How do I apply the ML inequality to this integral? The integral is
$$\int_{\gamma}f(z)dz, \quad f(z)=\frac{e^{iz}-1}{z(z^2-1)}, \quad \gamma=\{Re^{i\theta} : 0\leq\theta\leq\pi\},\quad R>0.$$
I need to show that the integral tends to zero as $R$ increases. I did the following:
$$\left|\int_{\gamma}f(z)dz\right|
=\left|\int_0^{\pi}\frac{e^{iRe^{i\theta}}-1}{Re^{i\theta}(R^2e^{2i\theta}+1)}iRe^{i\theta}d\theta\right|
=\left|\int_0^{\pi}\frac{e^{iR(\cos\theta+i\sin\theta)}-1}{(R^2e^{2i\theta}+1)}d\theta\right|\\
\leq R\pi\frac{\left|e^{iR(\cos\theta+i\sin\theta)}-1\right|}{\left|R^2e^{2i\theta}+1\right|}
\leq R\pi\frac{\left|e^{-R\sin\theta}-1\right|}{R^2-1}
\leq R\pi\frac{e^R+1}{R^2-1}\not\rightarrow 0.$$
When I looked at the answers, I saw
$$\left|\int_{\gamma}f(z)dz\right|
=\left|\int_0^{\pi}\frac{e^{iR(\cos\theta+i\sin\theta)}-1}{Re^{i\theta}(R^2e^{2i\theta}+1)}iRe^{i\theta}d\theta\right|
\leq \frac{2\pi R}{R(R^2-1)}\rightarrow 0.$$
This isn't very helpful as there isn't any working. Could somebody walk me through how to manipulate the inequalities into the answer given? In particular I am confused as to how there is still an $R$ in the denominator, a $2$ in the numerator and how to get rid of the $e^{-R\sin\theta}$.
 A: You got an extra factor $R$  because you applied the ML inequality to the parameterized integral $\int_0^\pi$ whose length is $\pi$ and not $R\pi$.
Also the estimate $|Re^{i\theta}| \le e^R + 1$ is too crude.
I would proceed as follows: For $z = Re^{i\theta}$ with $R > 1$ and $0 \le \theta \le \pi$ you have
$$
|f(z)| \le \frac{|e^{iz}| + 1}{|z| (|z|^2-1)} = \frac{e^{-R\sin\theta}+1}{R(R^2-1)} \le \frac{2}{R(R^2-1)} 
$$
because $-R\sin\theta \le 0$. That gives the desired estimate
$$
\left|\int_{\gamma}f(z)dz\right| \le R \pi \frac{2}{R(R^2-1)}  = \frac{2 \pi}{R^2-1} \to 0
$$
for $R \to \infty$.
Remark: Actually $|e^{-R\sin\theta}-1|  = 1 - e^{-R\sin\theta} \le 1$, but a constant factor is not important for this estimate.
A: Note that $\theta \in [0,\pi]$. Thus we have $\sin \theta \geq 0$. Hence $e^{-R \sin \theta}  \leq 1$. 
Just use this in your expression, and get    $$R\pi\frac{\left|e^{-R\sin\theta}-1\right|}{R^2-1} \leq R\pi\frac{\left|e^{-R\sin\theta}\right|+ \left|1\right|}{R^2-1} \leq R\pi\frac{2}{R^2-1}$$
