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

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.
• Where did I lose the $Re^{i\theta}$? – otah007 May 12 '19 at 22:05
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 $$R\pi\frac{\left|e^{-R\sin\theta}-1\right|}{R^2-1} \leq R\pi\frac{\left|e^{-R\sin\theta}\right|+ \left|1\right|}{R^2-1} \leq R\pi\frac{2}{R^2-1}$$