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I'm having some trouble with the following exercice:

Let $R>0$ and consider the following curve in the complex plane: $\gamma _R:[0,\pi]\to \mathbb C$ given by $\gamma_R(t)=Re^{it}$. What is the value of: $$\lim_{R\to\infty} \frac{2}{\pi i}\int_{\gamma_R} \frac{e^{-iz}}{z}dz$$

I tired evaluating this integral but I got such when I got to: $$\frac{2}{\pi}\int_0^\pi e^{-iRe^{it}} dt$$

How can this be done?

Thank you

Eduardo Magalhães
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1 Answers1

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$$ \frac{1}{2\pi i}\int_{|z|=R} \frac{e^{iz}}{z} \, dz $$ can be evaluated with Cauchy's integral formula or the residue theorem, it does not depend on $R$. Also $$ \frac{1}{2\pi i}\int_{|z|=R} \frac{e^{iz}}{z} \, dz = \frac{1}{2\pi i}\int_{\gamma_R} \frac{e^{iz}}{z} \, dz + \frac{1}{2\pi i}\int_{\gamma_R} \frac{e^{-iz}}{z} \, dz $$ and the first integral converges to zero for $R \to \infty$, see

Combining these facts you can compute the desired limit.

Martin R
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  • If we always choose the ccw direction, then $$\int_{|z| = R} \frac {e^{i z}} z dz = \int_{\gamma_R} \frac {e^{i z}} z dz + \int_{\gamma_R} \frac {e^{-i z}} z dz.$$ – Maxim Jul 09 '22 at 17:09
  • @Maxim: You are right, I made an error in the $z \mapsto -z$ substitution, thanks. – Martin R Jul 09 '22 at 17:42
  • why is $ \frac{1}{2\pi i}\int_{|z|=R} \frac{e^{iz}}{z} , dz = \frac{1}{2\pi i}\int_{\gamma_R} \frac{e^{iz}}{z} , dz + \frac{1}{2\pi i}\int_{\gamma_R} \frac{e^{-iz}}{z} , dz $? Should't it be $ \frac{1}{2\pi i}\int_{|z|=R} \frac{e^{iz}}{z} , dz = - \frac{1}{2\pi i}\int_{\gamma_R} \frac{e^{iz}}{z} , dz + \frac{1}{2\pi i}\int_{\gamma_R} \frac{e^{-iz}}{z} , dz $? – Eduardo Magalhães Jul 10 '22 at 18:59
  • @EduardoMagalhães: You split the integral into one over the upper semicircle and one over the lower semicircle. The upper semicircle integral is $\frac{1}{2\pi i}\int_{\gamma_R} \frac{e^{iz}}{z} , dz$. In the lower semicircle integral you substitute $z \mapsto -z$, and get $\frac{1}{2\pi i}\int_{\gamma_R} \frac{e^{-iz}}{z} , dz$. – Martin R Jul 10 '22 at 19:02
  • but if we substitute $z\mapsto -z$ then $\frac{e^{iz}}{z} \mapsto -\frac{e^{-iz}}{z}$ – Eduardo Magalhães Jul 10 '22 at 19:05
  • @EduardoMagalhães: $dz \mapsto -dz$ and $z \mapsto -z$, so the $(-1)$ factors cancel. Actually I had that wrong in the first version of my answer, and Maxim made me aware of the error. – Martin R Jul 10 '22 at 19:06
  • That's true, I forgot about the dz. Thank you. Just one more thing: I asked my teacher if we could perform substitutions when computing complex integrals and he told me that that's not always possible but he didn't elaborate any further. Why is it valid in this case to substitute $z \mapsto -z$? – Eduardo Magalhães Jul 10 '22 at 19:09
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    @EduardoMagalhães: It is allowed if you take into account that the integration path changes. If $\gamma: [a, b]\to D \subset \Bbb C$ is a path and $\phi: D \to \Bbb C$ a differentiable function then $\Gamma = \phi \circ \gamma$ is a new path, and $\int_\Gamma h(w) dw = \int_\gamma h(\phi(z)) \phi'(z) dz$. In our example $\phi(z) = -z$. – Martin R Jul 10 '22 at 19:18
  • I get it now. Thank you so much! – Eduardo Magalhães Jul 10 '22 at 19:20