I am trying to solve a PDE and need to solve the following integral:
$$\displaystyle \int\limits_{R^3} e^{-(x + y -z)^2}e^{-((a-x)^2 + (b-y)^2 + (c-z)^2)} \,dx\, dy\, dz.$$
I managed to solve similar integrals previously by separating the variables and integrating the Poisson's (Gauss') integral, but this one has coupled variables in the first exponend:
$$ 2xy-2yz-2xz$$
I tried working in spherical coordinates, but it doesn't seem to work. Also, the answer has no $\pi$'s.
Any hints?
The trick here is to rewrite the two exponents in a more convenient form. The combined exponents have a simple polynomial form $$-(x+y-z)^2 -(a-x)^2 -(b-y)^2 -(c-z)^2$$ we can now "extract" all $z$-dependent terms and write them in a square-like form $$- 2 \left( z - \frac{x +y +c}{2} \right)^2$$ which we know how to integrate. The remaining terms are in $x$ and $y$ only and we had to add a correction term $\frac{(x+y+c)^2}{2}$. In the same fashion we can now extract the $y$-dependent terms, etc.
If you do this correctly you find that the original exponent can be written as: $$ - 2 \left(z - \frac{x +y +c}{2} \right)^2 - \frac{3}{2} \left( y - \frac{2 b +c-x}{3} \right)^2 - \frac{4}{3} \left(x - \frac{3 a -b+c}{4}\right)^2 - \frac{(a+b-c)^2}{4} $$ Placing this back in the original problem, you see that the integral can be rewritten as: $$ e^{- \frac{(a+b-c)^2}{4}} \int d x~e^{- \frac{4}{3} \left(x - \frac{3 a -b+c}{4}\right)^2} \int d y~e^{- \frac{3}{2} \left( y - \frac{2 b +c-x}{3} \right)^2} \int d z ~e^{- 2 \left(z - \frac{x +y +c}{2} \right)^2} $$ which you now can easily evaluate.
Note that what we have done here is shift the integration variable $z$ with a constant that happens to be dependent on $x$ and $y$, but that the value of integral does not depend on them, because $$ \int_{-\infty}^\infty d z e^{-2(z-\gamma)^2} = \sqrt{\frac{\pi}{2}} $$ for any constant $\gamma$. In this particular case we have $\gamma=\frac{x+y+c}{2}$. In a similar fashion we can do the integral over $y$ and finally the integral over $x$.
A different way of looking at this approach is to realise that we made a change of variables and describe ${\mathbb R}^3$ not by $(x,y,z)$ but by the independent variables $(u,v,w) = (x - \frac{3 a -b+c}{4},y - \frac{2 b +c-x}{3},z - \frac{x +y +c}{2})$, where the transformation is chosen in a special fashion such that the cross terms "disappear".