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We are given $x_1,x_2 \in \mathbb{R}$ and we want to find two functions $v_1(t),v_2(t)$ such that:
$$x_1x_2 = \int_{-\infty}^{\infty} v_1(t)-v_2(t) dt$$ A very interesting restriction that we have is that the object generating $v_1(t)$ only knows $x_1$, while $v_2(t)$ is generated only by knowing $x_2$.

The application of this is like this. We have two ends of a wire with some component Y in between. We call one end as $1$ where $x_1$ is known and second end as $2$ where $x_2$ is known. We want to send a signal from both ends which gets aggregated at Y, but we want to choose two signals $v_1(t),v_2(t)$ such that when Y sums them up the sum of the two signals becomes equal to the multiplication of $x_1$ and $x_2$.

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    Just how does an *indefinite* integral end up as the product of two real numbers?2011-07-24
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    Maybe $\int_{-\infty}^\infty$ ?2011-07-24
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    :) yes its not an indefinite integral .. Andrea is correct ..2011-07-24
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    Given that the LHS is bilinear with respect to $x_1$ and $x_2$, it's hard to see how the integrand on the RHS could also be bilinear, given the particular form the integrand takes.2011-07-24

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It cannot be done. You are looking for two functions $u:\ (x,t)\mapsto u(x,t)$ and $v:\ (y,t)\mapsto v(y,t)$ such that $$x \cdot y\ \equiv \ \int_{-\infty}^\infty\bigl(u(x,t)-v(y,t)\bigr)\ dt$$ for all $(x,y)$ in some domain $\Omega\subset{\mathbb R}^2$. It follows that for any two $x_1\ne x_2$ and any $y$ we should have $$(x_1-x_2)y\ =\ \int_{-\infty}^\infty \bigl(u(x_1,t)-u(x_2,t)\bigr)\ dt\ .$$ This is impossible, as the RHS is constant with respect to $y$.

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    thank u so much .. ppl like me with lesser maths skills have these issues .. but now i understand ..2011-07-24
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    how does the $x_1-x_2$ in second equation get into the integral ..2011-07-24
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    Evaluate $x_1y$ using its integral definition, then evaluate $x_2$ similarly, and subtract the latter from the former. You're left with Christian's equation above.2011-07-24
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I've deleted my original answer, since Christian Blatter's nice and succint proof is better and works even without any assumptions about the existence and finiteness of the individual integrals $\int_{-\infty}^\infty v_1(t) \;dt$ and $\int_{-\infty}^\infty v_2(t) \;dt$. However, I've left in the following addendum:


What you can do, however, is have

$$x_1 x_2 = \exp \left( \int_{-\infty}^\infty v_1(t) \;dt - \int_{-\infty}^\infty v_2(t) \;dt \right) = \exp \left( \int_{-\infty}^\infty v_1(t) - v_2(t) \;dt \right),$$

with $v_1$ and $v_2$ chosen arbitrarily such that $\int_{-\infty}^\infty v_1(t) \;dt = \log x_1$ and $\int_{-\infty}^\infty v_2(t) \;dt = -\log x_2$. Of course, this particular solution only works for positive $x_1$ and $x_2$. Choosing appropriate functions $v_1$ and $v_2$ is left as an exercise, although obviously e.g. pulses with unit width and amplitudes $\log x_1$ and $-\log x_2$ will do.

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    what if $x_1,x_2$ are never zero ...2011-07-24
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    thank u so much .. ppl like me with lesser maths skills have these issues .. but now i understand ..2011-07-24
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    @Ilmari: Technically, you did not prove that $c_1$ and $c_2$ were finite.2011-07-24
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    this solution did come to me but I was not able to bring it home ... :) i will look at it now more deeply thank u again ..2011-07-24
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    @Didier: Right, I just assumed it. Edited to clarify.2011-07-24
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    so am I looking at something like this .. $x_1x_2=exp(c_1-c_2)=\int_{-\infty}^\infty v_1(t)-v_2(t) dt$ ..which defeats my requirement of distributed calculation of $v_1(t),v_2(t)$ because of $exp(c_1-c_2)$2011-07-24
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    No, I meant $x_1 x_2 = \exp(c_1 - c_2) = \exp \left( \int_{-\infty}^\infty v_1(t)-v_2(t) \;dt \right)$.2011-07-24
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    @Ilmari: But the point is that one does not need to assume these are finite to solve the problem (and that nothing guarantees they are finite since if they are, one can replace $v_1$ and $v_2$ by other functions such that they are not anymore). Note that some other solutions (including mine, which I now deleted) avoid this conundrum by taking care to deal only with *differences* like something-involving-$v_1$-minus-something-involving-$v_2$.2011-07-24
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    @Didier: You're right, yours and Christian's answers are/were better in that way. I'm not quite sure what to do about that, now that the OP has already accepted mine... :-/2011-07-24