Can I solve this integral with a squared sum in it?

Question

Title says it all. By now I have tried by hand and I think that it is indeed solvable, but I can't handle the very long terms.

I tried to run the thing through SAGEs integrator:

%var x, y, z, i, n
ai = function('ai')
xi = function('xi')
yi = function('yi')
zi = function('zi')

gauss(a,x,y,z) = a * e^(-(x^2+y^2+z^2)/2)
f(x,y,z) = (sum(  gauss(ai(i),x-xi(i), y-yi(i), z-zi(i))  ,i,1,n))^2

integral(integral(integral(f,x),y),z)

Which results in:

(x, y, z) |--> integrate(integrate(integrate(sum(ai(i)*e^(-1/2*x^2 - 1/2*y^2 - 1/2*z^2 + x*xi(i) - 1/2*xi(i)^2 + y*yi(i) - 1/2*yi(i)^2 + z*zi(i) - 1/2*zi(i)^2), i, 1, n)^2, x), y), z)

So I think it is telling me that it is not possible?

Funny thing is that if I remove the outer ^2 it has no problem with coming up with a result. Problem is that I need that to calculate the absolute area under the integral. Is there a better way to achieve this?

edit just to avoid confusion: I asked the same question on StackOverflow but it seemed to be a bad fit there.

Answer 1

Compared to the numeric solution this is the correct solution:

$\iiint\limits_{\rm I\!R^3} \left( \sum_{i=0}^n a_i e^{-\frac{(x-b_i)^2+(y-c_i)^2+(z-d_i)^2}{2}} \right)^2 = \frac{1}{8}\pi^{\frac{3}{2}} \sum_{i=0}^n \left( a_i^2 erf(x-b_i) erf(y-c_i) erf(z-d_i) \right) + \\ \frac{1}{4}\pi^\frac{3}{2} \sum_{i=0}^n \sum_{j=i}^n \left( a_i a_j erf(x - \frac{b_i+b_j}{2}) erf(y-\frac{c_i+c_j}{2}) erf(z-\frac{d_i+d_j}{2}) e^{ \left( - \frac{1}{4}(b_i-b_j)^2 - \frac{1}{4}(c_i-c_j)^2 - \frac{1}{4}(d_i-d_j)^2 \right) } \right) $

Jacks answer is not giving me the results I expect, but put me on the right track to solve this myself.

Answer 2

That is perfectly doable by hand. Just expand the square and exploit:

$$ \iiint_\mathbb{R^3} \exp\left(-\frac{(x-a_i)^2+(y-b_i)^2+(z-c_i)^2}{2}\right)\,d\mu = (2\pi)^{3/2},$$

$$ \iiint_\mathbb{R^3} \exp\left(-\frac{(x-a_i)^2+(x-a_j)^2+(y-b_i)^2+(y-b_j)^2+(z-c_i)^2+(z-c_j)^2}{2}\right)\,d\mu = \pi^{3/2}\exp\left(-\frac{(a_i-a_j)^2+(b_i-b_j)^2+(c_i-c_j)^2}{4}\right).$$ By setting $g(A,B,C)=\exp\left(-\frac{A^2+B^2+C^2}{4}\right)$ it follows that:

$$\iiint_{\mathbb{R}^3}\left(\sum_{i=1}^{n}a_i\cdot e^{-\frac{(x-b_i)^2+(x-c_i)^2+(x-d_i)^2}{2}}\right)^2\,d\mu =\\= (2\pi)^{3/2}\cdot\sum_{i=1}^{n}a_i^2+2(\pi)^{3/2}\cdot\sum_{1\leq i<j\leq n}a_i a_j\cdot g(b_i-b_j,c_i-c_j,d_i-d_j).$$