Consider the integrals of the form $$I_n=\int_0^1 \arcsin(\sqrt{1-x^n})\mathrm dx$$
With the help of calculators , I noticed:
$$I_{3}=\frac{3\Gamma{\left(\frac{11}{6}\right)}}{5\Gamma{\left(\frac{4}{3}\right)}}\sqrt{π}$$
$$I_{5}=\frac{5\Gamma{\left(\frac{17}{10}\right)}}{7\Gamma{\left(\frac{6}{5}\right)}}\sqrt{π}$$
$$I_{7}=\frac{7\Gamma{\left(\frac{23}{14}\right)}}{9\Gamma{\left(\frac{8}{7}\right)}}\sqrt{π}$$
$$I_{13}=\frac{13\Gamma{\left(\frac{41}{26}\right)}}{15\Gamma{\left(\frac{14}{13}\right)}}\sqrt{π}$$
So , based on the pattern, I thought that if $n$ is a positive integer greater than $3$, then :
$$I_{n}=\frac{n\Gamma{\left(\frac{3n+2}{2n}\right)}}{(n+2)\Gamma{\left(\frac{n+1}{n}\right)}}\sqrt{π}$$
Question:
How can we prove the above generalisation ? Can we extend it for even larger set of numbers ?
Best Answer
One integration by parts leads to $$\int\sin ^{-1}\left(\sqrt{1-x^n}\right)\,dx=x\sin ^{-1}\left(\sqrt{1-x^n}\right)+$$ $$\frac{n }{n+2}\,x \sqrt{x^n}\, _2F_1\left(\frac{1}{2},\frac{1}{n}+\frac{1}{2};\frac{1}{n}+\frac {3}{2};x^n\right)$$
$$I_n=\int_0^1\sin ^{-1}\left(\sqrt{1-x^n}\right)\,dx=\frac{\sqrt{\pi } \,\,\Gamma \left(\frac{1}{2}+\frac{1}{n}\right)}{2\, \Gamma \left(1+\frac{1}{n}\right)}$$ For large $n$ $$I_n=\frac \pi 2\left(1-\frac{2\log (2)}{n}+\frac{\pi ^2+12 \log ^2(2)}{6 n^2}+O\left(\frac{1}{n^3}\right)\right)$$
$n$ does not need to be an integer.