Calculating the expectation of $\frac{1}{|W \cap W_{tU}|}$ where W is a square with side $a$.

Question

I am dealing with a following problem: Calculate $$\mathbb{E}f(tU), t \geq 0$$ where U is a random variable with uniform distribution on the unit circle and $$f(h)=\frac{1}{|W \cap W_h|}$$ where $W$ is a square with side $a$ and $W_h$ is $W$ translated by the vector $h$. $|S|$ denotes the area of $S$. I would like to get an answer in closed form as a function of t. Is this possible? So far, I've gotten to $$ \mathbb{E}f(tU)=\frac{1}{2\pi}\int_{||u||=1}\frac{1}{|W \cap W_{tu}|} \mathrm{d}u = \frac{1}{2\pi}\int_{||u||=1}\frac{1}{(a-t|u_1|)(a-t|u_2|)} \mathrm{d}u_1\mathrm{d}u_2. $$ I assume that $a \geq 1$, since i will be probably using the result for $a=1$. So far, I've had no luck dealing with this integral and I am starting to think that it might not have a closed form. Would this maybe be possible for $W$ being a disk?

Answer 1

See comments under the question. Introducing $b=t/a$ in the first equality, $$\int_{-\pi}^{\pi}\frac{1}{(a-t|cos(\alpha)|)(a-t|sin(\alpha)|)} \mathrm{d}\alpha = \frac{1}{a^2} \int_{-\pi}^{\pi}\frac{1}{(1-b|cos(\alpha)|)(1-b|sin(\alpha)|)} \mathrm{d}\alpha = \\ = \frac{1}{a^2}4\int_0^{\pi/2}\frac{\mathrm{d}\alpha}{(1-\beta\cos\alpha)(1-\beta\sin\alpha)}.$$

Solving the last integral in Mathematica yielded the following for ∈(0,1): $$\frac{4}{a^2}\frac{2 \left(\left(b^2-1\right) \log (1-b)+\sqrt{1-b^2} \left(-\tan ^{-1}\left(\frac{b-1}{\sqrt{1-b^2}}\right)+\tan ^{-1}\left(\frac{b+1}{\sqrt{1-b^2}}\right)+\sin ^{-1}(b)\right)\right)}{b^4-3 b^2+2}.$$