Contour Integral Involving $e^z$, a Semicircle, and Triangle Inequality

Question

For context, part of this comes from a real improper integral I am trying to solve, which I am not entirely sure if I am even going in the right direction. But anyways, I am trying to evaluate

$$\oint_Cf(z)dz = \int_{-R}^{-\epsilon}f(z)dz + \int_{\gamma}f(z)dz + \int_{\epsilon}^{R}f(z)dz + \int_{\Gamma}f(z)dz,$$

where $C$ is a large semicircle of radius $R$, $\gamma$ is a small indented circular path of radius $\epsilon$ above $z=0$, and $\Gamma$ is the circumference above the real axis. I let

$$f(z) = \frac{e^{4z}-e^{2z}}{z(e^{2z}+1)^3}.$$

I can figure out how to deal with all the other integrals with no problem including the singularities, except for the integral over $\Gamma$. Basically, I'm trying to prove that

$$\lim_{R \to \infty}\int_{\Gamma}\frac{e^{4z}-e^{2z}}{z(e^{2z}+1)^3}dz = 0$$

by proving that

$$\lim_{R \to \infty}\left|\int_{\Gamma}\frac{e^{4z}-e^{2z}}{z(e^{2z}+1)^3}dz\right| = 0$$

and using the Squeeze Theorem.

Here is my work so far:

$$ \eqalign{ \left|\int_{\Gamma}\frac{e^{4z}-e^{2z}}{z\left(e^{6z}+3e^{4z}+3e^{2z}+1\right)}dz\right| &\leq \left|\frac{e^{4z}-e^{2z}}{z\left(e^{6z}+3e^{4z}+3e^{2z}+1\right)}\right|\left(\pi R\right) \cr &= \frac{\left|e^{4z}-e^{2z}\right|}{\left|z\right|\left|e^{6z}+3e^{4z}+3e^{2z}+1\right|}\left(\pi R\right) \cr &= \frac{\left|e^{4z}-e^{2z}\right|}{R\left|e^{6z}+3e^{4z}+3e^{2z}+1\right|}\left(\pi R\right) \cr &= \frac{\pi\left|e^{4z}-e^{2z}\right|}{\left|e^{6z}+3e^{4z}+3e^{2z}+1\right|} \cr &\leq \frac{\pi\left(e^{4R}+e^{2R}\right)}{\left|e^{6z}+3e^{4z}+3e^{2z}+1\right|}, } $$

where I used the inequality $\left|e^z\right| \leq e^R$. I got that from (and if someone can check if this is correct then that would be great)

$$\eqalign{ \left|e^z\right| &= \left|e^{Re^{i\theta}}\right| \cr &= \left|e^{e^{i\theta}}\right|^R \cr &= \left|e^{\cos{\theta}+i\sin{\theta}}\right|^R \cr &= \left|e^{\cos{\theta}}\right|^R \cdot \left|e^{i\sin{\theta}}\right|^R \cr &= \left|e^{\cos{\theta}}\right|^R \cr &\leq e^R, } $$ since $$\eqalign{ -1 \leq \cos{\theta} &\leq 1 \cr e^{-1} \leq e^{\cos{\theta}} &\leq e^1 \cr \left|e^{-1}\right| \leq \left|e^{\cos{\theta}}\right| &\leq \left|e^1\right| \cr \left|e^{\cos{\theta}}\right| &\leq e \cr \left|e^{\cos{\theta}}\right|^R &\leq e^R. \cr } $$

Question: How exactly can I deal with the bottom such that I can easily take the limit as $R$ goes to infinity? I want to make the bottom smaller and I tried using the fact that if $z$ is on $C$ then $|z| = R$. What I tried was using the Reverse Triangle Inequality to get that

$$\left|e^{2z}+1\right|^3 = \left|e^{2z}+1\right| \cdot \left|e^{2z}+1\right| \cdot \left|e^{2z}+1\right| \geq \left|\left|e^{2z}\right|-|1|\right| \cdot \left|\left|e^{2z}\right|-|1|\right| \cdot \left|\left|e^{2z}\right|-|1|\right| = \left|\left|e^{2z}\right|-|1|\right|^3,$$

but I don't know a way to express that last expression in terms of $R$. Intuitively, it "seems" like the denominator gets bigger faster than the numerator as the radius gets larger and larger. (I feel like I am just overthinking it and that this should be a lot easier than I expected.)

Answer 1

In THIS ANSWER, I showed that $|\cot(\pi z)|$ is bounded on the circle $|z|=(N+1/2)$ for $N\in \mathbb{N^+}$. Analogously, it is straightforward to show that $|\tanh(z)|$ is bounded on the circle $N\pi$, $N\in \mathbb{N^+}$.

Let $B$ represent an upper bound of $|\tanh(z)|$ on the circle $N\pi$, $C_N$ be the semi-circular arc with radius $N\pi$ in the upper-half plane, and $I_N=\int_{C_N}\frac{\sinh(z)}{z\cosh^3(z)}\,dz$. Then we have the estimates on the semi-circular arc of radius $N\pi$

$$\begin{align} |I_n|&=\left|\int_{C_N}\frac{\sinh(z)}{z\cosh^3(z)}\,dz\right|\\\\ &\le \int_{C_N}\frac{1}{N\pi}\left|\tanh(z)\right|\,\left|\text{sech}^2(z)\right|\,|dz|\\\\ &\le B \int_0^\pi \left|\text{sech}(N\pi e^{i\phi})\right|^2\,d\phi\\\\ &=8B \int_0^{\pi/2} \frac{|e^{-N\pi e^{i\phi}}|^2}{|1+e^{-2N\pi e^{i\phi}}|^2}\,d\phi\\\\ &\le \frac{8B}{\inf_{\phi\in[0,\pi], N\in\mathbb{N^+}} {|1+e^{-2N\pi e^{i\phi}}|^2}}\int_0^{\pi/2} e^{-2N\pi \cos(\phi)}\,d\phi\\\\ &\le \frac{8B}{\inf_{\phi\in[0,\pi], N\in\mathbb{N^+}} {|1+e^{-2N\pi e^{i\phi}}|^2}}\int_0^{\pi/2} e^{-2N\pi (1-2\phi/\pi)}\,d\phi\\\\ &\le \frac{8B}{\inf_{\phi\in[0,\pi], N\in\mathbb{N^+}} {|1+e^{-2N\pi e^{i\phi}}|^2}}\frac{1-e^{-2N\pi}}{4N} \end{align}$$

which clearly approaches $0$ as $N\to \infty$.

So, by judiciously selecting the radius of the semi-circle to be $N\pi$, we construct a sequence of integrals $I_N$ such that $\lim_{N\to\infty}I_N=0$.