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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.

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    $\mathbf{y}^TS\mathbf{y}$ is quadratic in the components of $\mathbf{y}$. I am pretty sure you can (rather) easily change variables and assume that $S$ is diagonal.2012-08-18
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    This integral doesn't converge if $S$ is positive definite, for example. Then you probably mean the Cauchy principle value2012-08-18
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    @Cocopuffs: why do you think so? E.g. $\int_{-\infty}^\infty{\exp(x^2)xdx} = 0$ because the integrand is antisymmetric with regard to $x=0$. I wish to see similar antisymmtry here...2012-08-18
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    breader: The integral in your comment is not zero because it does not exist. Your question makes sense when S is definite negative.2012-08-18
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    Yes, Riemann integral does not exist, but v.p. is still 0, isn't it?2012-08-18
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    Oh, all right, @Cocopuffs I agree with you. Sorry, I somehow missed the second part of your comment (shame on me.)2012-08-18
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    Does this integral end in $\vec{y}\,dV$, where $dV$ is a volume unit? Or does it end in $\vec{y}\cdot d\vec{y}$ , with the dot product of $\vec{y}$ and a differential vector? I would guess the latter. But the former integral yields a vector result, while the latter yields a scalar result. You indicate that you want to achieve $\vec{0}$, the vector. So I'm not sure.2012-08-18
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    breader: If indeed you had Cauchy p.v. in mind, your question should at least mention it, don't you think?2012-08-18
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    @did: I wasn't sure, sorry. Actually, I think I need Riemann integral. By the way, what's with your answer? I didn't have time to read it through yesterday, and today it seems like it's gone.2012-08-19
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    @alex.jordan I'm a bit confused here, sorry. Let me explain the background... I am working through a textbook and at some point there is an exercise to show that the probability distribution with the maximum entropy (with the given covariance and mean) is Gaussian. And in the multivariate case the mean is specified as $\int {p(\mathbf{x})\mathbf{x}d\mathbf{x}}$. I wasn't sure how to interpret this integral — I thought it's like having $D$ integrals, i.e. the result is a vector with $i$-th component equal to $\int{p(\mathbf{x})x_idx_1dx_2...dx_D}$2012-08-19
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    breader: I rewrote (and reposted) my answer, simplifying it and making it as unambiguous as possible.2012-08-19

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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, $v_i=0$.

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    I'm not sure that it affects your argument, but you have an integral with $dV$ as the differential, not $d\vec{y}$. Of course, if $d\vec{y}$ is meant to be the differential, then the end result is $0$, not $\vec{0}$. I'm not sure what OP wants.2012-08-18
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    @alex.jordan As clearly stated in my answer, I took $\mathrm d\vec y$ to mean $\mathrm dy_1\cdots\mathrm dy_D$, since the integral is over $\mathbb R^D$.2012-08-18
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    One can integrate over $\mathbb{R}^n$ with respect to $d\vec{y}=\langle dy_1,\ldots,dy_N\rangle$, dotting a vector-valued function (like $\exp(\vec{y}^TS\vec{y})\vec{y}$) with $d\vec{y}$ to obtain infinitesimal scalars, that sum to a total scalar. Since OP wrote $d\vec{y}$, I would at first assume that OP means $d\vec{y}=\langle dy_1,\ldots,dy_N\rangle$. If OP wants $dV=dy_1\cdots dy_N$, it should probably be written as $dV$. All this is moot thought since it seems any argument for either interpretation works both ways.2012-08-18
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    @alex.jordan Please note that I did not argue for any interpretation. I simply recalled which interpretation my answer is based on. In other words: *moot* indeed.2012-08-18
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    Oh my, it's _that_ simple. Thanks @did.2012-08-19
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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.

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    i agree with you, just use some orthogonal matrix to make it diagonal and the result can be easily seen.2012-08-18
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    i just thought out an easier way, just do variable change $y_1=-y$ to deduce the original integral changes to its opposite, so must be $0$. It's just like an odd function has integral 0 over a symmetric domain, if the integral exists.2012-08-18
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    @alex.jordan: I guess you want to replace $\vec y$ not in the exponential by $U\vec z$, not simply by $\vec z$?2012-08-18
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    @user18537: The change of variable $y_1=-y$ fails as soon as there are some nonzero off-diagonal elements $y_1y_k$ in $y^TSy$.2012-08-18
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    @did, Since $-1$ is a scalar, his change-of-variable $y_1=-y$ should work fine, no matter what square matrix $S$ is. The minus signs in $\vec{y}$ and $\vec{y}^T$ pass through $S$ and cancel each other.2012-08-18
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    @did the change of variable I make here is $\vec{z}=U^{-1}\vec{y}$, or $\vec{y}=U\vec{z}$, or $\vec{z}^T=\vec{y}^TU^{-1}$, or $\vec{z}^T=\vec{y}^TU^{T}$, all of which say the same. Note that if the last product is a dot product, then that $\vec{y}$ is really a $\vec{y}^T$.2012-08-18
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    @alex.jordan You mention $(y_1,y_2,\ldots,y_D)\to(-y_1,-y_2,\ldots,-y_D)$, I referred to what the OP mentioned (or what I understood was mentioned by the OP), namely $(y_1,y_2,\ldots,y_D)\to(-y_1,y_2,\ldots,y_D)$.2012-08-18
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    @alex.jordan The change of variable you make is clear (and there is no need to write four equivalent formulas for it, thank you). If you use it, some $\vec z$ should be replaced by $U\vec z$.2012-08-18
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    @did Sorry for all the equivalent transformations, but all those forms are directly used in the substitution, and I wanted to be clear to anyone who may not be too familiar with the properties of orthogonal matrices. I think here is what you might not be seeing: $\vec{y}\cdot d\vec{y} = (U\vec{z})\cdot d(U\vec{z})=(U\vec{z})\cdot (Ud\vec{z})=\vec{z}\cdot d\vec{z}$, by the orthogonality of $U$.2012-08-18
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    @alex.jordan Right, I seem to have forgotten what $\vec yd\vec y$ means to you and was referring to the $\vec y\,\mathrm dV$ interpretation. Sorry about the noise on this point.2012-08-19
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    Maybe i should have written it using another letter, what i meant is the variable change $w=-y$ with $w\in\mathbb{R}^D$, $y_1$ is just $w$, not the first coordinate of $y$.2012-08-19
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    @alex.jordan: Sorry, but what is "your" meaning of $\vec yd\vec y$ again? Assume for simplicity that $D=2$ and consider $I=\iint u(\vec y)\vec yd\vec y$ for some nice function $u:\mathbb R^2\to\mathbb R$, how do you suggest to define $I$? (If not as the point in $\mathbb R^2$ with coordinates $\iint u(y_1,y_2)y_1dy_1dy_2$ and $\iint u(y_1,y_2)y_2dy_1dy_2$, as I am doing.)2012-08-19