How to calculate the integral $\int_{-1}^{1}\frac{dz}{\sqrt[3]{(1-z)(1+z)^2}}$?

Question

The integral is $I=\displaystyle\int_{-1}^{1}\dfrac{dz}{\sqrt[3]{(1-z)(1+z)^2}}$. I used Mathematica to calculate, the result was $\dfrac{2\pi}{\sqrt{3}}$, I think it may help.

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{-1}^{1}{\dd z \over \root[3]{\pars{1 - z}\pars{1 + z}^{2}}} =\color{#c00000}{\int_{0}^{2}z^{-2/3}\pars{2 - z}^{-1/3}\,\dd z}:\ {\large ?}}$

\begin{align}&\int_{2}^{0}\!\!z^{-2/3}\expo{\pars{-2/3}0\ic} \pars{2 - z}^{-1/3}\expo{\pars{-1/3}2\pi\ic}\,\dd z +\int_{0}^{2}\!z^{-2/3}\expo{\pars{-2/3}0\ic} \pars{2 - z}^{-1/3}\expo{\pars{-1/3}0\ic}\,\dd z \\[3mm]&=\pars{-\expo{-2\pi\ic/3} + 1} \color{#c00000}{\int_{0}^{2}z^{-2/3}\pars{2 - z}^{-1/3}\,\dd z} =-2\pi\ic\,{\rm Res}_{z\ =\ 0}\bracks{% -\,{1 \over z^{2}}\,z^{2/3}\pars{2 - {1 \over z}}^{-1/3}} \\[3mm]&=2\pi\ic\,{\rm Res}_{z\ =\ 0}\bracks{{\pars{2z - 1}^{-1/3} \over z}} =2\pi\ic\,{\rm Res}_{z\ =\ 0} \bracks{{\verts{2z - 1}^{-1/3}\expo{\pars{-1/3}\pi\ic} \over z}} =2\pi\ic\expo{-\pi\ic/3} \end{align}

\begin{align} &\color{#00f}{\large% \int_{-1}^{1}{\dd z \over \root[3]{\pars{1 - z}\pars{1 + z}^{2}}}} =\color{#c00000}{\int_{0}^{2}z^{-2/3}\pars{2 - z}^{-1/3}\,\dd z} ={2\pi\ic\expo{-\pi\ic/3} \over -\expo{-2\pi\ic/3} + 1} \\[3mm]&=\pi\,{2\ic \over \expo{\pi\ic/3} - \expo{-\pi\ic/3}} ={\pi \over \sin\pars{\pi/3}} = {\pi \over \root{3}/2} =\color{#00f}{\large{2\root{3} \over 3}\,\pi}\ \approx {\tt 3.6276} \end{align}

Answer 2

Hint :

Putting $z=2t-1$ yields $$ \int_{-1}^{1}\dfrac{dz}{\sqrt[\Large3]{(1-z)(1+z)^2}}=\int_{0}^{1}\dfrac{dt}{\sqrt[\Large3]{(1-t)t^2}}=\int_{0}^{1} t^{\large-\frac23}(1-t)^{\large-\frac13}\ dt, $$ then use Beta function and Euler's reflection formula for the gamma function.

Answer 3

Rearrange the integral

$$\int \frac{1}{1+z}\sqrt[3]{\frac{1+z}{1-z}}dz$$

Now set $$t^3=\frac{1+z}{1-z}$$

so $$z=\frac{t^3-1}{t^3+1}=1-\frac{2}{t^3+1}$$

$$dz=\frac{6t^2}{(t^3+1)^2}dt$$

And the integral becomes

$$\int \frac{1}{1+z}\sqrt[3]{\frac{1+z}{1-z}}dz=\int \frac{t^3+1}{2t^3}t\frac{6t^2}{(t^3+1)^2}dt=\int \frac{3}{t^3+1}dt $$

Taking the limits into account, $$\int_{-1}^1 \frac{1}{1+z}\sqrt[3]{\frac{1+z}{1-z}}dz=\int_0^{\infty} \frac{3}{t^3+1}dt $$

Note that $$\frac{3}{t^3+1}=\frac{1}{t+1}-\frac{t-2}{t^2-t+1}= \frac{1}{t+1}-\frac{1}{2}\frac{2t-1}{t^2-t+1}+\frac{1}{2}\frac{3}{t^2-t+1}$$

Now $$\int_0^{\infty} \frac{1}{t+1}-\frac{1}{2}\frac{2t-1}{t^2-t+1}dt=\ln\left(\frac{t+1}{\sqrt{t^2-1+1}}\right) |_0^{\infty}=0$$

(how do you make the long line for the integral ?)

That leaves $$\int_0^{\infty} \frac{1}{2}\frac{3}{t^2-t+1}dt =\frac{3}{2} \int_0^{\infty} \frac{1}{(t-\frac{1}{2})^2+\frac{3}{4}}dt =\sqrt{3} \arctan \frac{2}{\sqrt{3}} (t-\frac{1}{2})|_0^{\infty}= \sqrt{3}(\arctan (\infty)-\arctan (-\frac{1}{\sqrt3})=\sqrt{3}(\frac{\pi}{2}+\frac{\pi}{6})=\frac{2\pi}{\sqrt{3}}$$

Answer 4

Perform the change of variable $z = \frac{1-t^3}{1+t^3}$ (where $t$ now ranges from $+\infty$ to $0$). The integral simplifies and becomes $I = 3\int_0^{+\infty} \frac{t\,dt}{1+t^3}$ and this is easy to compute (the antiderivative of $\frac{t}{1+t^3}$ is $\log\frac{\sqrt{1-t+t^2}}{1+t} + \sqrt{3}\arctan\frac{-1+2t}{\sqrt{3}}$). The same change of variables allows us to compute the original integral with arbitrary bounds.

But, more interestingly, how did I know which change of variable to perform to reduce this integral to a rational function? Well, consider the plane algebraic curve with equation $y^3 = (1-z)(1+z)^2$: it is a cubic curve with a singularity at $(z,y) = (-1,0)$, which implies that a straight line of varying slope through this point (i.e., a line with equation $y=t(1+z)$ where $t$ is the slope) will cut the curve at $(-1,0)$ with multiplicity $2$ and a third point, depending on the slope $t$, with multiplicity $1$, which parametrizes the curve rationally: so, in practice, substitute $y=t(1+z)$ inside $y^3 = (1-z)(1+z)^2$, factor out the $(1+z)^2$ and we are left with $z=\frac{1-t^3}{1+t^3}$ and $y=\frac{2t}{1+t^3}$ as parametrization of the curve $y^3 = (1-z)(1+z)^2$. But now this gives us the change of variable we need (the curve being parametrized by rational functions, the original integral of $\frac{dz}{y}$ will be the integral of a rational function, in this case $-3\frac{t\,dt}{1+t^3}$).