I see it now. Thank you very much!

@MartinHairer But in that case $$\int_{\mathbb{S}^1}\exp(x^\top[B-A]x)dx=\int_{\mathbb{S}^1}e^{-\lambda}dx=e^{-\lambda}$$ And \begin{align*} \frac{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Ax\right)dx}{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Bx\right)dx} &=\frac{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}(A-B)x\right)\exp\left(x^{\top}Bx\right)dx}{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Bx\right)dx}\\ &=\int_{\mathbb{S}^1}e^{\lambda}\frac{\exp(x^\top B x)}{\int_{\mathbb{S}^1}\exp(x^\top Bx)dx}dx\\ &=e^{\lambda} \end{align*} So the product is $1$.

@MattF. I am interested in "quotient times quotient". Sorry for the misleading title