Lorentzian Curve: Is It Normalized?

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A Lorentzian curve is defined such that its integral over all space equals one, indicating it is normalized. The specific form of the Lorentzian function is given by L(x) = (1/π) * (1/(2Γ)) / ((x - x0)² + (1/2Γ)²). This formulation confirms that the integral from negative to positive infinity results in a total area of one. Therefore, by definition, a Lorentzian curve is indeed normalized. The discussion clarifies the mathematical basis for this normalization property.
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Is a lorentzian curve by definition normalized? As far as I can tell it is such that ∫L(x) = 1.
 
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Sirben4 said:
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 L(x) = \frac{1}{\pi} \frac{\frac{1}{2} \Gamma}{(x-x_0)^2 + (\frac{1}{2} \Gamma)^2}, then \displaystyle \int_{-\infty}^{\infty} L(x) \ dx = 1.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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