How Can I Find the PDF Along One Axis for Exponential Decay in 2D Space?

[touqra]

I'll like to know the probability density function for one of the x or y axis, given that there is an exponential decay of a material in two-dimensional space. So, that means I have to marginalize, say y and keep x, but I couldn't solve the integration. I even tried with Mathematica and Matlab. Mathematica couldn't solve it. Matlab gives a Bessel function when x == 1, but when x != 1, it couldn't solve it. Please help.

$$PDF(x) = \int e^{-r} \, dy = \int e^{-\sqrt{x^2+y^2}} \, dy$$

 

[DrDu]

Something like $K_0(|x|)$, where K is the MacDonald function.
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.