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Area under the following curve (Lorentzian type)

  1. Sep 18, 2009 #1
    Hello all,
    I need to know the area under the following curve (Lorentzian type):
    [tex]f(x)=\frac{I(\Gamma/2) ^2}{(x-x_0)+(\Gamma/2) ^2}=\frac{I\Gamma ^2}{4(x-x_0)+\Gamma ^2}[/tex].
    In the above Loretzian function we all know that [tex]\Gamma[/tex] is fullwidth at half maximum, [tex]x_0[/tex] is peak's centre and [tex]I[/tex] is the height.
    What is the area under Lorentzian function shown above (for [tex]x_0=0[/tex])?
    [tex]\int f(x)=?[/tex].
    please help me!
    thanks
     
  2. jcsd
  3. Sep 18, 2009 #2

    LCKurtz

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    Re: Lorentzian_doubt

    [tex]\int \frac {k}{b(x-a) + c}\, dx = \frac {k \ln(bx - ba + c)}{b} +C[/tex]
     
  4. Sep 18, 2009 #3
    Re: Lorentzian_doubt

    Hello Kurtz,
    thanks for your help!..

    bye
     
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