# Area under the following curve (Lorentzian type)

1. Sep 18, 2009

### Rajini

Hello all,
I need to know the area under the following curve (Lorentzian type):
$$f(x)=\frac{I(\Gamma/2) ^2}{(x-x_0)+(\Gamma/2) ^2}=\frac{I\Gamma ^2}{4(x-x_0)+\Gamma ^2}$$.
In the above Loretzian function we all know that $$\Gamma$$ is fullwidth at half maximum, $$x_0$$ is peak's centre and $$I$$ is the height.
What is the area under Lorentzian function shown above (for $$x_0=0$$)?
$$\int f(x)=?$$.
thanks

2. Sep 18, 2009

### LCKurtz

Re: Lorentzian_doubt

$$\int \frac {k}{b(x-a) + c}\, dx = \frac {k \ln(bx - ba + c)}{b} +C$$

3. Sep 18, 2009

### Rajini

Re: Lorentzian_doubt

Hello Kurtz,