According to WolframMathworld, when we normalise the Lorentzian function, we get the following result:
$\dfrac{1}{\pi}\displaystyle \int_{-\infty}^\infty \dfrac{b}{(z-a)^2+b^2} dz = 1$
I'm trying to prove this using complex analysis, but I'm running into issues.
LHS = $\dfrac{1}{\pi}\displaystyle \int_{-\infty}^\infty \dfrac{b}{(z-a)^2+b^2} dz = \dfrac{b}{\pi}\displaystyle \int_{-\infty}^\infty \dfrac{1}{(z-a)-ib} \dfrac{1}{(z-a)+ib} dz $
With a region of absolute convergence $Re(z)>0$ and simple poles at $z=a \pm ib$, residues are:
$z=a-ib$: $\dfrac{1}{-2ib}$
$z=a+ib$: $\dfrac{1}{2ib}$
Then I have
$2\pi i \dfrac{b}{\pi} \displaystyle \sum residues = 0$
What am I doing wrong?
If your closed contour is a half-circle with the diameter on the real line and circular part in the upper-half plane it only contains the pole $a+ib$ assuming $a,b>0$ (as in this case $a-ib$ is in the lower half plane). So the integral becomes $2\pi i{b\over2ib}=\pi$.