triple integral $\iiint e^{-x^2-y^2-z^2+xy+yz+xz} \,dx\,dy \,dz$

Question

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.

Answer 1

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.

Answer 2

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 $A_q$, $$ \iiint_{\mathbb{R}^3}\exp\left[-q(x,y,z)\right]\,d\mu = \sqrt{\frac{\pi^3}{\det A_q}}.$$

Answer 3

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 $v$.)