Degenerate Gaussian Integral

I have an integral over a subspace of $$\mathbb{R}^n \times \mathbb{R}^n$$ with an integrand of the form $$\exp\left(-\frac{1}{2}\left[||u^2|| + \langle u, v \rangle + ||v||^2\right]\right)$$ The subspace is exactly the space for which $$u_{i} = u_{n-i}$$ (assume $$n$$ is even). In other words, $$v$$ is a true $$n$$-dimensional vector, whereas $$u$$ is two copies of an $$n/2$$ dimensional vector.

In order to evaluate this integral, I have been thinking about it as a density over ALL of $$\mathbb{R}^n \times \mathbb{R}^n$$ of two correlated Gaussians with covariance given by the block matrix $$\begin{bmatrix} A & B \\ B^t & C \end{bmatrix}$$ where $$A,C$$ are $$n \times n$$ identity matrices and $$B$$ is the block matrix $$\begin{bmatrix} I_{n/2} & I_{n/2} \\ 0 & 0 \end{bmatrix}$$ However, given that this is degenerate, I am having trouble finishing the computation. Would greatly appreciate any tips!

• Just want to make sure you don't actually mean inner product of u and v to have a coefficient of 2--that's how these integrals usually look because inner product of u+v with itself yields a coefficient of 2 on the middle term. This would make Iosef's answer even simpler. – Sheridan Grant Sep 15 at 22:34

$$\newcommand{\R}{\mathbb{R}}$$ Let $$U:=\{u\in\R^n\colon u_i=u_{n-i}\ \forall i\}$$ be your $$n/2$$-dimensional subspace. I am assuming that your integral is with respect to the product of the Lebesgue measures on $$U$$ and $$\R^n$$, and I will denote those measures by $$du$$ and $$dv$$, respectively. So, if $$\cdot$$ denotes the dot product, your integral is \begin{align} I&:=\int_U du\,\int_{\R^n}dv\,\exp\big(-(\|u\|^2+u\cdot v+\|v\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dv\,\exp\big(-(3\|u\|^2/4+\|v+u/2\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dw\,\exp\big(-(3\|u\|^2/4+\|w\|^2)/2\big) \\ &=(2\pi)^{n/2}\int_U du\,\exp\big(-3\|u\|^2/8\big) \\ &=(2\pi)^{n/2}\int_{\R^{n/2}} dt\,\exp\big(-3\|t\|^2/8\big) \\ &=(2\pi)^{n/2}(2\pi)^{n/4}(4/3)^{n/4}. \end{align} The penultimate equality here holds because both the Euclidean norm and the Lebesgue measure are rotation invariant, whereas the dimension of $$U$$ is $$n/2$$; in fact, this is how the Lebesgue measure on $$U$$ can/should be defined: by the condition that $$\begin{equation} \int_U du\,f(u)=\int_{\R^{n/2}} dt\,f(Tt) \end{equation}$$ for all nonnegative Borel-measurable functions $$f\colon U\to\R$$, where $$T\colon\R^{n/2}\to U$$ is a linear isometry.