Expected value of inverse of a random variable

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.

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.

I fed your formula into Wolfram Alpha to compute indefinite integrals. To shorten things, I wrote $x$ for $AB$.

First the indefinite integral over $\theta$:

$$\int(x^2+r^2-2xr\cos\theta)^{-2}\;\mathrm d\theta = f_1(r,\theta)+f_2(r,\theta)+\text{const.}\\ f_1(r,\theta)= \frac{2 (r^2+x^2)}{(r^2-x^2)^3}\arctan\frac{(r+x) \tan\frac\theta2}{r-x} \\ f_2(r,\theta)= \frac{2 r x \sin\theta (r^2-2 r x \cos\theta+x^2)}{(r-x)^2 (r+x)^2}$$

Since $f_2$ has the same value at your integration bounds, i.e. $f_2(0,r)=f_2(2\pi,r)$, these two cancel out. Therefore $f_2$ is irrelevant for the definite integral. $f_1$ is a different matter, since $\arctan$ is a multi-valued function so you have to keep track of which branch you are on. You have a removable singularity whenever $\theta=(2k+1)\pi$ for some $k\in\mathbb Z$.

So consider the value of

$$\arctan\left(\frac{r+x}{r-x}\tan\frac\theta2\right)$$

as you move $\theta$ from $0$ to $2\pi$. The value at the beginning and at the end will be zero (mod $\pi$), but in between the value of the $\tan$ will pass exactly once through infinity, indicating exactly one change to a neighboring branch of the $\arctan$. So the value will be $\pm\pi$. The factor $\frac{r+x}{r-x}$ has almost no influence on this, since it doesn't affect the initial or final value of zero, and since it doesn't affect the number of branch switches either. It does however determine the sign: $r<x$ and therefore $r-x<0$, so this is a negative factor. So the whole expression will move continuously (except for the removable singularity) from $0$ to $-\pi$, and therefore its value in the definite integral is $-\pi$.

Next you want to integrate the $r$-dependent part of $f_1$, and take the density factor of $r$ into account:

$$\int\frac{2 (r^2+x^2)}{(r^2-x^2)^3}\;r\,\mathrm dr = -\frac{r^2}{(r^2-x^2)^2}+\text{const.}$$

If you plug $0$ and $R$ into this and subtract, you get

$$-\frac{R^2}{(R^2-x^2)^2} +\frac{0^2}{(0^2-x^2)^2} = -\frac{R^2}{(R^2-x^2)^2}$$

Now let's look at all those constants I omitted along the way. You wrote $\frac{1}{2\pi}\frac{2r}{R^2}$. The $r$ I included with my second integral. The $2$ cancels in your terms already. The $\pi$ cancels with the $-\pi$ from $f_1(2\pi,r)-f_1(0,r)$ except for the sign. So all that's left is a division by $-R^2$ which makes the total integral equal to

$$(R^2-x^2)^{-2}$$

A quick Monte-Carlo simulation shows that this value seems reasonably close to be believable.

• Thanks MvG, your explanation is clear. I don't know Wolfram Alpha can give closed form for the integration. Nov 4 '14 at 20:50
• @Steven: Wolfram alpha can do lots of stuff, but has severe restrictions on computation time (unless you pay for it). It is my experience that indefinite integrals often work in cases where the definite ones time out.
– MvG
Nov 4 '14 at 20:53