Wigner function for a Lorentzian function I am calculating the Wigner function for a Lorentzian function, which is the Fourier transform of the exp(-|x|) damping function. But I am having a problem normalizing this function, as I am getting the Sinh Cosh function in the Wigner function for position coordinate. Please guide me on this, that is, how to normalize this Wigner function. My wavefunction is well-behaved (normalizable) in both position and momentum space.
 A: I assume you are familiar with the pathologies of the (Lorentz, Breit-Wigner) Cauchy distribution. But your hyperbolic functions do not seem right.
The basic integral you need is , of course,
$$
\int^\infty_{-\infty} dx \frac{1}{(1+x^2)^2}=\pi/2 ~.
$$
Your normalized wavefunction should be 
$$
\psi=\sqrt{\frac{2}{\pi}} \frac{1}{1+x^2}.
$$
  (cf. purple curve)

The real Wigner function ($\hbar=1$, upon nondimensionalization), automatically normalized (by virtue of the p integral collapse and the normalization of the wave function, above), is then 
$$
f(x,p)= \frac{1}{\pi^2}\int_{-\infty}^{\infty} dy ~ \frac{e^{iyp}}{(1+(x-y/2)^2)(1+(x+y/2)^2)} .
$$
Before evaluation, recall $f(x,p)=f(-x,p)=f(x,-,p)$, and, by above,  $~f(0,0)=1/\pi$.
$$
f(x,p)= \frac{2}{\pi^2}\int_{-\infty}^{\infty} dy ~\frac{e^{i2yp}}
{(y-x-i)(y-x+i)(y+x-i)(y+x+i)} ~.
$$
The poles are at $i\pm x, ~ -i\pm x$. 
Cauchy's theorem leads to contour-integrating in the upper-half complex plane for p>0, and the lower one for p<0.
For p>0,
$$
f=\frac{4i}{\pi} e^{-2p} \left (\frac{e^{i2px}}{8ix(x+i)} 
+ \frac{e^{-i2px}}{8ix(x-i)}\right)
=\frac{e^{-2p}}{\pi(1+x^2)} (\cos (2px) +\sin(2px) /x ).
$$
For p<0,
$$
f(x,p)=\frac{e^{2p}}{\pi(1+x^2)} (\cos (2px) -\sin(2px) /x ).
$$
Hence


*

*Hence, $$
f(x,p)=\frac{e^{-2|p|}}{\pi(1+x^2)} \left(\cos (2px) +\frac{\sin(2|p|x)} {x} \right),
$$
with the requisite symmetries and value at the origin specified in advance.

*In fact,
$$
f(0,p)= \frac{e^{-2|p|}}{\pi} (1+2|p|), \qquad f(x,0)= \frac{ 1}{\pi(1+x^2)}~,
$$
so the moment pathologies of the Cauchy distribution trail the p=0 Wigner function! 



Bonus aside on the δ potential : Would this look familiar? 
Well, if you interchanged x with p, your Fourier transform $\exp(-|x|)$ is the sole bound state of the attractive δ potential, for κ = 1; so f(p,x) instead of the above would be its Wigner Function!
