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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 $e^{-R\sin\theta}$.

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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.

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  • $\begingroup$ Where did I lose the $Re^{i\theta}$? $\endgroup$ – otah007 May 12 '19 at 22:05
  • $\begingroup$ @otah007: See updated answer. $\endgroup$ – Martin R May 13 '19 at 5:23
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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 $$R\pi\frac{\left|e^{-R\sin\theta}-1\right|}{R^2-1} \leq R\pi\frac{\left|e^{-R\sin\theta}\right|+ \left|1\right|}{R^2-1} \leq R\pi\frac{2}{R^2-1}$$

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