[Math] Evaluate by contour integration $\int_0^1\frac{dx}{(x^2-x^3)^{1/3}}$

Question

Evaluate by contour integration [i am learning complex analysis – calculus of residues]

$$\int_0^1\frac{dx}{(x^2-x^3)^{1/3}}$$

I tried by taking $x^3$ out from the denominator but that didnt work.

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \color{#c00000}{\int_{0}^{1}{\dd x \over \pars{x^{2} - x^{3}}^{1/3}}}& =\int_{\infty}^{1}{-\dd x/x^{2} \over \pars{1/x^{2} - 1/x^{3}}^{1/3}} =\int_{1}^{\infty}{\dd x \over x\pars{x - 1}^{1/3}} =\color{#c00000}{\int_{0}^{\infty}{x^{-1/3}\,\dd x \over x + 1}} \end{align} Use the following contour with $\ds{z^{-1/3} = \verts{z}^{-1/3}\expo{-\ic\phi\pars{z}/3}\,,\qquad 0 < \phi\pars{z} < 2\pi}$:

\begin{align} \color{#c00000}{\int_{0}^{\infty}{x^{-1/3}\,\dd x \over x + 1}} &=2\pi\ic\expo{-\pi\ic/3} -\int_{\infty}^{0}{x^{-1/3}\expo{-2\pi\ic/3}\,\dd x \over x + 1} =2\pi\ic\expo{-\pi\ic/3} +\expo{-2\pi\ic/3}\color{#c00000}{\int_{0}^{\infty}{x^{-1/3}\,\dd x \over x + 1}} \end{align}

\begin{align} \color{#c00000}{\int_{0}^{\infty}{x^{-1/3}\,\dd x \over x + 1}}&= 2\pi\ic\,{\expo{-\pi\ic/3} \over 1 - \expo{-2\pi\ic/3}} = 2\pi\ic\,{1 \over \expo{\pi\ic/3} - \expo{-\pi\ic/3}} ={2\pi\ic \over 2\ic\sin\pars{\pi/3}}={\pi \over \sin\pars{\pi/3}} \\[3mm]&={\pi \over \root{3}/2} = {2\root{3} \over 3}\,\pi \end{align}

$$ \color{#66f}{\large\int_{0}^{1}{\dd x \over \pars{x^{2} - x^{3}}^{1/3}} ={2\root{3} \over 3}\,\pi} \approx 3.6276 $$