# Prove $\int_0^1 \frac{dx}{(x-2) \sqrt[5]{x^2{(1-x)}^3}} = -\frac{2^{11/10} \pi}{\sqrt{5+\sqrt{5}}}$

$$\mbox{Prove}\quad \int_{0}^{1}{\mathrm{d}x \over \left(\,{x - 2}\,\right)\, \sqrt[\Large 5]{\,x^{2}\,\left(\,{1 - x}\,\right)^{3}\,}\,} = -\,{2^{11/10}\,\pi \over \,\sqrt{\,{5 + \,\sqrt{\,{5}\,}}\,}\,}$$

• Being honest I havent got a clue where to start. I dont think any obvious substitutions will help ($$x \to 1-x, \frac{1}{x}, \sqrt{x},$$ more).
• The indefinite integral involves hypergeometric function so some miracle substitution has to work with the bounds I suspect.
• Maybe gamma function is involved some how ??.

If anyone has an idea and can provide help I would appreciate it.

• I have a feeling this will be a very hot integral. I’m here before it happened. Aug 11, 2020 at 17:46

### Hint:

Substitute $$x \rightarrow\frac{1}{x-1}$$. We'll get: $$-\int_0^\infty \dfrac{x^{-3/5}dx}{(2x+1)}$$ Can you continue from here?

• Nice! What made you think of that substitution?
– user801111
Aug 11, 2020 at 18:05
• @MoQabar It results from the "obvious" substitutions that you mentioned applied multiple times. Aug 11, 2020 at 18:09
• @user376343 My integral reduces to $-\dfrac{π}{2^{2/5} \sin(2π/5)}$, since $\sin(2π/5)=\dfrac{\sqrt{(5+√5)}}{2√2}$. Both integrals are same. Aug 11, 2020 at 18:41
• Do you did $x\mapsto1/u$ then $u\mapsto 1-x$ then multiplied by $-1$? You have more dedication than me! Aug 11, 2020 at 18:55
• @user376343 \begin{aligned} I &= \int_2^{\infty} \frac{\frac{\mathrm{d}x}{(x-1)^2}}{\left(\frac{3-2x}{x-1}\right) \sqrt[5]{{\left(\frac{1}{x-1}\right)}^2{\left(\frac{x-2}{x-1}\right)}^3 }} \\ &= \int_2^{\infty} \frac{\mathrm{d}x}{\left(3-2x\right) {(x-2)}^{\frac{3}{5}}} \\ &\overset{x-2 \to x}= -\int_0^{\infty} \frac{x^{-\frac{3}{5}}}{2x+1} \mathrm{d}x \\ &\overset{2x \to x}= -2^{-\frac{2}{5}} \int_0^{\infty} \frac{x^{-\frac{3}{5}}}{x+1} \mathrm{d}x \end{aligned}
– Ty.
Aug 11, 2020 at 20:40

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ $$\ds{\bbox[10px,#ffd]{\int_{0}^{1}{\dd x \over \pars{x - 2}\root[\Large 5]{x^{2}\pars{1 - x}^{3}}}= -\,{2^{11/10}\,\pi \over \root{5 + \root{5}}}}:\ {\Large ?}}$$.

\begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{\dd x \over \pars{x - 2}\root[\Large 5]{x^{2}\pars{1 - x}^{3}}}} \,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\, \int_{\infty}^{1}{-\,\dd x/x^{2} \over \pars{1/x - 2}\pars{1/x}^{2/5} \pars{1 - 1/x}^{3/5}} \\[5mm] = &\ \int_{1}^{\infty}{\pars{x - 1}^{-3/5} \over 1 - 2x}\,\dd x \,\,\,\stackrel{x + 1\ \mapsto\ x}{=}\,\,\, -\,{1 \over 2}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x \end{align}
Lets consider $$\ds{\oint_{C}{z^{-3/5} \over z - 1/2}\,\dd z}$$ where $$\ds{C}$$ is a key-hole contour which "takes care" of the principal branch of $$\ds{\rule{0cm}{8mm}z^{-3/5}}$$ ( with a branch-cut along $$\ds{\left(-\infty,0\right]}$$ ). Namely,

\begin{align} 2\pi\ic\bracks{\pars{1 \over 2}^{-3/5}} & = \int_{-\infty}^{0}{\pars{-x}^{-3/5}\expo{-3\pi\ic/5} \over x - 1/2}\,\dd x + \int_{0}^{-\infty}{\pars{-x}^{-3/5}\expo{3\pi\ic/5} \over x - 1/2}\,\dd x \\[5mm] & = -\expo{-3\pi\ic/5}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x + \expo{3\pi\ic/5}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x \\[5mm] & = 2\ic\sin\pars{3\pi \over 5} \int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x \\[5mm] \implies & \bbox[10px,#ffd]{-\,{1 \over 2}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x = - 2^{3/5}\pi\csc\pars{3\pi \over 5}} \approx -2.5034 \end{align}

\begin{aligned} \text { Let } x&=\sin ^2 \theta, \quad \textrm{ then } d x=2 \sin \theta \cos \theta d \theta, \textrm{ and }\\ I&=\int_0^{\frac{\pi}{2}} \frac{2 \sin \theta \cos \theta d \theta}{\left(\sin ^2 \theta-2\right) \sqrt[5]{\sin ^4 \theta \cos ^6 \theta}}\\ &=2 \int_0^{\frac{\pi}{2}} \frac{\tan ^{\frac{1}{5}} \theta d \theta}{\cos ^2 \theta\left(\tan ^2 \theta-2 \sec ^2 \theta\right)}\\ &=-2 \int_0^{\infty} \frac{t^{\frac{1}{5}}}{t^2+2} d t ,\quad \textrm{ where }t=\tan \theta\\ &=-2\left(\frac{2^{\frac{1}{10}} \pi}{\sqrt{5+\sqrt{5}}}\right) \quad \textrm{(via beta function)} \\ &=-\frac{2^{\frac{11}{10}} \pi}{\sqrt{5+\sqrt{5}}} \end{aligned}