- #1

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[tex] PDF(x) = \int e^{-r} \, dy = \int e^{-\sqrt{x^2+y^2}} \, dy [/tex]

- Thread starter touqra
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- #1

- 287

- 0

[tex] PDF(x) = \int e^{-r} \, dy = \int e^{-\sqrt{x^2+y^2}} \, dy [/tex]

- #2

DrDu

Science Advisor

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so $$

f(x_0)=\int \exp(-r) dy|_{x=x_0}=\int dx dy \exp(-r) \delta(r \cos \phi -x_0)=$$

$$=\int dr d\phi r \exp(-r) \frac{1}{2\pi} \int dk \exp(ik (r \cos \phi -x))$$

Now we first integrate over ##\phi## using ##\int d\phi \exp(ikr\cos\phi)=2\pi J_0(kr)##:

$$

f(x)=\frac{1}{2\pi}\int dk \int dr r \exp(-r)2 \pi J_0(kr)\exp(ikx_0)

$$

The integral over r I found in Magnus Oberhettinger, Formeln und Saetze fuer die speziellen Funktionen der Physik, p33

##\int_0^\infty \exp(-at)J_\nu(bt)t^\nu dt=\frac{(2b)^\nu \Gamma(\nu+1/2)}{(a^2+b^2)^{\nu+1/2} \sqrt{\pi}}##

so that (setting x_0=x)

$$ f(x)= \int dk (k^2+1)^{-1/2} \exp(ikx)$$

The Fourier transform is standard and yields the MacDonald function.

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