Show $\int\limits_0^1\sqrt{1-t^3}\,dt = \frac{\sqrt{3}\cdot\Gamma(\frac{1}{3} )^3}{ 10\pi\sqrt[3]{2}}$

Question

Show $\int\limits_0^1\sqrt{1-t^3}\,dt = \frac{\sqrt{3}\cdot\Gamma(\frac{1}{3} )^3}{ 10\pi\sqrt[3]{2}}$

I know that $\Gamma(1/6)=\frac{\sqrt{3}}{\sqrt[3]{2}\sqrt{\pi}}\cdot\Gamma(\frac{1}{3} ) ^2$ so :

$\frac{\sqrt{3}\cdot\Gamma(\frac{1}{3} )^3}{ 10\pi\sqrt[3]{2}} =\frac{\Gamma(\frac{1}{3} )\Gamma(\frac{1}{6} )}{ 10\sqrt{\pi}} = \frac{\Gamma(\frac{1}{3} )\Gamma(\frac{1}{6} )}{ 10\Gamma(\frac{1}{2})} = \beta(1/3,1/6)=\frac{1}{10}\int_0^1t^{1/3-1}(1-t)^{1/6-1}dt$

and this is as far as I can get, I can't figure out how to get from $\frac{1}{10t^{2/3}(1-t)^{5/6}}$ to $\sqrt{1-t^3}$

Answer 1

The Beta function is defined as:

$B(x,y) = ∫_0^1 t^{x-1} (1-t)^{y-1} dt$

Let $k=t^{\frac{1}{3}}$

Then $3k^2 dk = dt$

$B(x,y) = ∫_0^1 k^{3(x-1)} (1-k^3)^{y-1} 3k^2 dk$

$B(x,y) = 3∫_0^1 k^{3x-1} (1-k^3)^{y-1} dk$

We need $3x-1=0$, so $x=\frac{1}{3}$. And similarly we need $y-1=\frac{1}{2}$, so $y=\frac{3}{2}$.

So we need to evaluate $\frac{1}{3} B(\frac{1}{3}, \frac{3}{2})$.

Using the common identity, we find that this is equivalent to:

$$\frac{1}{3} \frac{\Gamma(\frac{1}{3})+\Gamma(\frac{3}{2})}{\Gamma(\frac{1}{3}+\frac{3}{2})}$$

$$\frac{1}{3} \frac{\Gamma(\frac{1}{3})+\frac{\sqrt{π}}{2}}{\Gamma(\frac{11}{6})}$$