My problem is as follows:
I have a point $A$ and a circle with center $B$ and radius $R$. Points $A$ and $B$ are fixed, also $A$ is outside of the circle. A random point $C$ is picked with uniform distribution in the area of disk $B$. My question is how to calculate the expected value of $AC^{-4}$. I am working with the path loss in Wireless Communication so $AC^{-4}$ measures how much energy is dissipated along the distance $AC$
My approach is to first denote $\theta$ as the angle between AB and BC then $\theta$ is uniformly distributed between $[0,2\pi]$. Denote $r$ as the distance of BC then distribution of $r$ in $[0,R]$ is $\frac{2r}{R^2}$. Using the formula $AC^2 = AB^2 + BC^2 – 2AB\times BC \times \cos\theta$ , I have
\begin{align}
E[AC^{-4}] & = \int_0^{2\pi}\int_0^R (AB^2 + BC^2 – 2AB\times BC \times
\cos\theta)^{-2} f_\theta f_r \, dr \, d\theta \\
& = \int_0^{2\pi}\int_0^R (AB^2 + r^2 – 2AB\times r \times \cos\theta)^{-2} \frac{1}{2\pi} \frac{2r}{R^2} \, dr \, d\theta
\end{align}
However, I am unable to solve this integration. I want to ask if anyone know any method that can give me the closed-form of the above expected value. If not, then maybe an approximation method that can give a closed-form is also good. Thanks in advance.
Best Answer
We choose Cartesian coordinates initially in order to avoid issues with a non-uniform density over the disk, then compute the integral using an appropriate transformation. Let $B = (0,0)$ and without loss of generality, suppose $A = (a,0)$ with $a > R > 0$. Furthermore, let $C = (x,y)$ such that $x^2 + y^2 \le R^2$. Then $$(AC)^{-4} = ((x-a)^2 + y^2)^{-2}$$ and the resulting expected value is given by the double integral over the disk of radius $R$ $$\frac{1}{\pi R^2} \iint_R ((x-a)^2 + y^2)^{-2} \, dx \, dy.$$ Then with the coordinate transformation $$(x,y) = (r \cos \theta, r \sin \theta),$$ with $dx \, dy = r \, dr \, d\theta$, the above becomes $$\frac{1}{\pi R^2} \int_{\theta = 0}^{2\pi} \int_{r = 0}^R ((r \cos\theta - a)^2 + (r \sin\theta)^2)^{-2} r \, dr \, d\theta = \frac{1}{(a^2-R^2)^2},$$ according to Mathematica. At this time, I do not have a step-by-step evaluation of the given integrand.