Vtyjkyuk

Can the following integral be reduced to simpler terms?

$$int_-1^0mathrmdx_1 int_0^1mathrmdx_2 int_-1^1mathrmdx_3, delta(x_1+x_2+x_3) exp(a_1x_1 + a_2x_2 + a_3x_3 + cx_3^2)$$

where $delta(x)$ is Dirac's delta function, $a_1,a_2,a_3$ are real and $c$ is positive.

The identity you need here is
$$ int_Omegadelta(g(mathbf x))f(mathbf x)text dmathbf x = int_Omega^prime frac1nabla g(mathbf x) f(mathbf x) text dSigma $$

where $Omega^prime subset Omega$ is the set where $g(mathbf x) = 0$ for $xinOmega$, and $text dSigma$ 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 $ineq j$.

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 $mathbbR^3$, $text dSigma = | partial_x_1 mathbf r times partial_x_2 mathbf r | text dx_1 text dx_2$, which for our particular $mathbf r$, is $sqrtpartial_x_1 mathbf r + partial_x_2 mathbf r + 1 text dx_1 text dx_2 = sqrt3 text dx_1 text dx_2$.

Meanwhile, $|nabla g(mathbf x)| = |(1,1,1)| = sqrt3$. So, $frac1nabla g(mathbf x) text dSigma = fracsqrt3text dx_1 text dx_2sqrt3 = text dx_1 text dx_2$. 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.

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:

$$I = int_-1^0mathrmdx_1 int_0^1mathrmdx_2 , exp[(a_1-a_3)x_1 + (a_2-a_3)x_2 + c(x_1 + x_2)^2]$$

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):

$$exp(lambda x^2) = int_-infty^infty exp(-u^2/2 + sqrt2lambdau) , mathrmd u$$

to get rid of the squared term $(x_1 + x_2)^2$. We have:

$$beginaligned
I &= intfracmathrmdusqrt2pimathrme^-u^2/2int_-1^0mathrmd x_1int_0^1mathrmdx_2expleft[left(a_1-a_3right)x_1+left(a_2-a_3right) x_3+usqrtcleft(x_1+x_2right)right] \
&= int_-infty^inftyfracmathrmdusqrt2pimathrme^-u^2/2frac1-mathrme^-a_1+a_3-usqrtca_1-a_3+usqrtcfracmathrme^a_2-a_3+usqrtc-1a_2-a_3+usqrtc
endaligned$$

which is now a single integration. I still wonder if this can be simplified even more.