# Prove the integral of $f(z)=\frac{1-e^{2iz}}{4z^2}$ tends to $0$ on the curve $\gamma=Re^{it}$ as $R\rightarrow{\infty}$

I am asked to prove that the integral of $$f(z)=\frac{1-e^{2iz}}{4z^2}$$ on the curve $$\gamma=Re^{it}$$ $$0\leq t \leq \pi$$, when $$R\rightarrow{\infty}$$, goes to zero. To do so I tried $$\int_\gamma f(z)dz=\int_0^{\pi}f(\gamma)\gamma'dt=\int_0^{\pi}\frac{1-e^{2iRe^{it}}}{4(Re^{it})^2}Rie^{it}dt=\int_0^{\pi}\frac{1-e^{2iRe^{it}}}{4R^2e^{it}}idt$$ And now I need to see that $$\lim_{R\to\infty}\int_0^{\pi}\frac{1-e^{2iRe^{it}}}{4R^2e^{it}}idt=0$$ However I don't know how to solve that integral so I can't proceed.

First off, the denominator should be $$4Re^{it}.$$ It's actually not too hard to see that the magnitude of the integrand $$i\frac{1-e^{2iRe^{it}}}{4Re^{it}}$$ is bounded by a multiple of $$1/R$$. Indeed, first note that $$|e^{2iRe^{it}}|=|e^{2iR(\cos t+i\sin t)}|=e^{-2R\sin t}\leq 1,$$ for $$t\in [0,\pi],$$ and so $$\left|i\frac{1-e^{2iRe^{it}}}{4Re^{it}}\right|\leq\frac{1}{4R}+\frac{e^{-2R\sin t}}{4R}\leq\frac{1}{2R},$$ for $$t\in [0,\pi].$$ Using this to bound the integrand uniformly, $$\left|\int\limits_0^\pi i\frac{1-e^{2iRe^{it}}}{4Re^{it}}\ dt\right|\leq\frac{\pi}{2R}\rightarrow 0$$ as $$R\rightarrow\infty.$$
• what about $\pi < t < 2\pi$? the inequality doesn't work. May 18, 2019 at 19:58
• Thank you for your anwer, could you elaborate on why the inequality would not work for $\pi<t<2\pi$? May 18, 2019 at 20:29