Upper bound on |zf(z)| in a complex keyhole contour integral

Question

I have a problem understanding a technique that is being applied in virtually all of my excercises on solving integrals using the residue theorem. An example is the integral

$$\int^\infty_0dx \frac{x^\frac{1}{3}}{1 + x^2}$$

In class we solve this integral by instead integrating a complex integral $$\int dz \frac{z^\frac{1}{3}}{1 + z^2} $$ over a closed contour with a branch cut using the residue theorem, and then calculating each part of the contour separately as to be able to find the original integral.

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This is the contour I am using. I have already found the integral over the total contour using the residue theorem, but I'm having trouble showing that the integrals over the black and blue parts of the curve vanish. In class this is the method applied for the blue curve with a radius $r \rightarrow 0$.

$$|zf(z)| = \left | \frac{z^{4/3}}{1+z^2} \right| \leq \frac{r^{4/3}}{1-r^2}$$

Where $r$ is the modulus of z on the blue contour. Because our final term goes to zero then we have shown this integral vanishes. I didn't understand how the upper bound was chosen, and it was explained to me that we are trying to maximize the total fraction by separately maximizing the nominator and minimizing the denumerator. Then due to the fact that z can be anywhere on the blue contour, and the triangle inequality, the denominator will be smallest for $z$ that has a phase $\pi$. While I understand the logic behind $ 1 - r^2$ for r on the real axis being the smallest possible value on the contour, I don't understand how we acquire $-r^2$ instead of just $r^2$, I thought the modulus was an absolute value. And why would one have to maximize the total fraction to show it vanishes in the first place?

Additionally for the black curve the following explanation was given:

$$|zf(z)| = \left | \frac{z^{4/3}}{1+z^2} \right| \leq \frac{R^{4/3}}{R^2 - 1}$$

Where again the final term goes to 0, proving the vanishing of the integral. And this I don't understand at all because in what way is $R^2 - 1$ the minimal value of the modulus of z on the black curve if $R \rightarrow \infty$, and how would $|1|$ change signs all of a sudden?

This method of finding an upper bound for the modulus of an expression is used in every single one of my excercises concerning this topic (this being the easiest example), so I'm guessing the use of this method is required to show the vanishing. I would like some help understanding the reason behind using this method because I'm a bit confused on why we are looking for an upper bound in the first place, and of course how does one get the final terms?

Answer 1

Okay, so regarding the circular paths in keyhole contours, I usually parameterize the contours and use various inequalities to prove the integrals on them go to $0$. In your case, here is an example.

We have $$f(z)=\frac{z^\frac{1}{3}}{1 + z^2},\qquad 0<\arg(z)\le2\pi$$ Let the inner and outer paths be $\gamma$ and $\Gamma$ respectively. Then, we can parameterize them to get $$ \begin{alignat*}{5} \Gamma &:\text{ }z=Re^{i\theta},\qquad &\text{d}z&=iRe^{i\theta}\text{ d}\theta,\qquad &\theta&\in[\,\alpha,2\pi-\alpha\,]\\ \gamma &:\text{ }z=\epsilon e^{i\theta},\qquad &\text{d}z&=i\epsilon e^{i\theta}\text{ d}\theta,\qquad &\theta&\in[\,2\pi-\delta, \delta\,] \end{alignat*} $$ where $\alpha$ and $\delta$ are arbitrary angles the outer and inner arc endpoints make with respect to the $x$ axis, and will eventually go to $0$ as we take the $R$ and $\epsilon$ limits.

In this context, $R$ is the radius of the outer circle and $\epsilon$ is the radius of the inner circle.

Now we will show that the integrals along these paths go to zero. To do this, we will use the triangle inequality for integrals, and other variants of the triangle inequality shown below $$|a+b|\le|a|+|b|\Longleftrightarrow|a-b|\ge||a|-|b||\Longleftrightarrow\frac1{|a-b|}\le\frac1{||a|-|b||}$$

Let us start with $\Gamma$. We have \begin{align} \left|\int_{\Gamma}f(z)\text{ d}z\right|&\le \int^{2\pi-\alpha}_{\alpha}\frac{\left|R\right|^{\frac13}{\color{red}{\left|e^{i\theta}\right|^{\frac13}}}}{\left|R^2e^{2i\theta}+1\right|}\cdot{\color{red}{\left|i\right|}}\left|R\right|{\color{red}{\left|e^{i\theta}\right|}}\text{ d}\theta\\ &\le\int^{2\pi-\alpha}_{\alpha}\frac{R^{\frac13}}{\left||R|^2{\color{red}{\left|e^{2i\theta}\right|}}-|-1|\right|}\cdot R\text{ d}\theta\\ &=\int^{2\pi-\alpha}_{\alpha}\frac{R^{\frac43}}{R^2-1}\text{ d}\theta=\frac{(2\pi-2\alpha)R^{\frac43}}{R^2-1} \end{align} $$\implies\lim_{R\to\infty}\frac{R^{\frac43}}{R^2-1}=0$$

Similarly, we have for $\gamma$ that \begin{align} \left|\int_{\gamma}f(z)\text{ d}z\right|&\le\int^{\delta}_{2\pi-\delta}\frac{\left|\epsilon\right|^{\frac13}{\color{red}{\left|e^{i\theta }\right|^{\frac13}}}}{\left|\epsilon^2 e^{2i\theta}+1\right|}\cdot{\color{red}{\left|i\right|}}\left|\epsilon\right|{\color{red}{\left|e^{i\theta}\right|}}\text{ d}\theta\\ &\le\int^{\delta}_{2\pi-\delta}\frac{\epsilon^{\frac13}}{\left||\epsilon|^2{\color{red}{\left|e^{2i\theta}\right|}}-|-1|\right|}\cdot\epsilon\text{ d}\theta\\ &=\int^{\delta}_{2\pi-\delta}\frac{\epsilon^{\frac43}}{|\epsilon^2-1|}\text{ d}\theta=\frac{(2\delta-2\pi)\epsilon^{\frac43}}{|\epsilon^2-1|} \end{align} $$\implies\lim_{\epsilon\to+0}\frac{(2\delta-2\pi)\epsilon^{\frac43}}{|\epsilon^2-1|}=0$$

Usually, we can assume that the integral on any path that circles around a branch point will go to $0$, and we can assume that the integral on any circular limiting path going out to infinity will go to $0$ if $f(z)$'s numerator's growth rate is less than the denominator (which is also the case for this function).

I think this is quite a different method than yours, and I'm not sure if this is helpful so I apologize if it is of no use to you. On the other hand, if you need further explanation just leave a comment :)

Answer 2

The triangle inequality can be written $|a+b| \ge \big||a|-|b|\big|$ for all complex numbers $a$ and $b$. Applying this to $1$ and $z^2$ yields $|1+z^2| \ge 1-|z|^2 > 0$ if $|z|<1$, whereas $|1+z^2| \ge |z|^2-1 > 0$ if $|z|>1$.

Taking reciprocals, and using that the modulus of a quotient is the quotient of the modulus, you derive the displayed inequalities.