Computing a Fourier-Transform

Consider the $\ell^1$ and $\ell^\infty$ norms,
$$ \lVert v \rVert_1 = |\langle v, \mathbf{e}_1 \rangle| + |\langle v, \mathbf{e}_2 \rangle|$$
and
$$ \lVert v \rVert_\infty = \max(|\langle v, \mathbf{e}_1 \rangle|, |\langle v, \mathbf{e}_2 \rangle|) $$ We may write $$ \lVert Av \rVert_\infty = \frac{1}{2} \Big( |\langle Av, \mathbf{e_1} \rangle - \langle Av, \mathbf{e_2} \rangle| + |\langle Av, \mathbf{e_1} \rangle + \langle Av, \mathbf{e_2} \rangle| \Big) . $$ Using the fact that $\max(a,b) = \tfrac{1}{2} (|a-b| + |a+b|)$. Now applying linearity and the fact that $A$ is an orthogonal matrix preserving the inner product yields $$ \lVert Av \rVert_\infty = \frac{1}{\sqrt{2}} \Big ( |\langle Av, A\mathbf{e_1} \rangle| + |\langle Av, A\mathbf{e_2} \rangle| \Big) = \frac{1}{\sqrt{2}} \lVert v \rVert_1. $$ Since $M$ is the ball of radius $1/\sqrt{2}$ around $(0, \frac{1}{\sqrt{2}})$ in the $\ell^1$ norm, $AM$ is the ball of radius $1/2$ around $A(0, \frac{1}{\sqrt{2}}) = (1/2, 1/2)$. We see that $AM = [0,1]^2$, as $$\max(|x-1/2|, |y-1/2|) \leq 1/2 \implies x \in [0,1] \text{ and } y \in [0, 1].$$ Now we calculate $$ \widehat{\chi_M}(\xi) = \int_{\mathbb{R}^2} \chi_M(\mathbf{v}) \exp(2\pi i \langle \xi, \mathbf{v} \rangle) \, d\mathbf{v}. $$ Using a change of variables $v = A^{-1}u$, and calculating the Jacobian $\det A^{-1} = 1$ yields $$ \widehat{\chi_M}(\xi) = \int_{\mathbb{R}^2} \chi_M(A^{-1}\mathbf{v}) \exp(2\pi i \langle \xi, A^{-1}\mathbf{v} \rangle) \, d\mathbf{v} = \widehat{\chi_M}(\xi) = \int_{\mathbb{R}^2} \chi_{AM}(\mathbf{v}) \exp(2\pi i \langle A\xi, \mathbf{v} \rangle) \, d\mathbf{v}. $$ Using $\xi = (x,y)$, the integral is then $$ \widehat{\chi_M}(\xi) = \int_{0}^{1}\int_0^1 \exp(2\pi i \cdot \tfrac{x-y}{\sqrt{2}} \cdot s) \exp(2\pi i \cdot \tfrac{x+y}{\sqrt{2}} \cdot t) \, ds dt. $$ Splitting the integral as a product and evaluating yields $$ \widehat{\chi_M}(\xi) = \int_0^1 \exp(2\pi i \cdot \tfrac{x-y}{\sqrt{2}} \cdot s) ds \int_0^1 \exp(2\pi i \cdot \tfrac{x+y}{\sqrt{2}} \cdot t) dt, $$ which is trivial to compute. Specifically, we have

$$\widehat{\chi_M}(x,y) = \frac{(\exp(2\pi i \cdot \tfrac{x-y}{\sqrt{2}}) -1) (\exp(2\pi i \cdot \tfrac{x+y}{\sqrt{2}})-1)}{2\pi^2(x^2-y^2)}$$