10.4: Integrands with branch cuts

Example \(\PageIndex{1}\)

Compute

\[I = \int_{0}^{\infty} \dfrac{x^{1/3}}{1 + x^2} \ dx. \nonumber \]

Solution

Let

\[f(x) = \dfrac{x^{1/3}}{1 + x^2}. \nonumber \]

Since this is asymptotically comparable to \(x^{-5/3}\), the integral is absolutely convergent. As a complex function

\[f(z) = \dfrac{z^{1/3}}{1 + z^2} \nonumber \]

needs a branch cut to be analytic (or even continuous), so we will need to take that into account with our choice of contour.

First, choose the following branch cut along the positive real axis. That is, for \(z = re^{i \theta}\) not on the axis, we have \(0 < \theta < 2\pi\).

Next, we use the contour \(C_1 + C_R - C_2 - C_r\) shown in Figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\): Contour around branch cut: inner circle of radius \(r\), outer of radius \(R\). (CC BY-NC; Ümit Kaya)

We put convenient signs on the pieces so that the integrals are parametrized in a natural way. You should read this contour as having \(r\) so small that \(C_1\) and \(C_2\) are essentially on the \(x\)-axis. Note well, that, since \(C_1\) and \(C_2\) are on opposite sides of the branch cut, the integral

\[\int_{C_1 - C_2} f(z)\ dz \ne 0. \nonumber \]

First we analyze the integral over each piece of the curve.

On \(C_R\): Theorem 10.2.1 says that

\[\lim_{R \to \infty} \int_{C_R} f(z)\ dz = 0. \nonumber \]

On \(C_r\): For concreteness, assume \(r < 1/2\). We have \(|z| = r\), so

\[|f(z)| = \dfrac{|z^{1/3}|}{|1 + z^2|} \le \dfrac{r^{1/3}}{1 - r^2} \le \dfrac{(1/2)^{1/3}}{3/4}. \nonumber \]

Call the last number in the above equation \(M\). We have shown that, for small \(r, |f(z)| < M\). So,

\[|\int_{C_r} f(z)\ dz| \le \int_{0}^{2\pi} |f(re^{i \theta})||ire^{i \theta}| \ d\theta \le \int_{0}^{2\pi} Mr \ d\theta = 2\pi Mr. \nonumber \]

Clearly this goes to zero as \(r \to 0\).

On \(C_1\):

\[\lim_{r \to 0, R \to \infty} \int_{C_1} f(z)\ dz = \int_{0}^{\infty} f(x) \ dx = I. \nonumber \]

On \(C_2\): We have (essentially) \(\theta = 2\pi\), so \(z^{1/3} = e^{i2\pi /3} |z|^{1/3}\). Thus,

\[\lim_{r \to 0, R \to \infty} \int_{C_2} f(z)\ dz = e^{i2\pi /3} \int_{0}^{\infty} f(x) \ dx = e^{i 2\pi /3} I. \nonumber \]

The poles of \(f(z)\) are at \(\pm i\). Since \(f\) is meromorphic inside our contour the residue theorem says

\[\int_{C_1 + C_R - C_2 - C_r} f(z)\ dz = 2\pi i (\text{Res} (f, i) + \text{Res} (f, -i)). \nonumber \]

Letting \(r \to 0\) and \(R \to \infty\) the analysis above shows

\[(1 - e^{i2\pi /3}) I = 2\pi i (\text{Res} (f, i) + \text{Res} (f, -i)) \nonumber \]

All that’s left is to compute the residues using the chosen branch of \(z^{1/3}\)

\[\text{Res} (f, -i) = \dfrac{(-i)^{1/3}}{-2i} = \dfrac{(e^{i3\pi /2})^{1/3}}{2e^{i3\pi /2}} = \dfrac{e^{-i \pi}}{2} = -\dfrac{1}{2} \nonumber \]

\[\text{Res} (f, i) = \dfrac{i^{1/3}}{2i} = \dfrac{e^{i \pi /6}}{2e^{i \pi /2}} = \dfrac{e^{-i\pi /3}}{2} \nonumber \]

A little more algebra gives

\[(1 - e^{i2\pi /3}) I = 2\pi i \cdot \dfrac{-1 + e^{-i \pi /3}}{2} = \pi i (-1 + 1/2 - i \sqrt{3} /2) = -\pi i e^{i \pi /3}. \nonumber \]

Continuing

\[I = \dfrac{-\pi ie^{i \pi /3}}{1 - e^{i 2\pi /3}} = \dfrac{\pi i}{e^{i \pi /3} - e^{-\pi i/3}} = \dfrac{\pi /2}{(e^{i \pi /3} - e^{-i\pi /3})/2i} = \dfrac{\pi /2}{\sin (\pi /3)} = \dfrac{\pi}{\sqrt{3}}. \nonumber \]

Whew! (Note: a sanity check is that the result is real, which it had to be.)

Example \(\PageIndex{2}\)

Compute

\[I = \int_{1}^{\infty} \dfrac{dx}{x \sqrt{x^2 - 1}}. \nonumber \]

Solution

Let

\[f(z) = \dfrac{1}{z \sqrt{z^2 - 1}}. \nonumber \]

The first thing we’ll show is that the integral

\[\int_{1}^{\infty} f(x)\ dx \nonumber \]

is absolutely convergent. To do this we split it into two integrals

\[\int_{1}^{\infty} \dfrac{dx}{x\sqrt{x^2 - 1}} = \int_{1}^{2} \dfrac{dx}{x \sqrt{x^2 - 1}} + \int_{2}^{\infty} \dfrac{dx}{x \sqrt{x^2 - 1}}. \nonumber \]

The first integral on the right can be rewritten as

\[\int_{1}^{2} \dfrac{1}{x \sqrt{x + 1}} \cdot \dfrac{1}{\sqrt{x - 1}} \ dx \le \int_{1}^{2} \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{x - 1}} \ dx = \dfrac{2}{\sqrt{2}} \sqrt{x - 1} \vert_1^2. \nonumber \]

This shows the first integral is absolutely convergent.

The function \(f(x)\) is asymptotically comparable to \(1/x^2\), so the integral from 2 to \(\infty\) is also absolutely convergent.

We can conclude that the original integral is absolutely convergent.

Next, we use the following contour. Here we assume the big circles have radius \(R\) and the small ones have radius \(r\) (Figure \(\PageIndex{2}\)).

Figure \(\PageIndex{2}\): Copy and Paste Caption here. (CC BY-NC; Ümit Kaya)

We use the branch cut for square root that removes the positive real axis. In this branch

\[0 < \text{arg} (z) < 2\pi \ \ \ \text{and} \ \ \ 0 < \text{arg} (\sqrt{w}) < \pi. \nonumber \]

For \(f(z)\), this necessitates the branch cut that removes the rays \([1, \infty)\) and \((-\infty, -1]\) from the complex plane.

The pole at \(z = 0\) is the only singularity of \(f(z)\) inside the contour. It is easy to compute that

\[\text{Res} (f, 0) = \dfrac{1}{\sqrt{-1}} = \dfrac{1}{i} = -i. \nonumber \]

So, the residue theorem gives us

\[\int_{C_1 + C_2 - C_3 - C_4 + C_5 - C_6 - C_7 + C_8} f(z)\ dz = 2 \pi i \text{Res} (f, 0) = 2\pi. \nonumber \]

In a moment we will show the following limits

\[\lim_{R \to \infty} \int_{C_1} f(z)\ dz = \lim_{R \to \infty} \int_{C_5} f(z)\ dz = 0 \nonumber \]

\[\lim_{r \to 0} \int_{C_3} f(z)\ dz = \lim_{r \to \infty} \int_{C_7} f(z)\ dz = 0. \nonumber \]

We will also show

\[\begin{array} {rcl} {\lim_{R \to \infty, r \to 0} \int_{C_2} f(z) \ dz} & = & {\lim_{R \to \infty, r \to 0} \int_{-C_4} f(z) \ dz} \\ {} & = & {\lim_{R \to \infty, r \to 0} \int_{-C_6} f(z)\ dz = \lim_{R \to \infty, r \to 0} \int_{C_8} f(z)\ dz = I.} \end{array} \nonumber \]

Using these limits, Equation 10.5.23 implies \(4I = 2\pi\), i.e.

\[I = \pi /2. \nonumber \]

All that’s left is to prove the limits asserted above.

The limits for \(C_1\) and \(C_5\) follow from Theorem 10.2.1 because

\[|f(z)| \approx 1/|z|^{3/2} \nonumber \]

for large \(z\).

We get the limit for \(C_3\) as follows. Suppose \(r\) is small, say much less than 1. If

\[z = -1 + re^{i \theta} \nonumber \]

is on \(C_3\) then,

\[|f(z)| = \dfrac{1}{|z\sqrt{z - 1} \sqrt{z + 1}|} = \dfrac{1}{|-1 + re^{i \theta}| \sqrt{|-2 + re^{i \theta}|} \sqrt{r}} \le \dfrac{M}{\sqrt{r}}. \nonumber \]

where \(M\) is chosen to be bigger than

\[\dfrac{1}{|-1 + re^{i \theta}|\sqrt{|-2 + re^{i \theta}|}} \nonumber \]

for all small \(r\).

Thus,

\[|\int_{C_3} f(z) \ dz| \le \int_{C_3} \dfrac{M}{\sqrt{r}} \ |dz| \le \dfrac{M}{|\sqrt {r}|} \cdot 2 \pi r = 2 \pi M \sqrt{r}. \nonumber \]

This last expression clearly goes to 0 as \(r \to 0\).

The limit for the integral over \(C_7\) is similar.

We can parameterize the straight line \(C_8\) by

\[z = x + i \epsilon, \nonumber \]

where \(\epsilon\) is a small positive number and \(x\) goes from (approximately) 1 to \(\infty\). Thus, on \(C_8\), we have

\[\text{arg} (z^2 - 1) \approx 0 \ \ \ \text{and} \ \ \ f(z) \approx f(x). \nonumber \]

All these approximations become exact as \(r \to 0\). Thus,

\[\lim_{R \to \infty, r \to 0} \int_{C_8} f(z)\ dz = \int_{1}^{\infty} f(x)\ dx = I. \nonumber \]

We can parameterize \(-C_6\) by

\[z = x - i \epsilon \nonumber \]

where \(x\) goes from \(\infty\) to 1. Thus, on \(C_6\), we have

\[\text{arg} (z^2 - 1) \approx 2 \pi, \nonumber \]

so

\[\sqrt{z^2 - 1} \approx -\sqrt{x^2 - 1}. \nonumber \]

This implies

\[f(z) \approx \dfrac{1}{(-x)(-\sqrt{x^2 - 1})} = f(x). \nonumber \]

Thus,

\[\lim_{R \to \infty, r \to 0} \int_{C_2} f(z)\ dz = \int_{\infty}^{1} f(x) (-dx) = \int_{1}^{\infty} f(x)\ dx = I. \nonumber \]

The last curve \(-C_4\) is handled similarly.