Express $\int_0^1\frac{dt}{t^{1/3}(1-t)^{2/3}(1+t)}$ as a closed path integral enclosing the interval $(0,1)$

Question

From an old complex analysis qualifier:

Define $$I=\int_0^1\frac{dt}{t^{1/3}(1-t)^{2/3}(1+t)}.$$

1. Express $I$ as a closed path integral enclosing the interval $(0,1)$.

2. Evaluate $I$.

Ideas: At first I thought this was an exercise in the Schwarz-Christoffel integral, but now I'm not so sure. I'm thinking I'm just not aware of the method they're suggesting, and I was hoping someone could point me to it.

Thanks

Answer 1

Consider $$f(z) = e^{-1/3\mathrm{LogA} z} e^{-2/3\mathrm{LogB} (1-z)} \frac{1}{1+z}$$ where $\mathrm{LogA}$ is the logarithm with branch cut on the negative real axis and argument from $-\pi$ to $\pi$ and $\mathrm{LogB}$ has the branch cut on the positive real axis and argument from $0$ to $2\pi.$

Then we have continuity of $f(z)$ across the negative real axis because from above we have $$e^{-1/3\mathrm{LogA} z} e^{-2/3\mathrm{LogB} (1-z)} = e^{-1/3\log x - 1/3\times\pi i} e^{-2/3\log(1-x)-2/3\times 2\pi i} $$ and from below $$e^{-1/3\mathrm{LogA} z} e^{-2/3\mathrm{LogB} (1-z)} = e^{-1/3\log x + 1/3\times\pi i} e^{-2/3\log(1-x)}$$ because $$e^{- (1/3+4/3)\times\pi i} = e^{ 1/3\times\pi i}.$$ In this continuity argument we have used $x$ to denote the absolute value of $z$ on the negative real axis.

It follows by Morera's Theorem that $f(z)$ is actually analytic across the part of the cut on the negative real axis and $f(z)$ only has a cut on $[0,1].$

Now considering the counterclockwise handle-shaped contour enclosing $(0,1)$ we get that the integral above the cut is $$- e^{-2/3\times 2\pi i} I$$ where $I$ is the desired value and below the cut we get the value $I.$ Now the residue of $f(z)$ at $z=-1$ is $$e^{-1/3\times \pi i} e^{-2/3 \log 2 - 2/3\times 2\pi i} = 2^{-2/3} e^{-5/3\pi i}.$$ We thus obtain $$I = - 2\pi i\frac{2^{-2/3} e^{-5/3\pi i}}{1-e^{-2/3\times 2\pi i}} = -2\pi i 2^{-2/3} \frac{e^{-\pi i}}{e^{2/3\times \pi i}-e^{-2/3\times\pi i}}\\ = \pi 2^{-2/3} \frac{2i}{e^{2/3\times \pi i}-e^{-2/3\times\pi i}} = \pi \frac{2^{-2/3}}{\sin(2/3\pi)} = \pi\frac{\sqrt[3]{2}}{\sqrt{3}}.$$ There is no residue at infinity to consider because $f(z)$ is $O(1/R^2)$ there on a ray to infinity at distance $R$ from the origin.

This technique is documented at Wikipedia. In the present case we get that $$-\lim_{|z|\to\infty} z f(z) \sim -\lim_{|z|\to\infty} \frac{z}{|z|^2} = 0.$$ This concludes the argument.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{I \equiv \int_{0}^{1}{\dd t \over t^{1/3}\pars{1 - t}^{2/3}\pars{1 + t}}:\ {\large ?}}$

\begin{align} I&=\int_{\infty}^{1}{-\,\dd t/t^{2} \over t^{-1/3}\pars{1 - 1/t}^{2/3}\pars{1 + 1/t}} =\int^{\infty}_{1}{\dd t \over \pars{t - 1}^{2/3}\pars{t + 1}} =\int^{\infty}_{0}{t^{-2/3} \over t + 2}\,\dd t \\[3mm]&=2^{-2/3}\int^{\infty}_{0}{t^{-2/3} \over t + 1}\,\dd t\tag{1} \end{align}

Integrate $\ds{\int_{C}{z^{-2/3} \over z + 1}\,\dd z}$ where $C$ is a 'key-hole contour' and $z^{-2/3} = \verts{z}^{-2/3}\expo{-2\phi\pars{z}\ic/3}$ with $z \not= 0$ and $0 < \phi\pars{z} < 2\pi$: \begin{align} 2\pi\ic\expo{-2\pi\ic/3} &= \int^{\infty}_{0}{t^{-2/3} \over t + 1}\,\dd t + \int_{\infty}^{0}{t^{-2/3}\expo{-2\pars{2\pi}\ic/3} \over t + 1}\,\dd t = \pars{1 - \expo{-4\pi\ic/3}}\int^{\infty}_{0}{t^{-2/3} \over t + 1}\,\dd t \\[3mm]\mbox{Then,}& \int^{\infty}_{0}{t^{-2/3} \over t + 1}\,\dd t = 2\pi\ic\,{\expo{-2\pi\ic/3} \over 1 - \expo{-4\pi\ic/3}} = {2\pi\ic \over \expo{2\pi\ic/3} - \expo{-2\pi\ic/3}} = {\pi \over \sin\pars{2\pi/3}} = {\pi \over \root{3}/2} \\[3mm]&={2\root{3} \over 3}\,\pi \end{align} By replacing this result in $\pars{1}$, we find: $$\color{#0000ff}{\large% I=\int_{0}^{1}{\dd t \over t^{1/3}\pars{1 - t}^{2/3}\pars{1 + t}} = {2^{1/3}\root{3} \over 3}\,\pi} $$

Answer 3

My approach is similar to Marko Riedel's approach.

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

where $0 \le \arg(z), \arg(z-1) < 2 \pi$.

By omitting the line segment $[0,1]$, $f(z)$ is single-valued on the complex plane.

Then integrating around a dumbbell/dog-bone contour,

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

$$ = \ 2 \pi i \Big( \text{Res}[f(z),-1] + \text{Res}[f(z),\infty]\Big)$$

where

$$\begin{align} \text{Res}[f(z),-1] &= \lim_{z \to -1} \frac{1}{|z|^{1/3} e^{i/3 \arg z}|z-1|^{2/3} e^{2/3i \arg(z-1)}} \\ &= \frac{1}{e^{i \pi/3}2^{2/3}e^{2 \pi i/3}}\\ &= -2^{-2/3} \end{align}$$

and $$ \text{Res}[f(z),\infty] = 0$$

since the Laurent expansion of $f(z)$ at $\infty$ has no $ \displaystyle \frac{1}{z}$ term.

Therefore,

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

which implies

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