Is $\int_{\mathbb{R}^D}\exp(\mathbf{y}^\top\mathbf{S}\mathbf{y})  \mathbf{y}\,d\mathbf{y}$ equal to $\mathbf{0}$? 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.
 A: As Siminore states in a comment, make use of the fact that symmetric matrices are orthogonally diagonalizable. That is, $$S=UHU^{-1}$$ for an orthogonal matrix $U$ and a diagonal matrix $H$. Then you have 
$$\int_{\mathbb{R}^D}\exp(\vec{y}^TUHU^{-1}\vec{y})\vec{y}\,d\vec{y}$$
Now change variables so that $z=U^{-1}y$. Use the fact that $U^T=U^{-1}$, that an orthogonal transformation won't affect volumes or lengths, and that as $\vec{y}$ ranges over $\mathbb{R}^n$, so does $\vec{z}$.
$$\begin{align}\int_{\mathbb{R}^D}\exp(\vec{z}^TH\vec{z})\vec{z}\,d\vec{z}&=\int_{\mathbb{R}^D}\exp\left(\sum h_iz_{i}^2\right)\vec{z}\,d\vec{z}\\
\end{align}$$
Now if the integral converges at all, it would converge to zero, since every Riemann summand based at $\vec{z}_0$ can be paired with its negative, another Riemann summand based at $-\vec{z}_0$. 
As noted in comments, this integral does not converge unless all of the eigenvalues $h_i$ are negative. However if you are interested in redefining the integral as the limit of integrals over concentric spheres centered at the origin whose radii approach infinity, 
then the symmetry is still present to yield a result of zero.
A: Essentially, the trouble comes from some mistakes in indexations. With your notations, assuming $S$ is definite negative, one should rewrite the definition of $v_i$ for each $i\leqslant D$ (as it appears at its first occurrence in the post) as
$$
v_i=\iint_{\mathbb R^D}\varphi_i(y_1,\ldots,y_D)\cdot\mathrm dy_1\cdots\mathrm dy_D,
$$
where
$$
\varphi_i(y)=\exp\left(y^TSy\right)\cdot y_i,\qquad y=(y_1,\ldots,y_D).
$$
In particular, one cannot use $i$ both as the index of $v_i$ and as a running index in the product defining the function that $v_i$ is the integral of.
A different intepretation of $v_i$ is proposed on this page, which I am not sure to understand nor to agree with (but, as always, this is the OP's task to clarify the matter).
Now, to prove that each $v_i=0$ (with the definition of $v_i$ above), note that, for every $y$ in $\mathbb R^D$, $\varphi_i(-y)=-\varphi_i(y)$ hence, as the integral of an odd function, $v_i=0$.
