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Is this integral possible:

$$\int\left(f(a \cdot b,x)g(c \cdot d,x)\right)dx= h(a \cdot c,x)j(b \cdot d,x)$$

with $(a \cdot b\neq a \cdot c)$ and $(c \cdot d \neq b \cdot d)$ and $f \neq g \neq h \neq j$?

If this is possible, what are the functions $f$, $g$, $h$, and $j$?

PLEASE READ

I'm trying to find nontrivial instances of these functions such that they all include $x$, and that $(a\cdot b)$ isn't multiplied by $c \cdot d$ in the integral, and that $a \cdot c$ isn't multiplied by $b \cdot d$ in the result.

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I don't get this question right, I think. Can't we just let $f(\xi, x) = g(\xi, x) = h(\xi, x) = 1$, $j(\xi, x) = x$ for all $x, \xi$? –  martini Dec 7 '12 at 17:22
    
Sorry, I forgot to add a restriction on the functions to ensure they're not trivially equal. –  Matt Groff Dec 7 '12 at 17:27

1 Answer 1

up vote 1 down vote accepted

Assuming $b\neq 0$, take $f(ab,x)=\frac{ac}{b}x$, $g(cd,x)=(c+d)x$, $h(ac,x)=(ac^2+acd)x$ and $j(bd,x)=\frac{1}{b^2}\frac{x^2}{3}$. Obviously, \begin{equation}\int f(ab,x)g(cd,x)\, dx=h(ac,x)j(bd,x) \end{equation} Is this what you are asking?

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No. I'm trying to find nontrivial instances of the functions so that they all include $x$. –  Matt Groff Dec 7 '12 at 17:29
    
@MattGroff How about now? –  Nameless Dec 7 '12 at 17:32
    
Hmm. I'm trying to ensure that $a b$ and $c d$ aren't trivially multiplied by each other. Hopefully the latest version of the question captures this. Sorry. –  Matt Groff Dec 7 '12 at 17:34
    
@MattGroff This is my last offer. –  Nameless Dec 7 '12 at 17:38
    
I believe that your answer may work. Thanks for working through this with me! +1! –  Matt Groff Dec 7 '12 at 17:45

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