ \int_0^{\pi/2} t^3 \csc{t} \cot{t} \, dt$$ and by applying integration by parts yields $$ \left[ - t^3 \csc{t} \right]_0^{\pi/2} + \int_0^{\pi/2} 3t^2 \csc{t} \, dt = - \frac{\pi^2}{8} + \int_0^{\pi/2} 3t^2 \csc{t} \, dt$$ This produces one of the three special terms that you want. From here, I am not sure. We could calculate the power series of $\csc{t}$, from which the remaining integral can be readily expressed as an infinite series. Mathematica tells me that $$ \int_0^{\pi/2} \log| \csc{t} + \cot{t} | \, dt = 2G,$$ and that integral can fall out by integration by parts once more, but that seems a little too difficult to form a power series from.

Answer 4

Letting $\arcsin x \mapsto x$ yields $$ \begin{aligned} & I=\int_0^{\frac{\pi}{2}} x^3 \cot x \csc x d x=-\int_0^{\frac{\pi}{2}} x^3 d(\csc x) \\ & = \underbrace{ -\left[x^3 \csc x\right]_0^{\frac{\pi}{2}}}_{-\frac{\pi^3}{8}} +3 \int_0^{\frac{\pi}{2}} x^2 \csc x d x \end{aligned} $$ Using the identity:$e^{ix}=\cos x+i\sin x$ to expand the integrand into a power series, we have

$$ \begin{aligned} \int_0^{\frac{\pi}{2}} x^2 \csc x d x &=2 i \int_0^{\frac{\pi}{2}} \frac{x^2 e^{-x i}}{1-e^{-2 x i}} d x \\ &=-2 \sum_{n=0}^{\infty} \Im \int_0^{\frac{\pi}{2}} x^2 e^{-(2 n+1) x i} d x \\ &=2 \sum_{n=0}^{\infty} \int_0^{\frac{\pi}{2}} x^2 \sin (2 n+1) x d x \end{aligned} $$

Twice integration by parts with the last integral gives $$ \begin{aligned} & =-\frac{1}{2 n+1} \int_0^{\frac{\pi}{2}} x^2 d(\cos (2 n+1) x) \\ & =-\frac{1}{2 n+1}\left(-\frac{2}{2 n+1} \int_0^{\frac{\pi}{2}} x \cos (2 n+1) d x\right) \\ & =\frac{2}{(2 n+1)^2}\left([x \sin (2 n+1) x]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \sin (2 n+1) x d x\right) \\ & =\frac{2}{(2 n+1)^2}\left[\frac{\pi}{2}(-1)^n-\frac{1}{2 n+1}\right] \end{aligned} $$ Plugging back yields $$ \begin{aligned} I & =-\frac{\pi^3}{8}+6 \sum_{n=0}^{\infty} \frac{2}{(2 n+1)^2}\left[\frac{\pi}{2}(-1)^n-\frac{1}{2 n+1}\right] \\ & =-\frac{\pi^3}{8}+6 \pi \sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)^2}-12 \sum_{n=0}^{\infty} \frac{1}{(2 n+1)^2} \\ & =-\frac{\pi^3}{8}+6 \pi G-\frac{21}{2} \zeta(3) \end{aligned} $$

Answer 5

Letting $\arcsin x \mapsto x$ yields

\begin{aligned} & I=\int_0^{\frac{\pi}{2}} x^3 \cot x \csc x d x=-\int_0^{\frac{\pi}{2}} x^3 d(\csc x) \\ & = \underbrace{ -\left[x^3 \csc x\right]_0^{\frac{\pi}{2}}}_{-\frac{\pi^3}{8}} +3 \int_0^{\frac{\pi}{2}} x^2 \csc x d x \end{aligned}

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$$ \begin{aligned} \int_{0}^{\frac{\pi}{2}} x^{2}\csc x d x =& \int_{0}^{\frac{\pi}{2}} x^{2} d\left[\ln \left(\tan \frac{x}{2}\right)\right] \\ =& {\left[x^{2} \ln \left(\tan \frac{x}{2}\right)\right]_{0}^{\frac{\pi}{2}}-2 \int_{0}^{\frac{\pi}{2}} x \ln \left(\tan \frac{x}{2}\right) d x } \\ =&-8 \int_{0}^{\frac{\pi}{4}} y \ln (\tan y) d y \textrm{, where }y=2x\\ =&-8\left[-\frac{\pi}{4} G+\frac{7}{16} \zeta(3)\right] \\ =& 2 \pi G-\frac{7}{2} \zeta(3) \end{aligned} $$

where $ \int_{0}^{\frac{\pi}{4}} x \ln (\tan x) d x= -\frac{\pi}{4} G+\frac{7}{16} \zeta(3) $ from my post .

Now we can conclude that $$I= -\frac{\pi^3}{8}+6 \pi G-\frac{21}{2} \zeta(3)$$