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Integral of Cauchy / Lorentzian distribution in Wolfram|Alpha

Posted 9 months ago
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I am looking to do some line fitting of Lorentzian/Cauchy functions to a type of chromatogram. It is working fine, but I need an analytical expression to the area of the curve. The function is given as:

f(x) = I*y^2/((x-x0)^2+y^2)

where I,y,x0 are constants. Using Wolfram|Alpha to integrate, I get the following: F(x) = Iyatan((x-x0)/y) + C

The integral should be taken from -inf to inf. As atan approaches a constant value, the integral equals 0. Where am I going wrong in this? Is the integral undefined for this problem, or are my math skills just lacking? The integral should be dependent of I and y, but not x0.

POSTED BY: Bastian Andersen
2 Replies
Posted 9 months ago

I'm going to assume that I is not the imaginary number but a positive real constant. Using Mathematica I get the following (after changing I to i which might be the issue with Wolfram|Alpha):

Integrate[i*y^2/((x - x0)^2 + y^2), {x, -\[Infinity], \[Infinity]}, Assumptions -> {x0 \[Element] Reals, y > 0, i > 0}]
(* i \[Pi] y *)
POSTED BY: Jim Baldwin

Thanks for the reply! I found similar last night, when setting y=1 and I=1 and taking the numerical integral, thus getting pi, and then doubling y / I.

Now I also realized my mistake... Missing a minus sign from the result from Wolfram|Alpha.

F(x) = Iyatan((x-x0)/y) + C In interval -inf...inf A = Iy(atan(inf)-atan(-inf)) A = Iypi

POSTED BY: Bastian Andersen
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