# Fourier transform of Gamma function and Poisson distribution

• Apr 29th 2009, 07:41 AM
melies
Fourier transform of Gamma function and Poisson distribution
I would like to know what are the Fourier transform for the Gamma function and its logarithm.

if f(x)=gamma(x) and g(x)=log[gamma(x)]. What are the expression for F(w) and G(w)?

Furthermore, I would be interested in knowing how to translate density functions to Fourier domain. I guess the typical formula can be applied but I wonder if there is some basic bibliography or previous work that can be useful.:confused:

I am not mathematicien but enginner in astronomical problems. I am interested in creating a likelihood function in frequencial plane. In synthesis imaging in astronomy, sources are modeled like a poisson distribution for low emissions, so a likelihood function can be design to compare data obtained with a telescope and a model that is modified iteratively until convergence.

log L= SUMj[(-h + p*log(h) - log(p!)]

Where p are my datas and h my model in each iteration. Both p and h are 2D-vectors. Hence, looking for the maximums of that function you reach to a Richardson-Lucy scheme.

I can go from discrete poisson distribution to the continuous one by substituting the factorial at the denominator by a gamma function. I think this must be done because factorial is only defined for integers and, in Fourier domain, I would work with complex.

Finally, in order to work in the frequencial plane I think I should/could calculate the Fourier transform of the likelihood function.

This is a hard work and any help will be fruitful. I also think I could define some distribution directly in frequencial domain to describe my physical process (maybe an uniform, erlang, do not know, etc) but knowing how a poisson distribution looks like in Fourier domain will be helpful.

• Apr 30th 2009, 08:36 AM
chisigma
Given a function of real variable $f(t)$, its Fourier Transform is defined as...

$F(\omega)= \int_{-\infty}^{+\infty} f(t)\cdot e^{-i \omega t}\cdot dt$ (1)

Necessary condition for existence of the integral (1) is the existence of the integral...

$\int_{-\infty}^{+\infty}|f(t)|\cdot dt$ (2)

... i.e. $f(*) \in L_{1}$. However this is not the case of the function $\Gamma(*)$ so that (Shake)...

Kind regards

$\chi$ $\sigma$
• Apr 30th 2009, 09:37 AM
melies
Re: Fourier transform of Gamma function and Poisson distribution

However, from the following paper:
http://www.emis.de/journals/HOA/IJMM.../37-402091.pdf

I can see that the gamma function can be defined as the Fourier Transform of e^(s*x)exp(-e^x).

gamma(z)=F[e^(s*x)exp(-e^x)]
z belongs to Complex.

By the property F[F(f(x))]=2*pi*f(-x) -double Fourier Transform- Can I not say that F[gamma(z)]=2*pi*e^(-s*x)exp(-e^-x)?

In any case, in the Continuous Poisson distribution I want to define (in order to transform it to Fourier domain) there is no reason to have negative values, hence, my gamma function will not be defined for negative values of x or it can be considered zero for x<0. Remember that this distribution would describe the emission in the sky and this is supposed to be positive. In that case, would be this incomplete gamma function integrable for all x?

I do not know if all this things make sense for you.

Thanks again.

[IMG]file:///tmp/moz-screenshot.jpg[/IMG][IMG]file:///tmp/moz-screenshot-1.jpg[/IMG]
• Apr 30th 2009, 11:58 AM
chisigma
I'm afraid it has been some misundestanding from me... You [correctly] spoke about Representation of Gamma Function as Fourier Transform and I have intended Fourier Transform of Gamma Function... two completely different questions... very sorry! (Worried)...

Kind regards

$\chi$ $\sigma$
• Apr 30th 2009, 01:20 PM
albi
As far as I know sufficient condition for Fourier transform of f to exist is to f be exponentially bounded (it is f(x) < Cexp(ax))

When we define the function $f(\omega) = \Gamma(\sigma + i \omega)$

f is bounded, so the Fourier transform exist. And it is $\hat{f}(x) = e^{\sigma x} \exp(-e^x)$.

So the fourier transform of gamma exists, when we consider it along the axis which is paralell to imaginary axis...