# Integral over inersection of sheared and dilated cirlce with another circle.

I am currently working on a project relating to the topic of Wavelets and Shearlets. I am wanting to estimate a Shearlet transform of the indicator function of the ball of radius $$r$$ in $$\mathbb{R}^2$$.

This boils down to computing the following integral:

$$\int_{B_{a,s,t}} \psi_{a,s,t}(x)dx.$$ Where we have the following notation:

• $$a,s\in \mathbb{R}$$ and $$t \in \mathbb{R}^2$$.
• $$U_{\psi} = B_\rho$$ is the support of the shearlet $$\psi$$.
• $$B_{a,s,t} := (S_sA_a(B_\rho)+t)\cap B_r$$, where $$S_s = \begin{bmatrix} 1& s\\ 0 & 1\end{bmatrix} \ \ \ \text{and} \ \ \ A_a = \begin{bmatrix} a& 0\\ 0 & a^{1/2}\end{bmatrix}.$$
• $$\psi_{a,s,t}(x) = a^{-3/4}\psi(A_a^{-1}S_s^{-1}(x-t))$$

Somethings to know are:

• $$\int_{U_\psi}\psi(x)dx=0$$.
• We can parametrize $$\partial (S_sA_a(B_\rho)+t)$$ as the curve $$\gamma(\theta)= \begin{bmatrix} r(a\cos\theta + a^{1/2}s\sin\theta) +t_1\\ ra^{1/2}\sin\theta +t_2 \end{bmatrix} \ \ \ \theta\in[0,2\pi).$$ The approach I have tried is the following: Cauchy-Schwartz gives us $$\left|\int_{B_{a,s,t}} \psi_{a,s,t}(x)dx\right| \leq \lambda(B_{a,s,t})^{1/2}||\psi||_{L^2}.$$ Where $$\lambda$$ is the 2-dimensional Lebesgue Measure. Then since $$\partial B_{a,s,t}$$ can be seen as a simple closed piecewise smooth plane curve the Isoperimetric inequality tells us that $$\lambda(B_{a,s,t}) \leq \frac{1}{4\pi}l(\partial B_{a,s,t})^2$$ where $$l(\partial B_{a,s,t})$$ is the arc-length of the curve $$\partial B_{a,s,t}$$.

Intuitively and pictorially this approach is very simple, however it leads to having to solve a 2-dimensional quadratic equation with no coefficients zero. One could do this numerically but then one could also just compute the original integral numerically which would be easier. However I am not so much interested in exact numeric values as I am in analytic estimates.

Can you think of any other approach? Thanks in advance for any help.