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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? – J. M. Jul 24 '11 at 17:09
Maybe $\int_{-\infty}^\infty$ ? – Andrea Mori Jul 24 '11 at 17:13
:) yes its not an indefinite integral .. Andrea is correct .. – sheri Jul 24 '11 at 17:22
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. – John M Jul 24 '11 at 18:06
up vote 4 down vote accepted

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 .. – sheri Jul 24 '11 at 18:47
how does the $x_1-x_2$ in second equation get into the integral .. – sheri Jul 24 '11 at 18:51
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. – anon Jul 24 '11 at 18:56

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 ... – sheri Jul 24 '11 at 18:43
thank u so much .. ppl like me with lesser maths skills have these issues .. but now i understand .. – sheri Jul 24 '11 at 18:47
@Ilmari: Technically, you did not prove that $c_1$ and $c_2$ were finite. – Did Jul 24 '11 at 18:56
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 .. – sheri Jul 24 '11 at 18:56
@Didier: Right, I just assumed it. Edited to clarify. – Ilmari Karonen Jul 24 '11 at 19:20

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