Complex analysis: contour integration

Solution 1:

Draw a dumbbell contour $C$ about the branch cut $z \in [0,1]$. That is, two small circular segments about the branch points, and an upper and lower path connecting them above and below the real axis, respectively, as illustrated below:

enter image description here

We consider the integral

$$\oint_C dz \, f(z) = i 2 \pi \text{Res}_{z=0} \left [\frac{1}{z^2} f\left(\frac{1}{z}\right) \right ]$$

where

$$f(z) = z^{-2/3} (1-z)^{-1/3}$$

and the term on the right in the former equation is the residue at infinity. This residue may be computed by seeing that

$$\frac{1}{z^2} f\left(\frac{1}{z}\right) = \frac{(z-1)^{-1/3}}{z}$$

so that

$$\text{Res}_{z=0} \left [\frac{1}{z^2} f\left(\frac{1}{z}\right) \right ] = (-1)^{-1/3}$$

I will elaborate on this in a bit.

Now, we define

$$z^{-2/3} = e^{-(2/3) \log{z}}$$

such that $\arg{z} \in [-\pi,\pi)$. This definition is a result of the original branch cut of this factor being $(-\infty,0]$. Further define

$$(1-z)^{-1/3} = e^{-(1/3) \log{(1-z)}}$$

such that $\arg{(1-z)} \in [0,2\pi)$. This definition is a result of the original branch cut of this factor being $[1,\infty)$.

To summarize, on the lines above and below the real axis, $z=x \in [0,1]$ and therefore $\arg{z} = 0$. On the line above the real axis, however, $\arg{(1-z)} = 2 \pi$. Therefore above the real axis, $z^{-2/3} (1-z)^{-1/3} = x^{-2/3} (1-x)^{-1/3} e^{-i 2 \pi/3}$ Below the real axis, $z^{-2/3} (1-z)^{-1/3} = x^{-2/3} (1-x)^{-1/3}$ because there, $\arg{(1-z)} = 0$.

Further, it should be clear that the integrals about the small circular arcs of radius $\epsilon$ around the branch points vanish as $\epsilon^{1/3}$.

Therefore, we may write

$$\left ( 1-e^{-i 2 \pi/3}\right) \int_0^1 dx \: x^{-2/3} (1-x)^{-1/3} = i 2 \pi (-1)^{-1/3}$$

Because that residue was calculated from the $1-z$ term, then $-1=e^{i \pi}$ and we have

$$\int_0^1 dx \: x^{-2/3} (1-x)^{-1/3} = i 2 \pi \frac{e^{-i \pi/3}}{1-e^{-i 2 \pi/3}} = \frac{\pi}{\sin{(\pi/3)}} = \frac{2 \pi}{\sqrt{3}}$$

Solution 2:

Here is a way without contour integration. $$I = \int_0^1 \dfrac{dx}{(x^2-x^3)^{1/3}} = \int_0^1 \dfrac{dx}{x^{2/3}(1-x)^{1/3}}$$ Let $x=\sin^2(a)$. We then get that $$I = \int_0^{\pi/2} \dfrac{2 \sin(a) \cos(a) da}{\sin^{4/3}(a) \cos^{2/3}(a)} = 2 \int_0^{\pi/2} \sin^{-1/3}(a) \cos^{1/3}(a) da = \beta(1/3,2/3) = \dfrac{2 \pi}{\sqrt3}$$

Solution 3:

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

\begin{align}&\color{#c00000}{\int_{0}^{1}{\dd x \over \pars{x^{2} - x^{3}}^{1/3}}} =\int_{0}^{1}\pars{{1 \over x} - 1}^{2/3}\,{\dd x \over 1 - x} =\int_{\infty}^{1}{\pars{x - 1}^{2/3} \over 1 - 1/x}\,\pars{-\,{\dd x \over x^{2}}} \\[3mm]&=\int_{1}^{\infty}{\pars{x - 1}^{2/3} \over \pars{x - 1}x}\,\dd x =\color{#c00000}{\int_{0}^{\infty}{x^{2/3} \over x\pars{x + 1}}\,\dd x} =2\pi\ic\,{\expo{2\pi\ic/3} \over -1} -\int_{\infty}^{0}{x^{2/3}\expo{4\pi\ic/3} \over x\pars{x + 1}}\,\dd x \\[3mm]&=-2\pi\ic\,\expo{2\pi\ic/3} + \expo{4\pi\ic/3} \color{#c00000}{\int_{0}^{\infty}{x^{2/3} \over x\pars{x + 1}}\,\dd x} \end{align}

\begin{align}&\color{#c00000}{% \int_{0}^{1}{\dd x \over \pars{x^{2} - x^{3}}^{1/3}}} =\color{#c00000}{\int_{0}^{\infty}{x^{2/3} \over x\pars{x + 1}}\,\dd x} =-2\pi\ic\,{\expo{2\pi\ic/3} \over 1 - \expo{4\pi\ic/3}} =\pi\,{-2\ic \over \expo{-2\pi\ic/3} - \expo{2\pi\ic/3}} \\[3mm]&={\pi \over \sin\pars{2\pi/3}} \end{align}