multivariate normal moment derivation I am having trouble deriving the mean for a multivariate normal for $\mathbf{x} \sim \mathbb{N}(\mathbf{m},\Sigma)$:
$$
\mathbb{E}[\mathbf{x}]= \int_{R^d} \mathbf{x} \frac{1}{(2\Pi)^{d/2}(det(\Sigma))^{1/2}}exp(-1/2 \mathbf{x}^T \Sigma^{-1}\mathbf{x})d \mathbf{x}
$$
where I want to show that $\mathbb{E}[\mathbf{x}]= \mathbf{m}$. Tried to do it by a subustitution: $\mathbf{y}:= \mathbf{x} - \mathbf{m}$:
$$
\mathbb{E}[\mathbf{x}]= \int_{R^d} (\mathbf{y}+m) \frac{1}{(2\Pi)^{d/2}(det(\Sigma))^{1/2}}exp(-1/2 \mathbf{y}^T \Sigma^{-1}\mathbf{y})d \mathbf{y}\\
= \frac{1}{(2\Pi)^{d/2}(det(\Sigma))^{1/2}} \left(\int_{R^d}\mathbf{y} exp(-1/2 \mathbf{y}^T \Sigma^{-1}\mathbf{y})d \mathbf{y}+ \int_{R^d}\mathbf{m} \,  exp(-1/2 \mathbf{y}^T \Sigma^{-1}\mathbf{y})d \mathbf{y}\right)\\
$$
Next I set $\mathbf{y}:= \mathbf{V} \mathbf{z}$ with $\mathbf{V}$ being the matrix of eigenvectors (dimension $d x d$)
$$
= \frac{1}{(2\Pi)^{d/2}(det(\Sigma))^{1/2}} \left(\int_{R^d}\mathbf{V} \mathbf{z} \, exp(-1/2 \mathbf{V}^T \mathbf{z}^T \Sigma^{-1}\mathbf{V} \mathbf{z})d \mathbf{z}+ \int_{R^d}\mathbf{m} \,  exp(-1/2 \mathbf{V}^T \mathbf{z}^T \Sigma^{-1}\mathbf{V} \mathbf{z})d \mathbf{z}\right)\\
= \frac{1}{(2\Pi)^{d/2}(det(\Sigma))^{1/2}} \left(\int_{R^d}\mathbf{V} \mathbf{z} exp( -1/2 \sum_{k=1}^d\frac{1}{\lambda_k}z_k^2) d \mathbf{z}+\int_{R^d} \mathbf{m} \,  exp( -1/2 \sum_{k=1}^d\frac{1}{\lambda_k}z_k^2) d \mathbf{z}  \right)
$$
where $\lambda_k$'s are the eigenvalues of $\mathbf{V}$. I have no clue how to proceed from here. Maybe partial integration? Any help greatly appreciated.
 A: You're almost there. In your expression
$$
\frac{1}{(2\pi)^{d/2}(\operatorname{det}(\Sigma))^{1/2}} \left(\int_{R^d}\mathbf{y} \exp(-\frac12 \mathbf{y}^T \Sigma^{-1}\mathbf{y})d \mathbf{y}+ \int_{R^d}\mathbf{m} \,  \exp(-\frac12 \mathbf{y}^T \Sigma^{-1}\mathbf{y})d \mathbf{y}\right)\\
$$
the second integral equals $\mathbf m$ because you can pull the constant $\mathbf m$ out, leaving the integral of a $\mathbb N(0,\Sigma)$ density (the normalizing constant is still there), which equals 1.
As for the first integral, you want it to equal zero. You are right to substitute $\mathbf y=\mathbf V\mathbf z$. Leaving aside a constant multiplier, this leads to a integral of the form
$$ \int_{R^d} \mathbf z f(\mathbf z)\,d\mathbf z$$
where $f(\mathbf z)$ factors into the product of $d$ univariate densities, each depending on only one coordinate of $\mathbf z$: $$f(\mathbf z) = f_1(z_1) \cdot f_2(z_2) \cdots \cdot f_d(z_d).$$ Since each $f_k(z_k)=\exp\left( -z_k^2/2\lambda_k\right)$ is an even function of $z_k$, the integral $\int\mathbf zf(\mathbf z)\,d\mathbf z$ is zero.
