In order to prove the normalisation property of a Lorentzian function,

$L = \dfrac{1}{\pi}\displaystyle \int_{-\infty}^\infty \dfrac{b}{(z-a)^2+b^2} dz = 1$

we take a closed contour on the upper-half complex plane. This means we only consider the $z=a+ib$ pole when finding residues. I know this has to do with the winding number, but can you give a more physical explanation of why we do this?

Do we still use contours that cover the upper-half plane when $f(z)$ in

$F = \dfrac{1}{\pi}\displaystyle \int_{-\infty}^\infty \dfrac{b\,f(z)}{(z-a)^2+b^2} dz$

contains poles at say, $z=-c+id$ where $d=(2n+1)\pi i, n\in \mathbb{Z}$?


In your first example,

$$L=\frac1{\pi} \int_{-\infty}^{\infty} dx \frac{b}{(x-a)^2+b^2} = 1 $$

you could just as easily close a contour in the lower half-plane. In that case, though, the winding number is $-1$ rather than $1$. That is, assuming $b \gt 0$,

$$\int_{-\infty}^{\infty} dx \frac{b}{(x-a)^2+b^2} = - i 2 \pi \operatorname*{Res}_{z=a-i b} \frac{b}{(z-a)^2+b^2} = \pi$$

In your more general example, it depends on the behavior of $|f(z)|$ on the semicircular arc of radius $R$ as $R \to \infty$ in the upper half plane. For example, if $f$ is periodic on the real line, it tends to be exponential on the imaginary axis. If that exponential behavior is increasing rather than decreasing, then you cannot close in that half plane to determine the value of the integral.

  • $\begingroup$ Let's say $f(z)$ is purely real and goes to 0 at $\pm\infty$. It has infinitely many poles along the imaginary axis, i.e. $z=-c+i(2n+1)\pi$, $n\in\mathbb{Z}$, something like in en.wikipedia.org/wiki/Matsubara_frequency. Can we close in either of the half planes? $\endgroup$ – Medulla Oblongata Feb 8 '16 at 7:22
  • $\begingroup$ @MedullaOblongata: In the examples I am thinking of, yes, and they happen to align more or less with the weighting functions described in the linked text. (For example, $f(z) = \operatorname{sech}{z}$.) In this case, you already have the inverse-square dependence you need along the imaginary axis for convergence there. So, yes, I think you should be OK to close in either half-plane. $\endgroup$ – Ron Gordon Feb 8 '16 at 8:20

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