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

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    $\begingroup$ The Taylor series $$ \arcsin^3x=6\sum_{m=0}^\infty\frac{(2m+1)!!^2}{(2m+3)!}\sum_{k=0}^m\frac1{(2k+1)^2}x^{2m+3} $$ (see OEIS sequence A001824) might help. $\endgroup$ – joriki Feb 13 '13 at 9:54
  • $\begingroup$ Is this a homework or something just of interest? $\endgroup$ – Christopher A. Wong Feb 13 '13 at 10:19
  • $\begingroup$ @ChristopherA.Wong : Just for interest! I am reading "irresistible integrals". It has a similar one but simpler $\endgroup$ – Ryan Feb 13 '13 at 10:26
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    $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Feb 13 '13 at 14:48
  • $\begingroup$ @Ryan: Can you post your solution? $\endgroup$ – Mhenni Benghorbal Feb 13 '13 at 15:00
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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 $r\to1^-$.

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  • $\begingroup$ Nice solution! Can you have a look on my other question about $\int_0^{\frac{\pi}{2}}x^3\csc x\text{d}x$. That was the last part when I evaluating $\int_0^1\frac{\arcsin ^4x}{x^4}\text{d}x$ $\endgroup$ – Ryan Feb 14 '13 at 1:54
  • $\begingroup$ @Ryan: I posted an answer. $\endgroup$ – robjohn Feb 14 '13 at 5:51
  • $\begingroup$ Could you explain why we left imaginary parts apart? $\endgroup$ – bianco Apr 22 at 20:45
  • $\begingroup$ $\int_0^1\frac{\arcsin^3(x)}{x^2}\,\mathrm{d}x$ is real, so the complex part is $0$. We could check that by computing the imaginary part of the right hand side, and if zero, it would give us the identity $$\pi^2\sum_{k=0}^\infty\frac{(-1)^k}{8k+4 }=2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^3}$$ and indeed both are $\frac{\pi^3}{16}$. But since the imaginary part of the left side is $0$, we can just drop the imaginary part of the right side. $\endgroup$ – robjohn Apr 23 at 3:02
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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.

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