# For positive f and g, if $\int\int… f\, dx_1\,dx_2… =\infty$, $\int\int…g\, dx_1\,dx_2… =\infty$, is $\int\int… fg\, dx_1\,dx_2…= \infty$?

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

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

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Crossposted: mathoverflow.net/questions/63833 –  Zev Chonoles May 4 '11 at 4:59
What does $f(x,y)$ continous in $(-\infty, \infty)$ mean? –  Jose27 May 4 '11 at 5:16
It was wrongly edited. Sorry. –  user10455 May 4 '11 at 5:18
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. –  Jose27 May 4 '11 at 5:29

## 1 Answer

Firstly, if we have that f and g are continuous everywhere, then their integrals won't explode over a compact region. In fact, they will have a (finite) absolute max and minima over the region, and their magnitude is bounded by the area of the region times that absolute max/min (whichever is bigger in absolute value).

Secondly, it's not necessarily true. And instead of doing anything formal, I'll just speak of the intuition. Suppose that $f \equiv 0$ on half of the region $\Omega = [a,b] \times [c,d]$, and that $g \equiv 0$ on the other half of $\Omega$. Then the overall integral is $0$. As we consider larger and larger portions of the plane, we still get that the integral is $0$.

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@user: Although you have changed the question, the intuition behind this answer still applies. Simply let f and g be identically 0 on complementary domains. –  mixedmath May 4 '11 at 10:39