The integral is
$$\int_{-\infty}^{\infty} \frac {cos(z)}{(x^2+a^2)^2}dz $$
If I use an upper semi-circular contour, then there is an order-2 pole at $z=ia$.
I am trying to expand the integrand in a Laurent expansion, centered at the pole $z=ia$ but am having some trouble getting the expansion to come out in powers of $(z-ia)$ so that I can read off the coefficient of the negative power term, $\large \frac {a_{-1}}{z-ia}$.
My attempt is, using the known series for cos(z), and writing it as
$$\sum_{n=0}^{\infty} \frac {-1^n}{2n!}(z-ia)^{2n}$$
then multiplying by $\large \frac{1}{(z^2+a^2)^2}$ gives
$$\sum_{n=0}^{\infty} \frac {-1^n}{2n!}\frac{(z-ia)^{2n}}{(z^2+a^2)^2}$$
$$=\sum_{n=0}^{\infty} \frac {-1^n}{2n!}\frac{(z-ia)^{2n}}{(z-ia)^2(z+ia)^2}$$
$$=\sum_{n=0}^{\infty} \frac {-1^n}{2n!}\frac{(z-ia)^{2(n-1)}}{(z+ia)^2}$$
and this is where I am stuck. How can I simplify this to get it in powers of $(z-ia)$? The $(z+ia)$ factors in the denominator seem problematic... Thanks,
you may find it easier in this case to use the formula for a pole of order $n$: $$ Res_{z_0}f(z) = \lim_{z\to z_0} \frac1{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z) $$ with $z_0=ia$ and the pole of order 2 we therefore require
$$ r = \lim_{z\to ia}\frac{d}{dz}\frac{\cos z}{(z+ia)^2} $$ since $$ \frac{d}{dz}\frac{\cos z}{(z+ia)^2} = \frac1{(z+ia)^3}(-(z+ia)\sin z)-2\cos z) $$ the limit is: $$ r=-(4ia^3)^{-1}(ia\sin ia + \cos ia) \\ =\frac{i}{8a^3}(a(e^{-a}-e^a) +e^{-a}+e^{a})\\ =\frac{i}{8a^3}( (1-a)e^a +(1+a)e^{-a}) $$ (or something along these lines) giving: $$ \int_{-\infty}^{\infty} \frac {\cos z}{(x^2+a^2)^2}dz =2\pi r \\ = \frac{\pi}{4a^3}( (1-a)e^a +(1+a)e^{-a}) $$