Sirben4
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Is a lorentzian curve by definition normalized? As far as I can tell it is such that ∫L(x) = 1.
If we define it as [itex]L(x) = \frac{1}{\pi} \frac{\frac{1}{2} \Gamma}{(x-x_0)^2 + (\frac{1}{2} \Gamma)^2}[/itex], then [itex]\displaystyle \int_{-\infty}^{\infty} L(x) \ dx = 1[/itex].Sirben4 said:Is a lorentzian curve by definition normalized? As far as I can tell it is such that ∫L(x) = 1.