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It took into account the two previous remarks from the users who modified the page. I hope that this time I met their requirements. |
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and this completes the proof.
== A rigorous proof
First of all, <math>\displaystyle \int ^{\frac{\pi }{2}}_{0} \operatorname{Arcsin}( \sin\ t) \ dt\ =\int ^{\frac{\pi }{2}}_{0} t\ dt\
=\frac{\pi ^{2}}{8}
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<math>\begin{align}
& \int ^{\frac{\pi }{2}}_{0} \ (\sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \sin( t)^{2n+1}) dt\ & = & \ \ \sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \int ^{\frac{\pi }{2}}_{0} \sin( t)^{2n+1} dt\\
\\
& & = & \ \ \sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \frac{\left( n!\ 2^{n}\right)^{2}}{( 2n+1) !}\\
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