Estimation of the upper bound of the integral

Question

I studied the of @Ron Gordon answer to the question How to find inverse laplace transform of 2s√2s√+1 where is estimation of the magnitude of the integral over $C_2$ $$\oint_{C_2} dz \frac{e^{z t}}{1+2 \sqrt{z}}. \label{1} \tag{1}$$

$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.

I do not understand how exactly is derived following estimation from \eqref{1}: \begin{align} \frac{R}{2\sqrt{R}-1} \int_{\pi/2}^{\pi} d\theta \, e^{R t \cos{\theta}} = {} & \frac{R}{2\sqrt{R}-1} \int_{0}^{\pi/2} d\theta \, e^{-R t \sin{\theta}} \le \\ &\le \frac{R}{2\sqrt{R}-1} \int_{0}^{\pi/2} d\theta \, e^{-2 R t \theta/\pi} \le \frac{2 \pi}{2\sqrt{R}-1} \label{2} \tag{2} \end{align}

For example I derived something like this: \begin{align} \oint_{C_2} dz \frac{e^{z t}}{1+2 \sqrt{z}} \le {}&\int_{\frac{\pi}{2}}^{\pi} |d\theta| \frac{R |e^{ t R (\cos{\theta} +i sin{\theta})}|}{1+2 \sqrt{R}} \label{3} \tag{3} \end{align} Where is in \eqref{2} $\cos{\theta}, \sin{\theta}$ and why $2\sqrt{R}-1$ insted of $1+2 \sqrt{R}$ in denominator?

Any answers would be appreciated.

Answer 1

Line 1: $\cos \theta = -\sin(\theta + \pi/2)$. Equivalently, make the substitution $\theta = \phi + \pi/2$ so that $\cos \theta = \cos(\phi + \pi/2) = - \sin \phi$ using a quarter-period shift identity. Then switch the dummy variable from $\phi$ to $\theta$. It is worth remembering that this new $\theta$ is not the same as the old $\theta$, but this doesn't matter for dummy variables.

Line 2: $2\theta / \pi \leq \sin \theta$ on $[0,\pi/2]$.

The sine has negative second derivative on $(0, \pi/2)$, so the linear approximation through the endpoints must be below sine.

Then, do the easy integral.

Answer 2

If you want an exact value $$\frac{R}{2\sqrt{R}-1} \int_{\pi/2}^{\pi} d\theta \, e^{R t \cos{\theta}}=\frac{R}{2\sqrt{R}-1}\frac{\pi}{2} \, (I_0(R t)-\pmb{L}_0(R t))$$ where appear Bessel and Struve functions

Answer 3

Thanks hint of @Eric Towers. With his permission, I will write a more extended estimation. Despite the fact that the answer is still not completely agreed, the assessment is in any case true.

Remembering that $|e^{i \sin{\theta}}| = 1$ rewrite estimations: \begin{align} \oint_{C_2} dz \frac{e^{z t}}{1+2 \sqrt{z}} \le {}&\int_{\frac{\pi}{2}}^{\pi} |d\theta| \frac{R |e^{ t R (\cos{\theta} +i sin{\theta})}|}{1+2 \sqrt{R}} \le\\ &\le \frac{R}{1+2 \sqrt{R}} \int_{\frac{\pi}{2}}^{\pi} |d\theta| |e^{ t R \cos{\theta}}|\cdot | e^{i sin{\theta}}| \le\\ & \le \frac{R}{1+2 \sqrt{R}} \int_{\frac{\pi}{2}}^{\pi} |d\theta| |e^{ t R \cos{\theta}}| \label{4} \tag{4} \end{align} Introduce a dummy variable $$\phi = \theta - \frac{\pi}{2},$$ and $$\cos{\theta} = \cos{(\phi + \frac{\pi}{2})} = -\sin{\phi}.$$ Then $$\frac{R}{1+2 \sqrt{R}} \int_{\frac{\pi}{2}}^{\pi} |d\theta| |e^{ t R \cos{\theta}}| = \frac{R}{1+2 \sqrt{R}} \int_{0}^{\frac{\pi}{2}} |d\phi| |e^{ -t R \sin{\phi}}| \label{5} \tag{5}$$

Taking into account that $-2\phi / \pi \geq -\sin \phi$ new estimation will be: $$\frac{R}{1+2 \sqrt{R}} \int_{0}^{\frac{\pi}{2}} |d\phi| |e^{ -t R \sin{\phi}}| \leq \frac{R}{1+2 \sqrt{R}} \int_{0}^{\frac{\pi}{2}} |d\phi| |e^{ -t R 2 \phi/\pi}|\label{6} \tag{6}$$ Introduce dummy variable $u$: \begin{align} &u = -\frac{-tR2 \phi}{\pi} \\ &d \phi = -\frac{\pi}{-2tR } d u \\ &\int_{0}^{\frac{\pi}{2}} |d\phi| |e^{ - t R 2 \phi/\pi}| = -\frac{\pi}{-2tR } \int_{0}^{-tR} e^{u} du = \frac{\pi}{2tR }(1-e^{-tR}) \end{align} Thus \begin{align} \lim_{R \to \infty}\oint_{C_2} dz \frac{e^{z t}}{1+2 \sqrt{z}} \le {}&\lim_{R \to \infty}\int_{\frac{\pi}{2}}^{\pi} |d\theta| \frac{R |e^{ t R (\cos{\theta} +i sin{\theta})}|}{1+2 \sqrt{R}} \le \\ & \le \lim_{R \to \infty} \frac{\pi}{2t(1+2 \sqrt{R})}(1-e^{-tR}) = 0 \end{align}