Evaluate the triple integral problem Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x^2-2xy)e^{-Q}dxdydz$
, where $Q=3x^2+2y^2+z^2+2xy$.

\begin{align}
I:&=\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}dy\int_{-\infty}^{\infty}dz\left(x^2-2xy\right)\exp\left(-3x^2 - 2y^2 - z^2 - 2xy\right)\\
&=\sqrt{\pi}\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}dy\left(x^2-2xy\right)\exp\left(-3x^2-2y^2-2xy\right) \tag1\\
\end{align}
We split the integral in $(1)$ into two parts:
\begin{align}
I_1&=\sqrt{\pi}\int_{-\infty}^{\infty}dxx^2\exp\left(-3x^2\right)\int_{-\infty}^{\infty}dy\exp\left(-2y^2-2xy\right) \tag2
\end{align}
We complete the square:
\begin{align}
-2y^2-2xy&=-2\left(y^2+xy\right)\\
&=-2\left(\left(y+\frac{x}{2}\right)^2-\frac{x^2}{4}\right)\\
&=-2\left(y+\frac{x}{2}\right)^2+\frac{x^2}{2}
\end{align}
Thus, $(2)$ becomes
\begin{align}
I_1&=\sqrt{\pi}\int_{-\infty}^{\infty}dxx^2\exp\left(-\frac{5}{2}x^2\right)\int_{-\infty}^{\infty}dy\exp\left(-2\left(y+\frac{x}{2}\right)^2\right)\\
&=\frac{\pi}{\sqrt{2}}\int_{-\infty}^\infty dxx^2\exp\left(-\frac{5}{2}x^2\right)\\
&=\frac{\pi^{\frac{3}{2}}}{5\sqrt{5}} \tag3
\end{align}
Then, we look at the second integral:
\begin{align}
I_2&=2\sqrt{\pi}\int_{-\infty}^\infty dx x\exp\left(-3x^2\right)\int_{-\infty}^\infty dyy\exp\left(-2y^2-2xy\right)\\
&=2\sqrt{\pi}\int_{-\infty}^\infty dx x\exp\left(-2x^2\right)\int_{-\infty}^\infty dyy\exp\left(-2\left(y+\frac{x}{2}\right)^2\right)\\
&=2\sqrt{\pi}\int_{-\infty}^\infty dx x\exp\left(-2x^2\right)\int_{-\infty}^\infty du\left(u-\frac{x}{2}\right)\exp\left(-2u^2\right) \tag4
\end{align}
Splitting $(4)$ again, we have
\begin{align}
I_{2,1}&=2\sqrt{\pi}\int_{-\infty}^\infty dx x\exp\left(-2x^2\right)\int_{-\infty}^\infty duu\exp\left(-2u^2\right)\\
&= 0
\end{align}
\begin{align}
I_{2,2}&=\sqrt{\pi}\int_{-\infty}^\infty dx x^2\exp\left(-\frac{5}{2}x^2\right)\int_{-\infty}^\infty du\exp\left(-2u^2\right)\\
&=\frac{\pi}{\sqrt{2}}\int_{-\infty}^\infty dx x^2\exp\left(-\frac{5}{2}x^2\right)\\
&=\frac{\pi^{\frac{3}{2}}}{5\sqrt{5}} \tag5
\end{align}
Adding up $(3)$ and $(5)$, we have

\begin{align}
\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}dy\int_{-\infty}^{\infty}dz\left(x^2-2xy\right)\exp\left(-3x^2 - 2y^2 - z^2 - 2xy\right)=\frac{2\pi^{\frac{3}{2}}}{5\sqrt{5}}
\end{align}

