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My goal is to compute $$I=\int_{0}^{+∞}\frac{\cos{ax}}{1+x^2}dx$$ where $a>0$.

$$I=\frac{1}{2}\int_{-∞}^{+∞}\frac{\cos{ax}}{1+x^2}dx=\frac{1}{2}Re\bigg(\int_{-∞}^{+∞}\frac{e^{iax}}{1+x^2}dx\bigg)$$.

Let $f(z)=\frac{e^{iaz}}{1+z^2}$.

By Residue Theorem, $\int_{-R}^{R}\frac{e^{iax}}{1+x^2}dx+\int_{\gamma_R}\frac{e^{iaz}}{1+z^2}dz=2\pi Res(f,i)=\frac{e^{-a}}{2i}$, where $\gamma_R$ denotes the upper semi-circle centered at $O$ with radius $R$.

As $R$—>$+∞$,

$\int_{-R}^{R}\frac{e^{iax}}{1+x^2}dx$ —> $\int_{-∞}^{+∞}\frac{e^{iax}}{1+x^2}dx$

Now, I am stuck on how to prove $\int_{\gamma_R}\frac{e^{iaz}}{1+z^2}dz$ goes to $0$ as $R$ goes to infinity.

Anyone know how to do it? Many thanks.

ArsenBerk
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K.Yan
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    on the circle with radius $R$, $\bigg|\int_{|z|=R}\frac{e^{iaz}}{1+z^2}dz\bigg|\leq\frac{\pi R}{1+R^2}\rightarrow 0$ as $R\to\infty$ – daulomb Dec 23 '17 at 10:20
  • If you consider your question answered, you may accept the answer that fits you the most. – Rebellos Dec 24 '17 at 13:35

2 Answers2

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Recall Jordan's Lemma and your solution is done :

If the only singularities of $F(z)$ are poles, then : $$\lim_{R\to\infty}\int_{γR} e^{inz}F(z)dz$$ provided that $n>0$ and $|F(z)|\to 0$ as $R\to \infty$.

Now, you have :

$$\int_{\gamma_R}\frac{e^{iaz}}{1+z^2}dz$$

where :

$$F(z) = \frac{1}{1+z^2}$$

Trully, the only singularities of $F(z)$ are the poles $z=\pm i$ and this function satisfies Jordan's Lemma (easy to check!), so it will be :

$$\lim_{R\to\infty}\int_{\gamma_R}\frac{e^{iaz}}{1+z^2}dz=0$$

I have answered in detail and in good presentation a similar example that you can check here !

Rebellos
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Note that for $z=R(\cos(t)+i\sin(t))$ with $R>1$ and $t\in [0,\pi]$ $$\left|\frac{e^{iaz}}{1+z^2}\right|=\frac{e^{-aR\sin(t)}}{R^2-1} \leq \frac{1}{R^2-1}.$$ Hence, as $R\to +\infty$, $$\left|\int_{\gamma_R}\frac{e^{iaz}}{1+z^2}dz\right|\leq \frac{|\gamma_R|}{R^2-1}=\frac{\pi R}{R^2-1}\to 0.$$ P.S. This is a particular case of the Jordan Lemma.

Robert Z
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