Evaluate the triple integral problem

Question

Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x^2-2xy)e^{-Q}dxdydz$

, where $Q=3x^2+2y^2+z^2+2xy$.

Answer 1

Let $\Sigma$ be the symmetric $3 \times 3$ matrix which is specified by $\frac{1}{2}\mathrm{x}^{\mathsf{T}}\Sigma^{-1} \mathrm{x} = Q(\mathrm{x})$ for $\mathrm{x}\in \Bbb{R}^3$. Then $\Sigma$ satisfies

$$ \Sigma^{-1} = \begin{pmatrix} 6 & 2 & 0 \\ 2 & 4 & 0 \\ 0 & 0 & 2 \end{pmatrix}, \qquad \Sigma = \frac{1}{10} \begin{pmatrix} 2 & -1 & 0 \\ -1 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}, \qquad \det \Sigma = \frac{1}{40}. $$

Now introduce a random vector $X = (X_1, X_2, X_3)$ which has multivariate normal distribution $\mathcal{N}(0, \Sigma)$. Then its density is given by

$$ f_{X}(\mathrm{x}) = \frac{1}{\sqrt{(2\pi)^3\det \Sigma}} \exp\{ -\tfrac{1}{2}\mathrm{x}^{\mathsf{T}}\Sigma^{-1}\mathrm{x} \}. $$

From this, we can compute the integral as follows:

\begin{align*} \int_{\Bbb{R}^3} (x^2 - 2xy)e^{-Q(\mathrm{x})} \, d\mathrm{x} &= \sqrt{(2\pi)^3\det \Sigma} \int_{\Bbb{R}^3} (x^2 - 2xy) f_{X}(\mathrm{x}) \, d\mathrm{x} \\ &= \sqrt{(2\pi)^3\det \Sigma} \cdot \Bbb{E}[X_1^2 - 2X_1 X_2] \\ &= \sqrt{(2\pi)^3\det \Sigma} \cdot (\Sigma_{11} - 2\Sigma_{12}), \end{align*}

where $\Sigma_{ij} = \operatorname{Cov}(X_i, X_j)$ is the $(i,j)$-entry of the covariance matrix $\Sigma$. Computing everything, we obtain

$$ \int_{\Bbb{R}^3} (x^2 - 2xy)e^{-Q(\mathrm{x})} \, d\mathrm{x} = \frac{2\pi^{3/2}}{5\sqrt{5}}. $$

Answer 2

\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}