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Let $\mathbf{A}=\begin{bmatrix} f(x_1,x_1), & \ldots,& f(x_1,x_n)\\ \vdots&\ddots& \vdots \\f(x_n,x_1),&\ldots, &f(x_n,x_n) \end{bmatrix} $, where $f:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}$. I want to calculate $\int \mathbf{A}\mathrm{d}\mathbf{x}$, where $\mathbf{x}=\begin{bmatrix}x_1\\ \vdots\\x_n \end{bmatrix}$.

It would be nice if you can show some references.

Thanks :)

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2 Answers 2

up vote 1 down vote accepted

Consider the change

$$ {\rm d} {\boldsymbol y} = {\boldsymbol A}\, {\rm d} {\boldsymbol x} $$ $$ = \left[ \begin{matrix} f(x_1,x_1)\,{\rm d}x_1 + f(x_1,x_2)\,{\rm d}x_2 + \ldots \\ f(x_2,x_1)\,{\rm d}x_1 + f(x_2,x_2)\,{\rm d}x_2 + \ldots \\ \vdots \end{matrix} \right] $$

and its integral

$$ {\boldsymbol y} = \int {\boldsymbol A}\, {\rm d} {\boldsymbol x} $$

$$ = \left[ \begin{matrix} \int f(x_1,x_1)\,{\rm d}x_1 + \int f(x_1,x_2)\,{\rm d}x_2 + \ldots \\ \int f(x_2,x_1)\,{\rm d}x_1 + \int f(x_2,x_2)\,{\rm d}x_2 + \ldots \\ \vdots \end{matrix} \right] $$

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thx, this looks reasonable to me. –  han Jul 13 '12 at 14:51

Well, it seems the result should be the matrix whose $(i,j)$ coefficient is $$ \int_{\mathbb R^n} f(x_i,x_j)\mathrm dx_1\cdots\mathrm dx_n. $$ Be mindful though that the diagonal coefficients are undefined as soon as $n\geqslant1$ and the off-diagonal coefficients are undefined as soon as $n\geqslant2$, hence the whole notion seems rather dubious.

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