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I don't really understand how I solve ii) in the problem linked. I'm quite sure that it involves splitting up the integral in i) to parts over $C_R$ and $[-R,R]$, but I don't know how to solve the latter. Since it's just along the real axis, I can take the real part of the integral and get $$\int_{-R}^{R}\frac{e^{i\pi x}}{x^2+2x+2}dx.$$

Taking the limit as $R \rightarrow \infty$ (if what I'm thinking is correct) should give me the same answer as i), but I'm not sure how to evaluate this integral.

Thanks.

Problem

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user3255234
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2 Answers2

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Note that we can write for $R>2$

$$\begin{align} \left|\int_{C_R}\frac{e^{iz}}{z^2+2z+2}\,dz\right|&\overbrace{=}^{z=Re^{i\theta}}\left|\int_0^\pi \frac{e^{iRe^{i\theta}}}{(Re^{i\theta})^2+2Re^{i\theta}+2}\,iRe^{i\theta}\,d\theta\right|\\\\ &\le \int_0^\pi \frac{\left|e^{-R\sin(\theta)}e^{iR\cos(\theta)}\right|}{|(Re^{i\theta}+1)^2+1|}\,\,\left|iRe^{i\theta}\right|\,d\theta\\\\ &\le\int_0^\pi \frac{Re^{-R\sin(\theta)}}{||Re^{i\theta}+1|^2-1|}\,d\theta\\\\ &\le \int_0^\pi \frac{e^{-R\sin(\theta)}}{R-2}\,d\theta \\\\ &= \frac{2}{R-2}\int_0^{\pi/2} e^{-R\sin(\theta)}\,d\theta\\\\ &\le \frac{2}{R-2}\int_0^{\pi/2}e^{-2R\theta/\pi}\,d\theta\\\\ &=\frac{\pi(1-e^{-R})}{R(R-2)} \end{align}$$

which clearly approaches $0$ as $R\to \infty$. And we are done!

Mark Viola
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The solution for ii) doesn't rely on the solution for i). Actually, to the best of my understanding, the idea is to eventually use both parts combined to evaluate the integral on the real axis.

To solve ii), all you need is an estimate on the absolute value of the expression inside the integral. Note that the absolute value of the numerator is everywhere less than or equal to 1, whereas that of the denominator is more or less $R^2$.

Amitai Yuval
  • 19,308
  • So it's a simple application of the estimation lemma? That seems to make sense – user3255234 Apr 13 '17 at 03:15
  • @AmitaiYuval On the contour $C_R$, $z=Re^{i\theta}$ and $dz=iRe^{i\theta},d\theta$. Case closed. Why on earth do you think that the numerator (not "nominator") is uniformly bounded by $1$? It is not clearly. I've posted a rigorous way to estimate the integral of interest. – Mark Viola Apr 13 '17 at 03:58
  • The point is that without discussing this somewhat unconventional parameterization, the OP might not understand that the integral is $O(1/R^2)$ due to the $O(1/R)$ contribution from the integrand and the $O(1/R)$ contribution from the integral $\int_0^\pi e^{-R\sin(\theta)},d\theta$. From your post one might believe that the integral is $O(1/R^2)$ for the wrong reason. – Mark Viola Apr 13 '17 at 04:51