Ok so this a a classical application of the Residue Theorem: show that $\int_{0}^{\infty} \frac{\cos x}{(1+x^2)^{2}} dx = \frac{\pi}{2e} $.
I start out like this: $\int_{0}^{\infty} \frac{\cos x}{(1+x^2)^{2}} dx = 1/2 \int_{-\infty}^{\infty} \frac{\cos x}{(1+x^2)^{2}} dx $. Then I use the residue theorem and end up having to compute $ \pi i\lim_{z \to i} \frac{d}{dz} (z-i)^2\frac{\cos(z)}{(z^2+1)^2} = \frac{\pi}{4e}$. I can't figure out where I'm losing the factor of $2$. My computations are correct, so I must be losing the factor when I apply the Residue theorem. That is to say, Residue Theorem should tell me to compute $2 \pi i\lim_{z \to i} \frac{d}{dz} (z-i)^2\frac{\cos(z)}{(z^2+1)^2} = \frac{\pi}{2e}$ Could it be that the winding number is $2$ instead of $1$? I don't think it could, since I'm integrating over the upper half circle that only contains the pole at $i$ "once"... Or perhaps I need work with $e^{ix}$ and then take the real part, since $\cos z$ is not bounded.... Help please!
One of your ideas as to what's gone wrong was correct. We cannot simply compute the integral (including your external $\frac12$ factor) as $$\pi i\lim_{z\to i}\frac{d}{dz}\left(\frac{\cos z}{(z+i)^2}\right)=-\pi i\lim_{z\to i}\left(\frac{\sin z}{(z+i)^2}+\frac{2\cos z}{(z+i)^3}\right).$$We can't, because the usual semicircular contour allows $\Im z\to+\infty$ so$$\cos z=\frac{e^{-\Im z+(\Re z)i}+e^{\Im z-(\Re z)i}}{2}$$includes in the second term above a divergent contribution. We only want to keep $e^{-\Im z+(\Re z)i}=e^{iz}$, so the pole at $z=i$ contributes$$\pi i\lim_{z\to i}\frac{d}{dz}\left(\frac{e^{iz}}{(z+i)^2}\right)=\pi i\lim_{z\to i}e^{iz}\left(\frac{i}{(z+i)^2}-\frac{2}{(z+i)^3}\right)=\frac{\pi}{2e}.$$