I want to show that $\int_{\mathbb{R}^D}\exp(\mathbf{y}^\top\mathbf{S}\mathbf{y})\mathbf{y} \, d\mathbf{y}$, where $\mathbf{S}$ is a symmetric real matrix, is equal to $\mathbf{0}$ .
My intuition (although very immature) hints that it is true, but I can't show that — actually, it looks like the opposite is true.
$\int_{\mathbb{R}^D}\exp(\mathbf{y}^\top\mathbf{S}\mathbf{y})\mathbf{y}d\mathbf{y} = \int_{\mathbb{R}^D}\prod_{i,j}\exp(y_i y_j s_{ij})\mathbf{y} \, d\mathbf{y}=\mathbf{v}$;
$v_i=\int_{\mathbb{R}^D}\prod_{i,j}\exp(y_i y_j s_{ij})y_i \, d\mathbf{y} = \int_{-\infty}^\infty dy_{k\neq i,j}(\int_{-\infty}^\infty(\int_{-\infty}^\infty{\exp(y_i^2+2y_iy_j)}dy_i)\exp(y_j^2)dy_j)$.
$v_i=0\Leftarrow\forall{a}\;\int_{-\infty}^\infty\exp(x^2+2ax)x \, dx=0$.
Seems like the only hope would be for the function under the last integral to be antisymmetric with regard to some $x=b$, but it's clear that it is not for any $a\neq 0$.
Where's my mistake? I want to prove to myself that the title integral is zero...
[UPDATE]
I apologize to the comminuty for not clarifying what is meant by "$\mathbf{y}\;d\mathbf{y}$", which caused some misunderstanding between the commenters, the reason being my apparent ignorance. This question comes from me trying to derive multivariate Gaussian as the probability distribution with maximum entropy (with the given mean and variance); this is given as an exercise in the textbook I'm working through, and the mean is given by the integral $\int{p(\mathbf{x})\mathbf{x}d\mathbf{x}} = \mathbf{\mu}$ ($\mathbf{\mu}$ is a vector with $D$ elements.) I have a vague understanding of this notion, I supposed it means that the $i$-th component of this vector is given by the integral $\int{p(\mathbf{x})x_idx_1dx_2...dx_n}$. Seems like it's more a volume unit integral, not a dot product. Am I right here?
Also, there was some confusion on the Riemann integral vs. v.p. I don't think v.p. is what I need, after some consideration; please excuse me for misleading you.