An algebraic integral with a fifth root via the beta function

In the previous post, we showed that for \(a, b, p > 0\),

\[ \int_0^1 x^{a-1}(1-x)^{b-1} \frac{dx}{(x+p)^{a+b}} = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\frac{1}{(1+p)^{a} p^b}. \]

We can use this identity, together with Euler's reflection formula for the gamma function, to evaluate the algebraic integral

\[ \int_{0}^{1}\frac{1}{(2-x)\sqrt[5]{x^{2}(1-x)^{3}}}\,dx = \frac{2\pi \sqrt[10]{2}}{\sqrt{5+\sqrt{5}}}. \]

Matching the parametric form

Perform the reflection substitution \(t = 1-x\), so that \(dx = -dt\) and \(2-x = 1+t\). The integral becomes

\[ \begin{aligned} \int_0^1 \frac{1}{(2-x)\sqrt[5]{x^2(1-x)^3}}\,dx &= \int_0^1 \frac{1}{(1+t)\sqrt[5]{(1-t)^2 t^3}}\,dt \\ &= \int_0^1 t^{-\frac{3}{5}}(1-t)^{-\frac{2}{5}}\frac{dt}{1+t} \\ &= \int_0^1 t^{\frac{2}{5}-1}(1-t)^{\frac{3}{5}-1}\frac{dt}{(t+1)^{\frac{2}{5}+\frac{3}{5}}}. \end{aligned} \]

This is now the special case \(a = \frac{2}{5}\), \(b = \frac{3}{5}\), and \(p = 1\) of our parametric beta integral. Since \(a+b = 1\) and \(\Gamma(1) = 1\), we get

\[ \int_0^1 \frac{1}{(2-x)\sqrt[5]{x^2(1-x)^3}}\,dx = \frac{\Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right)}{2^{\frac{2}{5}}}. \]

Simplifying with Euler's reflection formula

By Euler's reflection formula, \(\Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin(\pi s)}\) for \(s \notin \mathbb{Z}\). Taking \(s = \frac{2}{5}\) gives

\[ \Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right) = \frac{\pi}{\sin\left(\frac{2\pi}{5}\right)}. \]

Recall the exact trigonometric value

\[ \sin\left(\frac{2\pi}{5}\right) = \frac{\sqrt{10+2\sqrt{5}}}{4} = \frac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}}. \]

Substituting this into our expression and combining the powers of \(2\) via \(\frac{2\sqrt{2}}{2^{2/5}} = 2^{1 + \frac{1}{2} - \frac{2}{5}} = 2 \cdot 2^{\frac{1}{10}} = 2\sqrt[10]{2}\), we arrive at

\[ \begin{aligned} \int_0^1 \frac{1}{(2-x)\sqrt[5]{x^2(1-x)^3}}\,dx &= \frac{\pi}{2^{\frac{2}{5}} \sin\left(\frac{2\pi}{5}\right)} \\ &= \frac{2\pi \sqrt[10]{2}}{\sqrt{5+\sqrt{5}}}. \end{aligned} \]