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$$\int_{0}^{\infty} \frac{\cos(x)}{(1+x^2)^2} dx$$

The main problem here is to choose the smart contour integral, but i don't see how. I think i am supposed to do this:

  1. note our integral is: $$\frac{1}{2} \int_{-\infty}^{\infty} \frac{\cos(x)}{(1+x^2)^2} dx$$

  2. if we use the contour of a half circle $[-R,R]$ and $\gamma_R = Re^{it}$ with $t \in [0,\pi]$ we see that $$\int_{-R}^{R} \frac{\cos(x)}{(1+x^2)^2} dx = \int_{-R}^{R} \frac{e^{iz}}{(1+z^2)^2} dz$$

  3. With the ML inequality the integral: $$\int_{gaR} \frac{e^{iz}}{(1+z^2)^2} dz = 0$$

Am I supposed to calculate the residuals and take the real part then?

spaceman
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Kees Til
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  • i got $-\frac{3}{8} e^{-1}$ using this tactic, don't know if it's true for sure though. – Kees Til Apr 15 '15 at 20:02
  • Please, take a look at: [1]: http://math.stackexchange.com/questions/1002255/what-is-wrong-in-this-calculation-int-infty-infty-frac-cos-x1x2dx?rq=1 – MeV_2015 Apr 15 '15 at 20:10
  • @KeesTil you have a typo -- the range of $t$ in the parametrization $Re^{it}$ of $\gamma_R$ should be $[0,\pi]$, not $[0,\pi i]$. – kobe Apr 15 '15 at 20:13

1 Answers1

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You're on the right track. Assume that $R > 1$ so that your contour (call it $C(R)$, say) contains the singularity at $z = i$ in the upper-half plane. You're not going to get $\int_{\gamma_R} \frac{e^{iz}}{(1 + z^2)^2}\, dz = 0$. Instead, $\lim_{R\to \infty} \int_{\gamma_R} \frac{e^{iz}}{(1 + z^2)^2}\, dz = 0$. This follows from the fact that $$\left|\int_{\gamma_R} \frac{e^{iz}}{(1 + z^2)^2}\, dz\right| \le \frac{\pi R}{(R^2 - 1)^2},$$

and this estimate follows from the ML-inequality. By the residue theorem, $\int_{C(R)} \frac{e^{iz}}{(1 + z^2)^2}\, dz$ is $2\pi iA$, where $A = \operatorname{Res}_{z = i} \frac{e^{iz}}{(1 + z^2)^2}$ (I'll leave it to you to compute $A$). On the other hand,

$$\int_{C(R)} \frac{e^{iz}}{(1 + z^2)^2}\, dz = \int_{-R}^R \frac{e^{ix}}{(1 + x^2)^2}\, dx + \int_{\gamma_R} \frac{e^{iz}}{(1 + z^2)^2}\, dz.$$

Taking the limit as $R \to \infty$ results in

$$2\pi i A = \int_{-\infty}^\infty \frac{e^{ix}}{(1 + x^2)^2}\, dx.$$

Taking real parts,

$$2\pi A = \int_{-\infty}^\infty \frac{\cos x}{(1 + x^2)^2}\, dx.$$

kobe
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  • hey kobe, can you explain how you do the $|e^ {iz}| \leq \pi R$ because suppose $z = a+bi$, the $|e^ {iz}| = |e^{-b}|$ – Kees Til Apr 15 '15 at 20:16
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    It's not that $|e^{iz}| \le \pi R$, but $|e^{iz}| \le 1$. For on $\gamma_R$, $z = Re^{it}$ for $t \in [0,\pi]$. Thus $|e^{iz}| = e^{-R\sin t}$, and since $\sin t \ge 0$ for $t\in [0,\pi]$, $e^{-R\sin t} \le 1$. Thus $|e^{iz}| \le 1$ on $\gamma_R$. – kobe Apr 15 '15 at 20:20
  • i see it now thanks :) – Kees Til Apr 17 '15 at 09:56