Question Number 222858 by MrGaster last updated on 09/Jul/25
$$\mathrm{Prove}: \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{arcsin}^{\mathrm{2}} {x}}{{x}}{dx}=\frac{\pi{i}}{\mathrm{6}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{36}}−\mathrm{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right)−\frac{\mathrm{1}}{\mathrm{3}}\zeta\left(\mathrm{3}\right) \\ $$
Commented by gabthemathguy25 last updated on 09/Jul/25
$$\mathrm{wait}\:\mathrm{im}\:\mathrm{still}\:\mathrm{answering}\:\mathrm{sir}. \\ $$
Answered by gabthemathguy25 last updated on 10/Jul/25
$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{arcsin}^{\mathrm{2}} {x}}{{x}}{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \frac{{t}^{\mathrm{2}} }{\mathrm{sin}\:{t}}{dt} \\ $$$$\Rightarrow\frac{\pi{i}}{\mathrm{2}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{18}}−\mathrm{Li}_{\mathrm{2}} \left({e}^{{i}\frac{\pi}{\mathrm{3}}} \right)\right)−\zeta\left(\mathrm{3}\right) \\ $$$${e}^{{i}\frac{\pi}{\mathrm{3}}} =\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)+{i}\mathrm{sin}\left(\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow\frac{\pi{i}}{\mathrm{2}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{18}}−\mathrm{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right)\right)\right)−\zeta\left(\mathrm{3}\right) \\ $$$$\left.\Rightarrow\frac{\pi{i}}{\mathrm{6}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{36}}−\mathrm{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right)−\frac{\mathrm{1}}{\mathrm{3}}\zeta\left(\mathrm{3}\right)\::\right) \\ $$