$$pv\int_{-\infty}^{\infty} \frac{x}{(x^2+4)(x^2+2x+2)}dx$$
I get an answer of $-\frac{\pi}{5}$ but wolframalpha disagrees by a factor of $2$ ($-\frac{\pi}{10}$):
http://www.wolframalpha.com/input/?i=integration+%28z%2F%28%28z^2%2B4%29%28z^2%2B2z%2B2%29%29%29++from+-+infinity+to+infinity
I cannot spot the error.
I get these residues:
$$Res(-1+i)=\frac{1+3i}{20}$$
$$Res(2i)=\frac{-2-i}{40}$$
which I add up and multiply by $2\pi i$.
What am I doing wrong?
On
I thought that it might be instructive to show another way to evaluate residues is to use L'Hospital's Theorem. We have
$$\begin{align} \text{Res}\left(\frac{z}{(z^2+4)(z^2+2z+2)},x=2i\right)&=\lim_{z\to 2i}\left(\frac{z(z-2i)}{(z^2+4)(z^2+2z+2)}\right)\\\\ &=\lim_{z\to 2i}\left(\frac{z+(z-2i)}{2z(z^2+2z+2)+(z^2+4)(2z+2)}\right)\\\\ &=\lim_{z\to 2i}\left(\frac{z}{2z(z^2+2z+2)}\right)\\\\ &=\frac{1}{4(-1+2i)}\\\\ &=\frac{-1-2i}{20} \end{align}$$
$$\text{Res}(2i)={x\over (x+2i)(x^2+2x+2)}\Big{|}_{x=2i}={2i\over4i(4i-2)}={1\over4(-1+2i)}={-1-2i\over20}$$