# Calculate an integral using contour integration

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.

• 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 $0 < x < 1$. Jun 6, 2022 at 21:02
• Have you ever used dogbone contours?
– J.G.
Jun 6, 2022 at 21:59

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

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),

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.

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

I will solve the more general integral

$$I(z):=\int_0^1\frac{dx}{x^{2/3}(1-x)^{1/3}(x+z)}$$

which is found to be given by

$$I(z)=\frac{2\pi}{\sqrt{3}}z^{-2/3}(1+z)^{-1/3}$$

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 $$x \to 1/x$$

$$I(z)=\int_1^\infty\frac{du}{(u-1)^{1/3}(1+zu)}$$

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

$$I(z)=\frac{3}{2}\int_0^{\infty}\frac{dt}{zt^{3/2}+z+1}=\frac{3}{2z^{2/3}(1+z)^{1/3}}\int_0^{\infty}\frac{da}{a^{3/2}+1}$$

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.

Hint:

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

• 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. Jun 7, 2022 at 16:35
• I was going to launch what I then found was your second solution, you beat me to it. +1 Mar 16 at 16:46