I want to calculate: $\int\limits_{-\infty}^{\infty} \frac{\cos(2x)\,dx}{(x^2+2x+2)^2}$
Firstly we notice that:
$\int\limits_{-\infty}^{\infty} \frac{\cos(2x)\,dx}{(x^2+2x+2)^2}=\operatorname{Re}\int\limits_{-\infty}^{\infty} \frac{e^{2ix}\,dx}{(x^2+2x+2)^2} $
The function has two poles of order two: $x_1=-1+i$ and $x_2=-1+i$. But the the winding number of $x_1$ is equal to zero, Jordan's lemma is satisfied so we are left with
$Re\int\limits_{-\infty}^{\infty} \frac{e^{2ix}\,dx}{(x^2+2x+2)^2}=\operatorname{Re}(2\pi i\operatorname{Res}(f,-1+i))$
Calculating the residue:
$\operatorname{Res}(f,-1+i)=\lim_{z \to -1+i} \frac{\partial }{\partial z} \frac{e^{2iz}}{(z+1+i)^2}=\frac{-3ie^{-2i-2}}{4} $
To sum up we have:
$\int\limits_{-\infty}^{\infty} \frac{\cos(2x)dx}{(x^2+2x+2)^2}=\operatorname{Re}(2\pi i(\frac{-3ie^{-2i-2}}{4}))=\frac{3\pi \cos(2)}{2e^2}$
But Wolfram says that the result is $\frac{\pi \cos(2)}{e^2}$. Can you help me spot the mistake?
You did nothing wrong. I've checked your computations, and they are correct. And when I ask Mathematica to compute that integral, I get $\frac{3 \pi \cos (2)}{2 e^2}$ as the answer.