Calculate improper integral using Euler's integral

Question

I have to evaluate the following integral

$$\int_0^2 \frac{dx}{\sqrt[5]{x^3(2-x)^2}}$$

Thanks in advance.

Answer 1

Using the substitution $u = x/2$, we get

\begin{align}\int_0^2 \frac{dx}{\sqrt[5]{x^3(2 - x)^2}} &= \int_0^1 \frac{2\, du}{\sqrt[5]{8u^3(2 - 2u)^2}}\, du\\ &= \int_0^1 \frac{du}{\sqrt[5]{u(1 - u)^2}}\\ &= \int_0^1 u^{-3/5}(1 - u)^{-2/5}\, du\\ &= B\left(\frac{2}{5}, \frac{3}{5}\right)\\ &= \frac{\Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right)}{\Gamma(1)}\\ &= \Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right)\\ &= \frac{\pi}{\sin \frac{2\pi}{5}}, \quad \text{by the Euler reflection formula.} \end{align}

Using the multiple angle formula

$$\sin 5x = 5\sin x - 20\sin^3 x + 16 \sin^5 x,$$

substituting $x = \pi/5$ and noting $r := \sin \pi/5 \neq 0$ shows that $\sin \pi/5$ is a root of the polynomial $16t^4 - 20 t^2 + 5$. Hence $r^2 = (5\pm \sqrt{5})/8$. Now since $r^2 < \sin^2(\pi/4) = 1/2$, $r^2 = (5 - \sqrt{5})/8$ and hence

$$r = \sqrt{\frac{5 - \sqrt{5}}{8}}$$

since $r > 0$. Now since

$$\cos \frac{\pi}{5} = \sqrt{1 - r^2} = \sqrt{1 - \frac{5 - \sqrt{5}}{8}} = \sqrt{\frac{3 + \sqrt{5}}{8}} = \sqrt{\frac{6 + 2\sqrt{5}}{16}},$$

we have

$$\sin \frac{2\pi}{5} = 2\sin \frac{\pi}{5}\cos \frac{\pi}{5} = 2\sqrt{\frac{(5 - \sqrt{5})(3 + \sqrt{5})}{64}} = \sqrt{\frac{10 + 2\sqrt{5}}{16}} = \sqrt{\frac{5 + \sqrt{5}}{8}}.$$

Hence

$$\frac{\pi}{\sin \frac{2\pi}{5}} = \pi \sqrt{\frac{8}{5 + \sqrt{5}}} = \pi \sqrt{\frac{8}{20}(5 - \sqrt{5})} = \pi \sqrt{\frac{10 - 2\sqrt{5}}{5}}.$$

Finally, we have

$$\int_0^2 \frac{dx}{\sqrt[3]{x^3(2 - x)^2}} = \pi \sqrt{\frac{10 - 2\sqrt{5}}{5}}.$$

Answer 2

By setting $x=2t$ we have:

$$ I = \int_{0}^{1} t^{-3/5}(1-t)^{-2/5}\,dt = \frac{\Gamma\left(\frac{2}{5}\right)\,\Gamma\left(\frac{3}{5}\right)}{\Gamma(1)} = \frac{\pi}{\sin\frac{2\pi}{5}}=\color{red}{\pi\,\sqrt{2-\frac{2}{\sqrt{5}}}}.$$