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Evaluate $$\int_0^\infty \frac{\cos x}{(x^2+1)^2} \, dx \tag{1}$$

It's just setting this one up that's the hardest part. Finding the right contour is what is most tricky to me. My first move is to note that the above integrand is even. So it suffices to instead calculate

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

Now, I will construct a semicircular arc consisting of the real interval from $-R$ to $R$ for $R>0$ and the arc $\gamma$ given by the top half of the circle $|z|=R$. Call this contour $C_R$. This is the trickiest step. From here, can I directly write $(2)$ as

$$\frac{1}{2} \int_{C_R} \frac{\cos z}{(z^2+1)^2} \, dz \hspace{0.3cm} ? \tag{3}$$

If so, then from here I observe that $z= \pm i$ are each poles of order $2$ and by Cauchy's limit formula for residues, I can find the appropriate residues and then ultimately evaluate $(3)$ via the residue theorem.

Hence, my post begs the question is the substitution from $(2)$ to $(3)$ valid? If not, what steps do I need to take instead?

Aram Nazaryan
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    $(3)$ and $(2)$ are not equivlant. Specifically the integral over the semicircular arc $\gamma$ doesn't $\to 0$ as $R \to \infty$. It is instead a better idea to consider $\text{Re } \int_{-\infty}^\infty \frac{e^{ix}}{(x^2+1)^2} , dx$. – Sahaj Jun 02 '24 at 13:48
  • @Sahaj Are you utilizing $\cos x = \frac{1}{2}(e^{ix} + e^{-ix})$? – Aram Nazaryan Jun 02 '24 at 13:51
  • Yes, or more simply here that $\text{Re } e^{ix} = \cos(x)$. – Sahaj Jun 02 '24 at 13:52
  • @Sahaj Okay. Is Re $\int_{- \infty}^\infty \frac{e^{ix}}{(x^2+1)^2} , dx$ calculable via residues? Your insistence that I integrate over a line begs my initial question, namely that I'm struggling to make the connection between integration over the real line versus a contour – Aram Nazaryan Jun 02 '24 at 13:56
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    https://math.stackexchange.com/q/3228057/1118406 – Sine of the Time Jun 02 '24 at 14:00
  • @Sahaj I think Grigor is saying that you should first consider the function $f(z) = \frac{e^{iz}}{(z^2 + 1)^2}$ and then write $\int_{-R}^{-\epsilon} f(x) , dx + \int_{\gamma^{+}{\epsilon}} f(z) , dx + \int{\epsilon}^{R} f(x) , dx + \int_{\gamma^{+}{R}} f(z) , dz$ for some suitable chosen semicircles $\gamma^{+}{R}$ and $\gamma^{+}_{\epsilon}$ in $\mathbb{C}$ to properly invoke Cauchy's theorem. Then proceed with further analysis. Right? – Trouble Is Back Jun 02 '24 at 14:00
  • When you construct the integral $\oint_{C_R}\frac{e^{iz}}{(z^2+1)^2}dz$, you rewrite it as $\int_{-R}^{R}\frac{e^{iz}}{(z^2+1)^2}dz + \int_{\gamma}\frac{e^{iz}}{(z^2+1)^2}dz$, then apply $R \to \infty$ and $\frac{1}{2}\Re$ afterward on both sides to get the integral you want. In this case, the integral over $\gamma$ vanishes as $R \to \infty$, but this isn't obvious IMO. You have to take the extra steps to prove that. You can see a related way to prove this here. – Accelerator Jun 02 '24 at 15:13
  • @TroubleIsBack Technically, this would work, but it's unnecessary. There's no pole at the origin, so there's no need to do the $\epsilon$ stuff. – Accelerator Jun 02 '24 at 15:20
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    @Accelerator Yes that is correct, my apologies. – Trouble Is Back Jun 02 '24 at 18:25

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