Simplify an integration Can the following integral be reduced to simpler terms?
$$\int_{-1}^0\mathrm{d}x_1 \int_{0}^1\mathrm{d}x_2 \int_{-1}^1\mathrm{d}x_3\, \delta(x_1+x_2+x_3) \exp(a_1x_1 + a_2x_2 + a_3x_3 + cx_3^2)$$
where $\delta(x)$ is Dirac's delta function, $a_1,a_2,a_3$ are real and $c$ is positive.
 A: The identity you need here is 
$$ \int_{\Omega}\delta(g(\mathbf x))f(\mathbf x)\text d\mathbf x = \int_{\Omega^\prime} \frac{1}{\|\nabla g(\mathbf x)\|} f(\mathbf x) \text d\Sigma $$
where $\Omega^\prime \subset \Omega$ is the set where $g(\mathbf x) = 0$ for $x\in\Omega$, and $\text d\Sigma$ is the surface measure on $\Omega^\prime$.
Here, $\mathbf x = (x_1,x_2,x_3)$, $g(\mathbf x) = x_1+x_2+x_3$, $\Omega = (-1,0)\times(0,1)\times(-1,1)$.

To evaluate, let $\mathbf r(\mathbf u)$ be a parameterization of $\Omega^\prime$. If $g(\mathbf x) = 0$ can be rearranged to say $x_i = h(\mathbf x)$ so that $h$ is independent of $x_i$ for some $i$, then we can choose $r_i = h(\mathbf x)$ and $r_j = x_j$ for $i\neq j$.
In our example, $g(\mathbf x) = x_1+x_2+x_3 = 0$ can be written as $x_3 = -(x_1+x_2) = h(\mathbf x)$ independent of $x_3$, so our parameterization of the surface is simply $\mathbf r(x_1,x_2) = (x_1,x_2,-(x_1+x_2))$. In $\mathbb{R}^3$, $\text d\Sigma = \| \partial_{x_1} \mathbf r \times \partial_{x_2} \mathbf r \| \text dx_1 \text dx_2$, which for our particular $\mathbf r$, is $\sqrt{\partial_{x_1} \mathbf r + \partial_{x_2} \mathbf r + 1} \text dx_1 \text dx_2 = \sqrt{3} \text dx_1 \text dx_2$.
Meanwhile, $\|\nabla g(\mathbf x)\| = \|(1,1,1)\| = \sqrt{3}$. So, $\frac{1}{\|\nabla g(\mathbf x)\|} \text d\Sigma = \frac{\sqrt{3}\text dx_1 \text dx_2}{\sqrt{3}} = \text dx_1 \text dx_2$. Thus, the integral simplifies to 
$$ \int_{-1}^0 \text dx_1 \int_0^1 \text dx_2 f(x_1,x_2,-(x_1+x_2)) $$
or, plugging in for $f$ and rearranging,
$$ \int_{-1}^0 \text dx_1 \int_0^1 \text dx_2 \exp((a_1-a_3)x_1+(a_2-a_3)x_2+c(x_1+x_2)^2) $$
This is not closed form integrable, so where you take it from here is up to you.
A: This is not a full answer, just too long to be a comment.
One can at least perform the integration on $x_3$ simply, since in the region of integration given, for any $x_1,x_2$ there is a $x_3$ such that $x_1 + x_2 + x_3 = 0$. Therefore one just needs to replace $x_3$ with $-x_1-x_2$, obtaining:
$$I = \int_{-1}^0\mathrm{d}x_1 \int_{0}^1\mathrm{d}x_2 \, \exp[(a_1-a_3)x_1 + (a_2-a_3)x_2 + c(x_1 + x_2)^2]$$
I don't know what else to do from here.
Update: I had an idea that lets me convert this to a single integration. Use the identity (Hubbard-Stratonovich transform):
$$\exp(\lambda x^2) = \int_{-\infty}^\infty \exp(-u^2/2 + \sqrt{2\lambda}u) \, \mathrm{d} u$$
to get rid of the squared term $(x_1 + x_2)^2$. We have:
$$\begin{aligned}
I &= \int\frac{\mathrm{d}u}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2}\int_{-1}^{0}\mathrm{d} x_{1}\int_{0}^{1}\mathrm{d}x_{2}\exp\left[\left(a_{1}-a_{3}\right)x_{1}+\left(a_{2}-a_{3}\right) x_{3}+u\sqrt{c}\left(x_{1}+x_{2}\right)\right] \\
  &= \int_{-\infty}^{\infty}\frac{\mathrm{d}u}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2}\frac{1-\mathrm{e}^{-a_{1}+a_{3}-u\sqrt{c}}}{a_{1}-a_{3}+u\sqrt{c}}\frac{\mathrm{e}^{a_{2}-a_{3}+u\sqrt{c}}-1}{a_{2}-a_{3}+u\sqrt{c}}
\end{aligned}$$
which is now a single integration. I still wonder if this can be simplified even more.
