I have to calculate the following integral using contour integration: $$\int_0^1 \frac{dx}{(x+2)\sqrt[3]{x^2(x-1)}}$$

I've tried to solve this using the residue theorem, but I don't know how to calculate the residue of the function $$f(z) = \frac{1}{(z+2)\sqrt[3]{z^2(z-1)}}$$ Then I tried to make a substitution in the real integral, so that I would get a function whose residue I know how to calculate, but I couldn't figure out what substitution would do the trick. I would really appreciate if someone could help.

  • $\begingroup$ A quick question: Is $z\mapsto\sqrt[3]{z}$ a principal complex cube root or real cube root? I am asking this because we have $x^2(x-1) < 0$ if $0 < x < 1$. $\endgroup$ Jun 6, 2022 at 21:02
  • $\begingroup$ Have you ever used dogbone contours? $\endgroup$
    – J.G.
    Jun 6, 2022 at 21:59

2 Answers 2


I will instead compute

$$ I = \int_{0}^{1} \frac{\mathrm{d}x}{(x+2)\sqrt[3]{x^2\bbox[color:red;padding:3px;border:1px dotted red;]{(1-x)}}}. $$

You will have no problem converting this to your case, depending on which branch of $\sqrt[3]{\,\cdot\,}$ is used.

1st Solution. Let $\sqrt[3]{z} = \exp(\frac{1}{3}\log z)$ be the principal complex cube root. Also, let $f(z)$ be the holomorphic function defined on $\mathbb{C} \setminus [0, 1]$ by

$$ f(z) = \frac{1}{(z+2) z \sqrt[3]{1 - z^{-1}}}. $$

Then consider the integral

$$ J = \int_{|z|=R_0} f(z) \, \mathrm{d}z, $$

where $R_0 > 2$ so that $|z| = R_0$ encloses all the singularities of $f$. Now we will compute $J$ in two ways. On one hand, by noting that $|f(z)| = \mathcal{O}(|z|^{-2})$, we get

$$ J = \lim_{R\to\infty} \int_{|z|=R} f(z) \, \mathrm{d}z = 0. $$

On the other hand, by "shrinking" the contour $|z| = R_0$ (blue circle in the figure below), we obtain a small circle around the pole $-2$ of $f$ and the dogbone contour around $[0, 1]$:


In this limit, noting that $|f(z)| = \mathcal{O}(|z|^{-2/3})$ as $z \to 0$ and $|f(z)| = \mathcal{O}(|z-1|^{-1/3})$ as $z \to 1$, we obtain

$$ J = 2\pi i \mathop{\mathrm{Res}}_{z=-2} f(z) + (e^{i\pi/3} - e^{-i\pi/3}) I. $$

In this step, we utilized the observation that, for $0 < x < 1$,

\begin{align*} \lim_{\varepsilon \to 0^+} \sqrt[3]{1-\frac{1}{x+i\varepsilon}} &= e^{i\pi/3} \sqrt[3]{\frac{1-x}{x}}, \\ \lim_{\varepsilon \to 0^+} \sqrt[3]{1-\frac{1}{x-i\varepsilon}} &= e^{-i\pi/3} \sqrt[3]{\frac{1-x}{x}}. \end{align*}

Finally, since $J = 0$, solving the above equality for $I$ gives

$$ I = -\frac{2\pi i}{e^{i\pi/3} - e^{-i\pi/3}} \left( \mathop{\mathrm{Res}}_{z=-2} f(z) \right) = \frac{\pi}{\sin(\pi/3)} \frac{1}{\sqrt[3]{12}} $$

2nd Solution. The integrand has two branch points, namely $0$ and $1$. So it would be easier if we can send one to $\infty$. This can be done, for example, by invoking the substitution

$$ t = \frac{x}{1-x}, \qquad \text{i.e.,} \qquad x = \frac{t}{1+t}. $$

Indeed, the above substitution yields

$$ I = \int_{0}^{\infty} \frac{\mathrm{d}t}{t^{2/3}(3t+2)}. $$

Now this integral can be tackled by a fairly standard manner. For example, choosing the branch cut of $\log$ as $[0, \infty)$ and using the Hankel contour (or more precisely, keyhole contour followed by limit),

Hankel contour

we get

\begin{align*} \left(1 - \frac{1}{e^{4\pi i/3}} \right) I &= \int_{\text{Hankel}} \frac{1}{z^{2/3}(3z+2)} \, \mathrm{d}z \\ &= 2\pi i \left( \mathop{\mathrm{Res}}_{z=-2/3} \frac{1}{z^{2/3}(3z+2)} \right) = \frac{2\pi i}{3 e^{2\pi i/3}(2/3)^{2/3}}. \end{align*}

Solving this for $I$ gives the same answer.

  • $\begingroup$ +1. Very nice presentation. What tool do you use to draw the picture? $\endgroup$
    – user1046533
    Jun 6, 2022 at 22:51
  • 4
    $\begingroup$ @user1046533, I used TikZ. The hardest part was to figure out how to systematically export the output in PNG format. $\endgroup$ Jun 6, 2022 at 22:53
  • $\begingroup$ Thank you for helping me! $\endgroup$ Jun 7, 2022 at 16:34

I will solve the more general integral


which is found to be given by


Note that no matter your choice of branch of the cubic root function, your integral will be proportional to $I(2)$ (specifically $-I(2)$ or $e^{i\pi/3}I(2)$ for the two most common definitions of the function as mentioned in the comments)

To prove the statement, perform the change of variables $x \to 1/x$


Now it is easy to see that the integral will become elementary by setting $t=(u-1)^{2/3}$ which yields the form


The remaining integral is standard and can be done using complex analysis, with an appropriate contour (hint below) for the result advertised in equation 2.


Use a pizza slice contour centered at the origin of angle $2\pi/r$. $$\int_0^\infty\frac{da}{a^r+1}=\frac{\pi}{r\sin\pi/r}, r\in \mathbb{R}$$

  • 1
    $\begingroup$ Thank you for helping me! I am happy I received three different solutions. I like your solution equally as the solutions provided by Sangchul Lee. I only accepted their answer, because they provided two different solutions. I am happy that you still received many upvotes. $\endgroup$ Jun 7, 2022 at 16:35
  • $\begingroup$ I was going to launch what I then found was your second solution, you beat me to it. +1 $\endgroup$ Mar 16 at 16:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .