TITLE: Evaluate by contour integration $\int_0^1\frac{dx}{(x^2-x^3)^{1/3}}$
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: $\newcommand{\+}{^{\dagger}}
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\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}

$$
\color{#66f}{\large\int_{0}^{1}{\dd x \over \pars{x^{2} - x^{3}}^{1/3}}
={2\root{3} \over 3}\,\pi} \approx 3.6276
$$
{"set_name": "stack_exchange", "score": 6, "question_id": 843852}