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Let $C_R$ be the path parametrized by $z = R e^{i\theta}$ with $\theta \in [0,\pi]$ on the complex plane. I want to show that, with $\xi < 0$,

$$\lim_{R\to \infty} \int_{C_R} \dfrac{e^{-2\pi i z\xi}}{z^2+a^2}dz = 0$$

in a rigorous way. The non rigorous way would be to write $z = x+iy$ and notice that

$$e^{-2\pi ix\xi}=e^{2\pi y\xi}e^{-2\pi i x\xi}.$$

In that case when $\xi < 0$ we have $2\pi \xi < 0$ and as $R\to \infty$, $y\to \infty$ and the exponential goes to zero, and since it is the exponential, the integral goes to zero.

I'm interested in the rigorous version of the proof of this limit. How can I show that this integral goes to zero as $R\to \infty$ rigorously?

EDIT: Thinking a little bit more I think I have it: we have

$$f(z) = \dfrac{e^{2\pi y\xi}e^{-2\pi ix\xi}}{z^2+a^2}$$

Hence we have

$$|f(z)| = \dfrac{e^{2\pi y\xi}}{|z^2+a^2|},$$

but $|z^2+a^2|\geq ||z|^2+a^2| = |R^2+a^2|,$ thus we have

$$|f(z)|\leq \dfrac{e^{-2\pi R\sin \theta |\xi|}}{R^2+a^2}$$

but we have $\theta \in [0,\pi]$ so that $2\pi R |\xi|\sin \theta \geq 0$. With this we have $e^{-2\pi R\sin\theta |\xi|} \leq e^{0} = 1$, and we have

$$|f(z)|\leq \dfrac{1}{R^2+a^2}.$$

By a standard theorem on complex analysis we then have

$$\left|\int_{C_R} f(z)dz\right|\leq \dfrac{\pi R}{R^2+a^2},$$

and now this last thing obviously goes to zero as $R\to \infty$, proving the limit. Is this the way to do it?

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  • $\begingroup$ Search on Google Jordan's Lemma. There you will find a lot of ways to prove what you are looking for. $\endgroup$ – HBR Dec 9 '16 at 19:24
  • $\begingroup$ I'm assuming now you aren't familiar with the Residue theorem, but this is another way to rigorously compute the integral. $\endgroup$ – Mark Dec 9 '16 at 19:26
  • $\begingroup$ I'm trying to pove this limit exactly to use it in the context of the Residue theorem, to compute one integral over the real line. Anyway, I think I found a proof. Is my attempt correct? $\endgroup$ – user1620696 Dec 9 '16 at 19:36
  • $\begingroup$ Since $\Bigl\lvert \frac{1}{z^2+a^2}\Bigr\rvert \sim R^{-2}$, in this particular case it suffices to note that for $\xi < 0$ the exponential factor $e^{-2\pi i \xi z}$ is bounded in the upper half-plane, the estimation lemma then yields the result. If the function multiplied with the exponential decays slower, see e.g. here for how to show it. $\endgroup$ – Daniel Fischer Dec 9 '16 at 19:46
  • $\begingroup$ the denominator in your bound should be $R^2 - a^2$ instead of $R^2 + a^2$ (assume $a$ is real). $\endgroup$ – achille hui Dec 9 '16 at 19:50

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