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Edited:

Given that $f(x_1,x_2,...)=f_1(x_1)f_2(x_2)...$ and $g(x_1,x_2,...)=g_1(x_1)g_2(x_2)...$ are always positive.

Also, $f$ and $g$ are continuous everywhere.

If

$[\displaystyle\prod_{n=1}^∞\int_{-1}^{1}dx_n] f(x_1,x_2,...) =∞$ and

$[\displaystyle\prod_{n=1}^∞\int_{-1}^{1}dx_n] g(x_1,x_2,...) =∞$,

is it necessarily true that

$[\displaystyle\prod_{n=1}^∞\int_{-1}^{1}dx_n] f(x_1,x_2,...)g(x_1,x_2,...) =∞$ ?

For more tricky case, if $g$ is not always positive or negative and

$[\displaystyle\prod_{n=1}^∞\int_{-1}^{1}dx_n] g(x_1,x_2,...) =±∞$ (oscillating),

is it true that

$[\displaystyle\prod_{n=1}^∞\int_{-1}^{1}dx_n] f(x_1,x_2,...)g(x_1,x_2,...) =±∞$?

  • 1
    Crossposted: http://mathoverflow.net/questions/638332011-05-04
  • 0
    What does $f(x,y)$ continous in $(-\infty, \infty)$ mean?2011-05-04
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    It was wrongly edited. Sorry.2011-05-04
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    Then either $[a,b]$ or $[c,d]$ is infinite (otherwise none of the integrals you're asking about can be infinite), but then, variants of the functions $f(x,y)=1/x=g(x,y)$ (for $x$ away form $0$) give that the integral of $fg$ isn't necesarily infinite.2011-05-04

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