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I'm having some trouble with the following exercise:

Let $C_R$ be a curve in the complex plane parametizied as: $\gamma(t)= Re^{it},0\leq t\leq\pi$. Let $$I(R):=\int_{C_R} \frac {e^{iz}}{z}dz$$What is the value of $\lim_{R\to \infty} I(R)$?

I tried two things:

1 - Evaluating the integral using the definition of line integral:

$$I(R)=i\int_0^\pi \frac {e^{iRe^{^{it}}}}{Re^{it}}Re^{it}dt=i\int_0^\pi e^{iRe^{it}} dt$$

But I couldn't continue from here.

2 - My second approach was using the fact that:

$$\left|\int_\gamma f\right| \leq Kl(\gamma)$$

Where $K=\sup\{|f(z)|,z\in C_R\}$

In this particular case, $l(\gamma) = R\pi$ and I got that: $$\left|\frac {e^{iz}}{z}\right| \leq \frac{1}{R}$$

So I got that: $$|I(R)|\leq \pi$$

and this doesn't allow me to conclude anything when $R\to \infty$

How can I solve this exercise?

Eduardo Magalhães
  • 4,878

1 Answers1

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$$\int_\gamma \frac{e^{iz}}{z}dz=\int_0^\pi \frac{e^{iRe^{it}}}{Re^{it}}iRe^{it}dt=\int_0^\pi e^{iRe^{it}}idt.$$ Now take absolute value, so $$\left|\int_0^\pi e^{iRe^{it}}idt\right|\leq\int_0^\pi |e^{iRe^{it}}|dt=\int_0^\pi |e^{-R\sin t}|dt=\int_0^\pi e^{-R\sin t}dt=2\int_0^\frac{\pi}{2} e^{-R\sin t}dt\\ \leq2\int_0^\frac{\pi}{2} e^{-2Rt/\pi}dt=\frac{-\pi}{R}(e^{-R}-1).$$ So $$\left|\int_\gamma \frac{e^{iz}}{z}dz\right|\leq\frac{-\pi}{R}(e^{-R}-1),$$ which clearly goes to zero as $R$ tends to infinity.

Guillermo García Sáez
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