I am reading an article where the authors say that $$\int Q^{p}(y) e^{-\frac{y\cdot z}{|z|}}dy = \int Q^{p}(x) e^{-x_1}dx$$ since $Q$ is a radially symmetric function. Note here $x=(x_1,x_2,\cdots,x_n)\in \mathbb{R}^{n}$ is the usual vector in the canonical basis and $Q:\mathbb{R}^{n}\to \mathbb{C}.$
I can intuitively understand why this is true. In some sense, we orient our axis in the direction of $\frac{z}{|z|}$ but I am not sure how to justify this mathematically. Perhaps someone could give a rigorous explanation of this equality?
Let $x = x_i\hat{x}_i$. Pick a rotation matrix $R \in O(n)$ that takes $\hat{x}_1$ to $\frac{z}{|z|}$ (both have unitary norm), and let $x = R^Ty$. Then $$\frac{y \cdot z}{|z|} = Rx \cdot R\hat{x}_1 = x \cdot \hat{x}_1 = x_1$$ and, by radial symmetry, $Q^p(y) = Q^p(Rx) = Q^p(x)$.
It follows that $$\int Q^p(y)e^{-y \cdot \frac{z}{|z|}}\text{d}y = \int Q^p(x)e^{-x_1}\text{d}(Rx) = \int Q^p(x)e^{-x_1}\text{d}x.$$