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

Use a dogbone contour with a single branch around [0,1].

Consider $$f(z)=\frac{1}{z^{2/3}(1-z)^{1/3}}$$ and use the branches

$\displaystyle -\pi\leq arg(z)<\pi, \;\ 0\leq arg(1-z)< 2\pi$

Arg above the cut: $arg(z)=0, \;\ arg(1-z)=2\pi$

Arg below the cut: $arg(z)=0, \;\ arg(1-z)=0$

$$f(z)=\frac{1}{|z|^{2/3}e^{2i\cdot arg(z)/3}|1-z|^{1/3}e^{i\cdot arg(1-z)/3}}$$

$$\int_{0}^{1}\frac{1}{x^{2/3}e^{2i (0)/3}(1-x)^{1/3}e^{i(2\pi )/3}}dx-\int_{0}^{1}\frac{1}{x^{2/3}e^{2i(0)/3}(1-x)^{1/2}e^{i(0)/3}}dx=2\pi i Res(f,\infty)$$

the residue at infinity is $e^{2\pi i /3}$, (I have to admit I just done this with Mathematica to save some time, or one may use the Laurent expansion) so:

$$(e^{-2\pi i/3}-1)\int_{0}^{1}\frac{1}{x^{2/3}(1-x)^{1/3}}dx=2\pi i e^{2\pi i/3}$$

$$\int_{0}^{1}\frac{1}{x^{2/3}(1-x)^{1/3}}dx=\frac{2\pi i e^{2\pi i/3}}{e^{-2\pi i/3}-1}=\frac{2\pi}{\sqrt{3}}$$

EDIT:

To find the residue at infinity, one can multiply by $-1/x^{2}$ and let $x\to 1/x$ to obtain the series expansion about 0:

$\displaystyle \frac{-1}{x^{2}}\cdot \frac{1}{\frac{1}{x^{2/3}}(1-\frac{1}{x})^{1/3}}=\frac{e^{2\pi i/3}}{x}+\frac{e^{2\pi i/3}}{3}+\frac{2e^{2\pi i/3}}{9}x+\cdot\cdot\cdot $

The coefficient of the 1/x term is the residue.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}

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