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I'm interested in evaluating $\displaystyle \int_{0}^{1} \frac{1}{(x^2-x^3)^{1/3}} \ dx $ using contour integration and the residue at infinity.

But I'm not sure how to define $\displaystyle f(z) = \frac{1}{(z^2-z^3)^{1/3}}$ so that it is well-defined on the complex plane if the line segment $[0,1]$ is omitted.

And then once defined, how do you determine the value of $f(z)$ just above the branch cut and just below the branch cut?

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Let us analyze the function for each of the branch points $z=0$ and $z=1$ separately:

  • Around $z=0$, the function $f_0(z) = z^{-2/3} + O(z^{1/3}).$ We choose the branch cut such that it extends to $z>0$. Thus, we have $f_0(z) = |z|^{-2/3} e^{-i (2/3)\arg z} + O(z^{1/3})$ with $0 <\arg z < 2\pi$. Just above the the branch cut (at $z=0^+ + i 0^+$), we have $\mathop{\rm arg} z = 2\pi$ and thus $f_0(z) = |z|^{-2/3} e^{-i 4\pi/3} = |z|^{-2/3} e^{i 2\pi/3}$. Just below the branch cut (at $z=0^+ - i 0^+$), we have $\arg z =0$ and thus $f_0(z) = |z|^{-2/3} .$

  • At $z=1$, we have a somewhat simpler structure. We can write the function $f_1(z) = |z^2- z^3|^{-1/3} e^{-i(1/3) [\arg (z^2 -z^3) + 2\pi n]}$ which has a branch cut along the real line for $z<1$. Here, $n\in\mathbb{Z}$ will be chosen below such that the two expressions $f_0$ and $f_1$ coincide for $z$ close to 0.

So, let us see what happens at $z=0$:

  • Just above the branch cut, the function $f_1(z)$ assumes the form $f_1(x+i 0^+) = |x^2 - x^3|^{-1/3} e^{-i 2\pi n/3} $

  • Just below the branch cut, the function $f_1(z)$ assumes the form $f_1(x-i 0^+) = |x^2 - x^3|^{-1/3} e^{-i 2\pi/3 - i 2\pi n/3} $

We observe that for $n=-1$ we have found a consistent branch cut structure, where the branch cut only extends between $z=0$ and $z=1$.

The function just above the branch cut reads $f(x + i 0^+) = | x^2 -x^3|^{-1/3} e^{i 2\pi /3} .$ Just below the branch cut, we have $f(x - i 0^+) = | x^2 -x^3|^{-1/3} .$ For $x>1$, we have $f(x)= f_1(x) = |x^2 -x^3|^{-1/3} e^{-i(1/3)(\pi -2\pi)} = |x^2 -x^3|^{-1/3} e^{i \pi/3}.$ For $x<1$, we have $f(x)= |x^2 - x^3|^{-1/3} e^{-i 2\pi/3}.$

In general for $|z| \to \infty$ the function assumes the form $f(z) = |z|^{-1} e^{-i\arg (z)+i \pi/3} = \frac{e^{i \pi/3}}{z}$ which gives you the residue $e^{i\pi/3}$ at infinity.

Thus, we have $\int_{0}^{1} [f(x - i 0^+) - f(x + i 0^+) ] \ dx = -2 \pi i e^{i \pi/3}$ or in other words $\int_{0}^{1} \frac{1}{(x^2-x^3)^{1/3}} (1- e^{i 2\pi/3}) \ dx = -2 \pi i e^{i \pi/3}$ from which you obtain (e.g., taking the imaginary part) the result $\int_{0}^{1} \frac{1}{(x^2-x^3)^{1/3}} = \frac{2\pi}{\sqrt{3}}.$

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    @N3buchadnezzar: I am very bad at drawing by hand and even worse drawing with a computer. Thus, I do not think a drawing of mine would help the answer.2012-03-26
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We can write $f(z) = \alpha z^{-1} (1 - 1/z)^{-1/3}$ where $\alpha$ is one of the cube roots of $-1$, and use the principal branch of $w \to w^{-1/3}$: this has a branch cut for $w$ on the negative real axis, which corresponds to $z \in (0,1)$.

Now for $z = s + i \epsilon$ with $0 < s < 1$ and $\epsilon \to 0+$, $1-1/z \sim 1-1/s + i \epsilon/s^2$ approaches $1-1/s$ from the upper half plane, and $(1-1/z)^{-1/3} \to (1/s-1)^{-1/3} e^{-\pi i/3}$. If you want real values for $f(z)$ as $z$ approaches the branch cut from the upper half plane, you therefore want to take $\alpha = e^{\pi i/3}$. On the other side, for $z = s - i \epsilon$, $1 - 1/z \sim 1-1/s - i \epsilon/s^2$, $(1-1/z)^{-1/3} \to (1/s-1)^{-1/3} e^{+\pi i/3}$, and $f(z) \to (1/s-1)^{-1/3} e^{2\pi i/3}$.

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    For example if $s = 1/2$, $1/z = 1/(1/2 + i\epsilon) = 2/(1 + 2 i \epsilon) \approx 2 (1 - 2 i \epsilon) = 2 - 4 i \epsilon$ so $1 - 1/z \approx -1 + 4 i \epsilon$. This is in the upper half plane, so for principal-branch powers the $-1$ is treated as $e^{i\pi-}$, and $(1-1/z)^{-1/3} \approx (e^{i\pi-})^{-1/3} \approx e^{-i\pi/3}$.2012-03-26
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$ I=\int_0^1\frac{dx}{(x^2-x^3)^{1/3}}. $ The function on hand is \begin{gather*} f(z)=\frac{1}{z^{2/3}(1-z)^{1/3}} =\frac{e^{-\frac{2}{3}(\log |z|+ i (\arg z+2\pi k))}}{(1-z)^{1/3}}\\ =\frac{1}{z}\sqrt[\displaystyle3]{\frac{z}{1-z}} =\frac{e^{(\log|z|+ i (\arg z+2\pi k))/3}}{z(1-z)^{1/3}};\\ 0\le\arg z\le2\pi \; \; (\!\Rightarrow-1=e^{ i \pi}), \quad k=-1,0,1. \end{gather*}

Off the segment $\,[0,1]\,$ (or off any continuum containing the points $\,0,1$) this formula defines three different functions $\,f_{-1},f_0, f_1,\,$ - the branches of $\,f$. The points $\,0,1\,$ are third order branch points of $\,f.\,$ Circumventing each of these points changes the branch, returning to the initial one after the third rotation. Circumventing simultaneously the pair of the branch points keeps the branch unchanged, because the summary increase of the argument will be $\,\pm6\pi,\,$ which divided by $\,3\,$ amounts to $\,2\pi.\, $ Since passing from one branch to the next one occurs when only one of the branch points is circumvented, we only consider windings about $\,0,\,$ hence only $\,\arg z\,$ figures in the above expansions, and for which reason we're not interested in $\,\arg(1-z)\,$ (which value also doesn't change as a result of a full revolution around only $\,0$).

Hence the question of finding res$[f,\infty]\,$ has no meaning. What you can ask is finding the residue of the given fixed branch of the function.

The branches can be identified by their values at an arbitrary fixed point off the segment $\,[0,1],\,$ say, at $\,-1.\,$ We have, $\,f_k(-1)=2^{-1/3}e^{-2 i \pi(1+2k)/3}, k=-1,0,1.\,$ It's easy to see that $\,(C_r=\{|z|=r\})$, $ \int_{C_r}f_k(z)dz=\int_0^{2\pi}\sqrt[\displaystyle3] {\frac{re^{i\varphi}} {1-re^{i\varphi}}}\frac{i re^{i\varphi}\,d\varphi} {re^{i\varphi}} \to 2\pi i \{(-1)^{1/3}\}_k, \quad r \to \infty, $ and since the functions are holomorphic, taking the limit is the same as putting equality sign. Hence res$[f_k,\infty]=-e^{i\pi(1+2k)/3}, \; k=-1,0,1$.

Denoting by $\,I_k^-, I_k^+\,$ the integral over the segment $\,[0,1]\,$ for the values of $\,f_k\,$ accordingly on the lower and upper edges of the branch cut over $\,[0,1],\,$ we see that the value of the integral is the same for all three branches: \begin{gather*} 2\pi i e^{ i \pi(1+2k)/3}=\int_{C_r}f_k(z)\,dz=I_k^--I_k^+\\[-1.5ex] =\int_0^1\frac{dx}{x^{2/3}e^{4(1+k)\pi i /3}(1-x)^{1/3}} -\int_0^1\frac{dx}{x^{2/3}e^{4k\pi i /3}(1-x)^{1/3}}\\ =(e^{-4(1+k)\pi i /3}-e^{-4k\pi i /3})I =-e^{-4k\pi i /3}(1-e^{2 i \pi/3})I.\\ \Rightarrow I=-\frac{2\pi i e^{ i \pi(6k+1)/3}}{1-e^{2\pi i /3}} =-\frac{2\pi i }{e^{-\pi i /3}-e^{\pi i /3}} =\frac{\pi}{\sin\frac{\pi}{3}}; \\ \Rightarrow \boxed{ I = \frac{2\pi }{\sqrt{3}} \quad (k=-1,0,1)}. \end{gather*}

We may also use the following formula. Assume that $\,p\,$ is an integer and the rational function $\,Q\,$ has the finite poles $\,a_1,a_2,\ldots,a_n,\,$ none of which is zero or a positive number. Then \begin{equation} \int_0^\infty x^pQ(x)\,dx=-\sum_{k=1}^n{\rm res}[z^p{\rm Log} z\cdot Q(z); a_k], \tag{1} \end{equation} where $\,{\rm Log}\,$ is the branch of the logarithm, corresponding to $\,0\le\arg z\le2\pi\,$ (see multiple text-or-problem-books on Complex Analysis, say, the Russian edition (after second, 1972(?)) of the Сборник задач по ТФКП by L.I. Volkovyskii, G.L. Lunts, I.G. Aramanovitch (problems 4.169, 4.175), or the Spanish translation published in Moscow in 1977 (problems 711, 717). The Dover Publications edition, NY, 1965, ''A Collection of Problems on Complex Analysis'' is a translation from the first Russian edition and contains the incorrect versions of statements (problems 877, 883)). \begin{gather*} I=\int_0^1\sqrt[\displaystyle3]{\frac{x}{1-x}}\frac{dx}{x}. \\ {\rm Denote } \;\; t=\sqrt[\displaystyle3]{\frac{x}{1-x}}, \Rightarrow x=\frac{t^3}{1+t^3};\quad dx=\frac{3t^2}{1+t^3}; \; \Rightarrow I=3\int_0^{\infty}\frac{dt}{1+t^3} \end{gather*} We have $\,p=0\,$ in the situation of (1). The function $\,Q(z)=\frac{1}{1+z^3}\,$ has (simple) poles at $\,-1, e^{\pm i \pi/3}\,$ with residues $\,{\rm res}[{\rm Log}\cdot Q; -1]=\frac{ i \pi}{3},$ $\;{\rm res}[{\rm Log}\cdot Q; e^{ i \pi/3}]=\frac{ i \pi}{9}e^{-2 i \pi/3},$ $\; {\rm res}[{\rm Log}\cdot Q; e^{- i \pi/3}]=\frac{5 i \pi}{9}e^{2 i \pi/3}.\,$ According to (1): \begin{gather*} I=-3\{\frac{ i \pi}{3} +\frac{2 i \pi}{9}\frac{e^{-2 i \pi/3}+e^{2 i \pi/3}}{2} +\frac{4 i \pi}{9}e^{2 i \pi/3}\}\\ =-3\{\frac{ i \pi}{3}-\frac{2 i \pi}{9}\sin\frac{\pi}{6} +\frac{4 i \pi}{9}[-\sin\frac{\pi}{6}+ i \cos\frac{\pi}{6}]\} \Rightarrow \boxed{ I=\frac{2\pi}{\sqrt{3}}={\rm B}\left(\frac{1}{3},\frac{2}{3}\right)}. \end{gather*}