Generalising $I_n=\int_0^1 \arcsin(\sqrt{1-x^n})\mathrm dx$.

Question

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 ?

Answer 1

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.

Answer 2

Substituting $$x = \sin^{2 / n} \theta, \qquad dx = \frac{\sin^{2 / n} \theta \cos \theta \,d\theta}{n \sin \theta}$$ transforms the integral to $$\frac{1}{n} \int_0^{\pi / 2} \underbrace{(\pi - 2 \theta)}_u \,\underbrace{\sin^{(2 - n) / n} \theta \, \cos \theta \,d\theta}_{dv} .$$ Applying integration by parts with $u, dv$ as indicated gives $$\int_0^{\pi / 2} \sin^{2 / n} \theta \,d\theta = \mathrm{B}\left(\frac12 + \frac1n, \frac12\right) = \frac{\Gamma\left(\frac12 + \frac1n\right) \Gamma\left(\frac12\right)}{\Gamma\left(1 + \frac1n\right)} = \boxed{\frac{\sqrt\pi \,\Gamma\left(\frac12 + \frac1n\right)}{\Gamma\left(1 + \frac1n\right)}} ,$$ where $\mathrm{B}$ is the Beta function. We can see that this lattermost expression is equivalent to yours by taking $z = \frac12 + \frac1n$ in the Gamma function identity $\Gamma(z + 1) = z \Gamma(z)$. (We assume $n > 0$; if $n = 0$ the original integrand is identically zero, and if $n < 0$ the original integral is not defined.)