TITLE: normalization factor for restricted density
QUESTION [5 upvotes]: Bounty update: this can be solved by change of basis, but I'm intrigued by David's solution relying on Fourier Transform of Dirac Delta function, so the bounty is for whoever finds a way to fix his solution to give the right result.
Suppose I have a non-negative real-valued function over $d$-dimensional real vectors as follows
$$f(\mathbf{x})=\exp(-\mathbf{x}' A \mathbf{x})$$
Where $A$ is some symmetric positive definite $d\times d$ matrix. What is the normalization factor to turn this into a valid density over the following set?
$$S_d=\{(x_1,\ldots,x_d)\in \mathbf{R}^d | \sum_i x_i=0 \}$$
Below, David Bar Moshe gives a general solution to computing that integral over space orthogonal to some vector $v$, but I suspect it has a mistake because the answer depends on the norm of $v$.
In particular, suppose $A$ is $d$-by-$d$ identity matrix. Let $v$ be a vector of all ones. Because of symmetry, integrating over space orthogonal to $v$ should be the same as as $d-1$ dimensional Gaussian integral, ie $\pi^{(d-1)/2}$, whereas David's solution gives
$$\frac{\pi^{(d-1)/2}}{d}$$

REPLY [3 votes]: This is a correction of the answer of David Bar Moshe edit: and generalizing it for the case, when $\mathbf{A}$ is degenerate. Using formulas for Fourier and inverse Fourier transforms we can write
$$f(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k\int_{-\infty}^{\infty} \mathrm{d}y f(y)e^{iky}. (1)$$
Suppose $S_d(y)=$"$S_d$ shifted by $(y,0,\dots,0)$" and
$f(y)=\int_{\mathbf{x} \in S_d(y)} \mathrm{d}\mathbf{x} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x}).$
Then using the formula (1) we will get
$$N^{-1}=f(0) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k\int_{-\infty}^{\infty} \mathrm{d}y f(y)e^{iky}  =$$
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k\int_{-\infty}^{\infty} \mathrm{d}y \int_{\mathbf{x} \in S_d(y)}\mathrm{d}\mathbf{x} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x} + iky) =$$
$$\frac{\|\mathbf{v}\|}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k \int_{\mathbf{x} \in \mathbb{R}^d}\mathrm{d}\mathbf{x} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x} + ik \mathbf{v}^T \mathbf{x}),$$
where $\mathbf{v}$ is vector, orthogonal to the plane we are interested in. For example, we can take $\mathbf{v}=(1,\dots,1)$. Note $\|\mathbf{v}\|$ in the last integral, appearing from change of coordinates from $y$ and coordinates in $S_d$ to coordinates in $\mathbb{R}^d$.
Now $N^{-1}=$
$$\frac{\|\mathbf{v}\|}{2\pi}\int_{-\infty}^{\infty} \mathrm{d}k \int_{\mathbf{x} \in \mathbb{R}^d}\mathrm{d}\mathbf{x} \exp(-(\mathbf{x}+\frac{i}{2}k\mathbf{A}^{-1}\mathbf{v})^T \mathbf{A}(\mathbf{x}+\frac{i}{2}k\mathbf{A}^{-1}\mathbf{v}) - \frac{k^2}{4} \mathbf{v}^T \mathbf{A}^{-1}\mathbf{v})=$$
$$\frac{\|\mathbf{v}\|}{2\pi} \frac{\pi^{d/2}}{\sqrt{\det{A}}} \frac{2\sqrt{\pi}}{\sqrt{\mathbf{v}^T \mathbf{A}^{-1}\mathbf{v}}}=\frac{\|\mathbf{v}\| \pi^{(d-1)/2}}{\sqrt{\det{A}}\sqrt{\mathbf{v}^T \mathbf{A}^{-1}\mathbf{v}}}=\frac{\|\mathbf{v}\| \pi^{(d-1)/2}}{\sqrt{\mathbf{v}^T \mathbf{C}\mathbf{v}}},$$
where $\mathbf{C}$ is adjugate matrix to $\mathbf{A}$. Initial integral and the answer both are continuous in $\mathbf{A}$ when $A$ restricted to $S_d$ is not degenerate. From this we can conclude, that equation
$N^{-1}=\frac{\|\mathbf{v}\| \pi^{(d-1)/2}}{\sqrt{\mathbf{v}^T \mathbf{C}\mathbf{v}}}$
is true whenever $\mathbf{A}$ restricted to $S_d$ is not degenerate (i.e. even if $\mathbf{A}$ is degenerate).