Prove that $\int_\mathbb{R^n} |(\lambda, x)|^pe^{-\pi |x|^2}dx= |\lambda|C_p$

Question

I was trying to prove that $$\int_\mathbb{R^n} |(\lambda, x)|^pe^{-\pi |x|^2}dx= |\lambda|C_p$$ where $p\in (0,+\infty)$, $\lambda \in \mathbb{R^n}$ and $C_p$ is a constant independent of n. I have already proved the equality for $n=1$ but I couldn't prove the general case. Any idea?

Answer 1

Assume that $\lambda\ne 0$ (otherwise equality is clear) and let $\{e_1,\ldots,e_n\}$ be a new orthonormal basis in $\mathbb{R}^n,$ such that $e_1=\frac{\lambda}{|\lambda|}.$ Then in the new coordinates $y_i=(x,e_i),$ $1\leq i\leq n,$ the integral transforms to $$ |\lambda|^p\int_{\mathbb{R}^n}|y_1|^pe^{-\pi(y^2_1+\ldots+y^2_n)}dy_1\ldots dy_n= $$ $$ =|\lambda|^p\int_\mathbb{R} |y_1|^pe^{-\pi y^2_1}dy_1\bigg(\int_\mathbb{R} e^{-\pi y^2}dy\bigg)^{n-1} $$ In the brackets there is a Gauss integral which is known to be equal to $1$

Answer 2

Essentially the idea from @Georgii Riabov: Let $A$ be orthonormal matrix which switch $\lambda$ to $e_{1}$, then \begin{align*} \int|(\lambda,x)|^{p}e^{-\pi|x|^{2}}dx&=\int|(\lambda,A^{-1}x)|^{p}e^{-\pi|x|^{2}}dx\\ &=\int|(A\lambda,x)|^{p}e^{-\pi|x|^{2}}dx\\ &=\int|(e_{1},x)|^{p}e^{-\pi|x|^{2}}dx\\ &=\int|x_{1}|^{p}e^{-\pi|x|^{2}}dx, \end{align*} now use Fubini and Gaussian.

Note that the Jacobian of unitary matrix is $1$ and $A^{\ast}=A^{-1}$.