# Integrating the product of normal and exponential distributions

I'm trying to model the foraging behaviour of pollinating insects as a function of range. The basic idea that I have is to use a normal dispersal kernel to quantify the probability of an individual visiting a flower as a function of range, parameterised by $\sigma_s$ which is specific to any species. I would then like to integrate over all species, and give a each species a weight which is taken from an exponential distribution, parameterised by $\lambda$. Integrating over $\sigma$ gives me a kernel representing the aggregate over the whole community, parameterised simply by $\lambda$.(Basically a rank abundance curve).

This gives me the integral:

$$\int_0^\infty \frac{1}{\lambda\sigma\sqrt{2\pi}}e^{-\frac{r^2}{2\sigma^2}-\frac{1}{\lambda}\sigma} \ \ d\sigma$$

I have tried integration by parts, using the normal and exponential components of the integral, but haven't been able to something I can integrate. I have a access to maxima.

Can anyone point me in the right direction? I may also wish to use this as a general method, and try different functions for the species kernel and community probability weighting. A general method of tackling these problems would be the ideal solution. It's a bit beyond any maths that I have studied previously.

• If the variance of your Gaussian is exponentially distributed, then you end up with a Laplace distribution (double sided exponential). Having the standard deviation being exponentially distributed is more complicated. You might try finding the moment generating function, for which Fubini's theorem may be helpful. Then you can either look for the distribution that goes with your MGF, or do a countour integral for the inverse transform form the MGF back to the probability distribution. I didn't try, so don't know how hard that would be to do. – jdods Apr 24 '16 at 15:26

We need to take the integral:

$$I(a,b)=\int_0^\infty \frac{1}{t}e^{-a^2/t^2-b t} dt$$

I've got rid of all unnecessary constants and introduced new parameters:

$$a^2=\frac{r^2}{2},~~~b=\frac{1}{\lambda}$$

Mathematica gives the following 'closed form' for the above integral:

$$I(a,b)=\frac{1}{2\sqrt{\pi}} G_{3,0}^{0,3}\left( 0,0, \frac{1}{2} \left| \frac{(ab)^2}{4} \right.\right)$$

Here $G$ is the Meijer G-function. I wasn't able to find if this case of G-function simplifies to something more useful.

For example, there is a similar form of G-function, which simplifies to a modified Bessel function:

$$K_{0 }(x)=\frac {1}{2}G_{0,2}^{2,0} \left( \begin{array}( - \\ 0, 0 \end{array} \left| \frac{x^2}{4} \right.\right)$$

$$K_{1 }(x)=\frac {1}{2}G_{0,2}^{2,0} \left( \begin{array}( - \\ \frac{1}{2}, -\frac{1}{2} \end{array} \left| \frac{x^2}{4} \right.\right)$$

So, this might be something similar, but more complicated.