I have to solve the integral $\int_{-∞}^\infty \frac{cos(x)dx}{(x^2+a^2)(x^2+b^2)}$ using residues.
Here's my attempt to the point where I am stuck:
Noting that $\cos(x)=\text{Re}(e^{ix})$. We can write
$$\int_{-\infty}^\infty \frac{\cos(x)}{(x^2+a^2)(x^2+b^2)}\,dx=\operatorname{Re}\left(\int_{-\infty}^\infty \frac{e^{ix}}{(x^2+a^2)(x^2+b^2)}\,dx\right)$$
By the residue theorem, the integral around the semi-circular contour is just $2\pi i$ times the sum of f's residues at its poles in the upper half plane. So, if we show that as R goes to infinity, the integral on the big semicircle goes to 0, then we'll just have the integral on the real axis of $f(z)$.
However, I am stuck at calculating the residues specifically, since we are dealing with upper half plane I assume we need to calculate residues when $z=ia$ and $z=ib$, however when I take the limit and plug directly I get $0$ in the denominator. I don't know what's wrong here. I'd highly appreciate if someone could help me arrive at the full solution.
Best Answer
It seems you forgot that we should consider $\lim_{z\to\zeta}(z-\zeta)f(z)$ where $\zeta$ is the simple pole. You seem to have only calculated $\lim_{z\to\zeta}f(z)$, which does not exist (precisely because $\zeta$ is a pole). N.B. this method doesn't work for poles of order higher than one. So, if $a=b$ then this approach will not work and you'll have to much more carefully compute the residue, since the pole will be of second order.
Tip: $$\frac{z-ia}{z^2+a^2}\to\frac{1}{2ai}$$By L'Hopital's rule. Alternatively consider $z^2+a^2=(z-ia)(z+ia)$ so if you remove the $z-ia$ factor you obtain $z+ia\to2ai$.