The improper integral $ \int_{0}^{1}\frac{\sqrt{x}}{\sqrt{1-x^6}} dx $

Question

I've ran into this improper integral:

$$ \int_{0}^{1}\frac{\sqrt{x}}{\sqrt{1-x^6}} dx $$

I've tried the substitution $t=\sqrt x$ without success.

Any help will be greatly appreciated.

Thank you, Yaron.

Answer 1

Substitute $u = x^6$, $x=u^{1/6}$, $dx = (1/6) u^{-5/6} du$. Then look up what a beta function is. http://en.wikipedia.org/wiki/Beta_function

I get after substitution

$$\frac16 \int_0^1 du \, u^{-5/6} u^{1/12}(1-u)^{-1/2}$$

which takes the value

$$\frac16 \frac{\Gamma\left( \frac14 \right ) \Gamma \left ( \frac12 \right)}{\Gamma\left ( \frac{3}{4} \right)} = \frac{\Gamma^2\left(\frac14\right)}{6 \sqrt{2\pi}}$$

Answer 2

Substitute $t=x^6$ and you get $$ \begin{align} \int_0^1\frac{\sqrt{x}}{\sqrt{1-x^6}}\,\mathrm{d}x &=\frac16\int_0^1t^{-3/4}(1-t)^{-1/2}\,\mathrm{d}t\\ &=\frac16\mathrm{B}\left(\frac14,\frac12\right)\tag{$\ast$}\\ &=\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac12\right)}{6\,\Gamma\left(\frac34\right)}\\ &=\frac{\sqrt\pi}{6}\frac{\Gamma\left(\frac14\right)}{\Gamma\left(\frac34\right)}\\ &=\frac{\Gamma\left(\frac14\right)^2}{6\sqrt{2\pi}} \end{align} $$ The last equation is due to the reflection formula: $\displaystyle\Gamma(x)\,\Gamma(1-x)=\frac\pi{\sin(\pi x)}$

$(\ast)$ follows from $\displaystyle\mathrm{B}(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}\,\mathrm{d}t=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$