# An approach to finding the upper bound of $\int_{0}^{\pi}\frac{\sqrt{Re^{i \theta}}}{4+ (Re^{i \theta})^2}iRe^{i \theta}\,\mathrm d\theta$

$\def\d{\mathrm{d}}$I'm having trouble understanding how ML estimates were used to find the upper bound of the contour ntegral defined in (2). My understanding of the example detailing the integral in question can be followed through from (1).

If $f$ is a complex-valued, continuous function on the contour $\Gamma$ and if its absolute value $| f(z) |$ is bounded by a constant $M$ for all $z$ on $\Gamma$, then we have the following inequality

$$\left|\int_{\Gamma} f(z) \,\d z\right| \leq |f(z)|\int_{a}^{b} \gamma'(t) \,\d t.$$

$$z(\theta)=z_{o}+Re^{i \theta}, \quad a \leq \theta \leq \beta. \tag{0}$$

$$I = \int_{0}^{\infty}\frac{x^{\frac{1}{2}}}{4+x^{2}} \,\d x. \tag{1}$$

The problem in the examples, comes to evaluating the integral in (1) via complex methods. I made the following observations in

$$I = \int_{0}^{\infty}\frac{x^{\frac{1}{2}}}{4+x^{2}} \,\d x \implies \oint_C \frac{z^{\frac{1}{2}}}{4+z^{2}} \,\d z\\ \implies \oint_{C}^{}\frac{re^{i(\theta + 2 \pi n)^\frac{1}{2}}}{4+ (Re^{i \theta})^{2}}dz \implies \oint_{C}^{}\frac{r^\frac{1}{2}e^{i(\theta + 2 \pi n)}}{4+ (Re^{i \theta})^{2}} \,\d z. \tag{2}$$

Remark: Following the observations in (2) our integral can now be defined on our contour denoted by $C$, depicted in the figure below. The author in (3) parameterizing the integral on the RHS side, to show the direction $f(z)$ is being evaluated on $\gamma_{r}$ this is done by utilizing definition (0). For our parametrization this is shown in (4).

$$z(\theta)=Re^{i \theta} \ (0 \leq \theta < \pi), \quad \d z=iRe^{i \theta}\,\d\theta.$$

$$\int_{\gamma_{R}}\frac{r^\frac{1}{2}e^{i(\theta + 2 \pi n)}}{4+ (Re^{i \theta})^{2}} \,\d z = \int_{0}^{\pi}\frac{\sqrtRe^{i \theta}}{4+ (Re^{i \theta})^2}iRe^{i \theta}\,\d\theta.$$

From (4) the author makes use of the ML Inequality to estimate the upper bound of the integral in (4) on the RHS side. My observations can be seen in (5).

Applying the ML inequality to the RHS side of (4), yields the following:

$$\left|\int_{0}^{\pi}\frac{\sqrtRe^{i \theta}}{4+ (R^{2}e^{i \theta})}iRe^{i \theta}\,\d\theta\right| \leq \max \left|\frac{\sqrtRe^{i \theta}}{4+ (R^{2}e^{i 2\theta})}\right| \int_{\gamma}\frac{\d z}{\d \theta}\left|\frac{\sqrtRe^{i \theta}}{4+ (R^{2}e^{i 2\theta})}\right|.$$

Remark:

Within (5) the attempt made by the author to construct a bound can be observed as follows within (5.1).

$$\max_{f \in \gamma} \left| \frac{\sqrtRe^{i \theta}}{4+ r^{2}e^{ia \theta}} \right| < \frac{\left|\sqrtRe^{i \theta}\right|}{\left|4+ (R^{2}e^{i 2\theta})\right|} < \frac{R^{\frac{1}{2}}}{R^{2}-4}. \tag{5.1}$$

Following from (5.1), in (5.2) the author lets contour cover the upper half of the $\mathbb{C}$ $$\left|\int_{\gamma_{R}} f(z) \,\d z\right| < \frac{\pi R^{\frac{3}{2}}}{R^{2}-4} \rightarrow 0, \quad R \, \rightarrow \infty$$

$$\lim_{x\to\infty} \int_{\gamma_r}f(z)\,\d z=0.$$

The crux of question lies between the steps taken in (5.1) and (5.2) I'm trouble verifying the construction the upper bound in (5.1) and the notion that the integral vanishes in (5.2).

• Notice that: $$\left|\frac a{b+c}\right|=\frac{|a|}{|b+c|}\le\frac{|a|}{\min(|b|+|c|,|b|-|c|)}$$and also use $|a\cdot b|=|a|\cdot|b|$ and $|e^{i\theta}|=1$. – Simply Beautiful Art May 1 '17 at 0:32
• $$|z_1-z_2|\ge ||z_1|-|z_2||\implies \frac{1}{|z_1-z_2|}\le \frac{1}{||z_1|-|z_2||}$$ – Mark Viola May 1 '17 at 3:01
• Does the notion in $(5.2)$ follow from the Residue Theorem ? – Zophikel May 1 '17 at 16:10