Prove that the integral of $sin(z)/(z^2+4z+5)$ from negative to positive infinity is $-\pi sin(2)/e$

Question

I think I've made the problem a lot nastier than it supposed to look. Here's what I have so far.

First notice that $(z^2+4z+5)$ is equivalent to $(z^2+4z+4)+1$ so our singularities are -2-i and -2+i but the only one we examine is the latter.

To find the line integral, we do $2\pi i$Res(f,-2+i) where f(z)=$e^{iz}/(z^2+4z+4)+1$. Now from here, when you take the Residue it seems like the "+1" term at the end of the denominator makes this problem a bit messy. Am I on the right track or did I misinterpret something?

Answer 1

You are almost there - it is not so messy as you fear: You correctly stated that

$z^2 + 4z + 5 = [z - (-2 + i)][z - (-2-i)]$.

Then the residue is \begin{equation} \text{Res}_{z = -2 + i}f(z) = \left.\frac{e^{iz}}{z - (-2-i)}\right|_{z = -2 + i} = \frac{e^{-1 -2i}}{2i}, \end{equation} so $2\pi i \text{Res}_{z = -2 + i}f(z) = \pi e^{-1 - 2i}$.

By taking the imaginary part, we end up with the desired result.