Why $\int_{\mathbb R^d}\frac{(e^{-2i\pi \xi\cdot z}-1)^2}{|z|^{d+1}}dz=\int_{\mathbb R^d}\frac{(e^{-2i\pi|\xi|e_1\cdot z}-1)^2}{|z|^{d+1}}dz$

Question

I was trying to compute the integral on the LHS, and in my solution it's written that $$\int_{\mathbb R^d}\frac{(e^{-2i\pi \xi\cdot z}-1)^2}{|z|^{d+1}}dz=\int_{\mathbb R^d}\frac{(e^{-2i\pi|\xi|e_1\cdot z}-1)^2}{|z|^{d+1}}dz,$$ but I don't understand why. Any explanation ?

Answer 1

For any unitary transform (rotation) $T$ so that $T\!\left(\frac{\xi}{|\xi|}\right)=e_1$, $$ \begin{align} \int_{\mathbb{R}^d}\frac{\left(e^{-2i\pi\xi\cdot z}-1\right)^2}{|z|^{d+1}}\,\mathrm{d}z =\int_{\mathbb{R}^d}\frac{\left(e^{-2i\pi|\xi|e_1\cdot T(z)}-1\right)^2}{|T(z)|^{d+1}}\,\mathrm{d}T(z) \end{align} $$