triple integral $\iiint e^{-x^2-y^2-z^2+xy+yz+xz} \,dx\,dy \,dz$ I need to solve this
triple integral $$\iiint e^{-x^2-y^2-z^2+xy+yz+xz} \,dx\,dy \,dz$$ 
where $V$ is all $\Bbb R^3$ 
I've spent on this task a few hours,
ok firstly
$$e^{-x^2-y^2-z^2+xy+yz+xz}=e^{-1/2[(x-y)^2+(y-z)^2+(z-x)^2]}$$
I tried to use theorm of the change of variables, but
this substitution doesn't make sense
$$x-y=a$$
$$y-z=b$$
$$z-x=c$$
also
$$e^{-x^2-y^2-z^2+xy+yz+xz}=e^{(x-1/2y^2-1/2z^2)^2+((\sqrt{3}/2)y-
(\sqrt3/2)z)^2}$$
Maybe here change of variables'll be OK.
Please, help me. It's my homework, so important and I have no idea what I should do next.
 A: Since:
$$ \iiint_{\mathbb{R}^3} \exp\left[-\left(a^2 x^2+b^2 y^2+c^2 z^2\right)\right]\,d\mu = \frac{\pi^{3/2}}{|abc|}$$
by Fubini's theorem, in order to compute our integral it is enough to find the product of the eigenvalues of the matrix associated with the quadratic form $q(x,y,z)=x^2+y^2+z^2-xy-yz-xz$, i.e.
$$ \det\begin{pmatrix}1 & -\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & 1 & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} & 1\end{pmatrix}=\color{red}{0} $$
to be able to state that the original integral $\color{red}{\text{is not}}$ converging. 
In the general case, if $q(x,y,z)$ is a positive definite quadratic form associated with a matrix $A_q$,
$$ \iiint_{\mathbb{R}^3}\exp\left[-q(x,y,z)\right]\,d\mu = \sqrt{\frac{\pi^3}{\det A_q}}.$$
A: To pursue the change-of-variables method further, note that with $a$ and $b$ defined as originally done, we have
$$
x^2 + y^2 + z^2 - xy - xz - yz = \frac{1}{2} \left[ (x - y)^2 + (y - z)^2 + (z - x)^2 \right] = \frac{1}{2} \left[ a^2 + b^2 + (a + b)^2 \right],
$$
since $a + b = x - z$.  Rearranging, we then get
$$
\frac{1}{2} \left[ a^2 + b^2 + (a + b)^2 \right] = \frac{1}{4} \left[ 3 (a + b)^2 + (a - b)^2 \right],
$$
since $a^2 + b^2 = \frac{1}{2}(a + b)^2 + \frac{1}{2}(a - b)^2$.  We can therefore use a different set of coordinates:
$$
u = a + b = x - z, \qquad v = a - b = x - 2y + z, \qquad w = z
$$
(Really, $w$ can be pretty much anything—see below.)  The integral then becomes, up to an overall constant coming from the Jacobian,
$$
\iiint e^{-\frac{3}{4}u^2 - \frac{1}{4} v^2} \, du \, dv\, dw.
$$
Since there is no $w$-dependence in the integrand, and since its range of integration will run from $-\infty$ to $\infty$, we conclude that the integral does not converge.
The dummy coordinate $w$, by the way, can be chosen to be any coordinate that linearly dependent on $x, y, z$ and that is not degenerate with $u$ and $v$ (i.e., one that doesn't cause the Jacobian matrix to have zero determinant, which basically means something that is not a linear combination of $u$ and $v$.) 
A: I think we can do it with simple calculus, in addition to what has already been written, to clarify things. Starting with using Fubini's lemma to "separate" the integrals order.
I will also use this notation:
$$\int_{-\infty}^{+\infty} = \int$$ 
For the sake of simplicity.
$$\iiint e^{-x^2 - y^2 - z^2 + xy + xz + zy}\ \text{d}x\ \text{d}y\ \text{d}z = \int e^{-x^2}\ \text{d}x \int e^{-y^2 + xy}\ \text{d}y \int e^{-z^2 - z(x+y)}\ \text{d}z$$
Now we use an important result from elementary calculus which states
$$\int_{-\infty}^{+\infty} e^{-ax^2 + bx}\ \text{d}x = \sqrt{\frac{\pi}{a}} e^{b^2/4a}$$
Thence applying that to our integral (with $a = 1$ and $b = x+y$) we get:
$$\iiint e^{-x^2 - y^2 - z^2 + xy + xz + zy}\ \text{d}x\ \text{d}y\ \text{d}z = \int e^{-x^2}\ \text{d}x \int e^{-y^2 + xy}\ \text{d}y \left(\sqrt{\pi} e^{(x+y)^2/4}\right)$$
Arranging and we have
$$\sqrt{\pi}\int e^{-x^2 + x^2/4}\ \text{d}x \int e^{-y^2 + xy + y^2/4 + xy/2}\ \text{d}y$$
Using the same procedure we get:
$$\sqrt{\pi}\int e^{-x^2 + x^2/4} \left(2 e^{3x^2/4 + 3xy/2}\right)\ \text{d}x$$
namely 
$$2\frac{\pi}{\sqrt{3}}\int e^{-3x^2/4 + 3x^2/4}\ \text{d}x = 2\frac{\pi}{\sqrt{3}}\int_{-\infty}^{+\infty} 1 \text{d}x = \infty$$
Thence it does not converge. 
