# Do we have a general form for this integral?

Is there a general formula or recursion for this integral? $$\int_0^1\left(\frac{\arcsin x}{x}\right)^n\text{d}x,\ \ n\in\mathbb{N}$$

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Why the dirichlet-series tag? – Gerry Myerson Feb 14 '13 at 5:25
What is $\sin^{-1}x$, $\arcsin x$ or $\frac 1{\sin x}$ ? – Kaster Feb 14 '13 at 5:32
@Kaster, $\arcsin x$ – Ryan Feb 14 '13 at 5:39

The generating function would be $$g(t) = \sum_{n=0}^\infty \int_0^1 \left( \frac{\arcsin(x)}{x} t \right)^n \ dx = \int_0^1 \dfrac{x}{x - t\; \arcsin(x)}\ dt$$ but I doubt that this has a closed form. The first few terms $n = 0$ to $6$ are \eqalign{a_0 =&1\cr a_1 =& \frac12\,\pi \,\ln \left( 2 \right) \cr a_2 =& 4\,{\it Catalan}-\frac14\,{\pi }^{2}\cr a_3 = &-\frac1{16}\,\pi \, \left( {\pi }^{2}-24\,\ln \left( 2 \right) \right)\cr a_4 = &-\frac1{48}\,{\pi }^{4}-\frac12\,{\pi }^{2}+8\,{\it Catalan}+{\pi }^{2}{\it Catalan}-8\,{\it Im} \left( {\it polylog} \left( 4,i \right) \right) \cr a_5 =& -{\frac {1}{384}}\,\pi \, \left( 3\,{\pi }^{4}-160\,{\pi }^{2}\ln \left( 2 \right) +40\,{\pi }^{2}+720\,\zeta \left( 3 \right) -960\, \ln \left( 2 \right) \right)\cr a_6 =&{\frac {9}{32}}\,{\pi }^{4}{\it Catalan}-{\frac {1}{320}}\,{\pi }^{6}+15\,{\pi }^{2}{\it Catalan}-{ \frac {11}{64}}\,{\pi }^{4}+12\,{\it Catalan}-\frac34\,{\pi }^{2}\cr &-\frac32\,{ \it Im} \left( 9\,{\pi }^{2}{\it polylog} \left( 4,i \right) -72\,{ \it polylog} \left( 6,i \right) +80\,{\it polylog} \left( 4,i \right) \right) }