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I know it's not correct to write:

$$\int_{a}^{b}f(x)g(x) dx = \int_{a}^{b}f(x)dx\int_{a}^{b}g(x)dx$$

This result seems obvious, but I can't think of a way to prove that $\int_{a}^{b}f(x)g(x) dx$ can't be expressed as a function of the form :$$F\left(\int_{a}^{b}f(x)dx;\int_{a}^{b}g(x)dx\right)$$

Is it something simple I'm missing ? How can one prove that a definite integral can't be represented as mentioned before ?

andynash
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2 Answers2

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Just find two pairs of functions $f_1, g_1$ and $f_2,g_2$ such that $$\int f_1 = \int f_2, \quad\quad \int g_1 = \int g_2$$ but $$\int f_1 g_1 \neq \int f_2 g_2$$

Probably any two pairs you come up with will work.

fonini
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    For example, $f_1 = f_2 = g_1$ can be some bump function, and $g_2$ can be a translate of that same bump function so that its support is disjoint from the other guys. Then the first integral is positive but the second integral is zero. – Qiaochu Yuan May 21 '15 at 21:21
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Hint:

Let: $$ \int_a^x f(t)dt=F(x) \qquad \int_a^x g(t)dt=G(x) $$ From the fundamental theorem of calculus we have $$ \dfrac{d}{dx} F(x)=f(x) \qquad \dfrac{d}{dx} G(x)=g(x) $$

and, using the product rule for derivative:

$$ \dfrac{d}{dx}\left[\int_a^x f(t)dt\int_a^x g(t)dt\right]=\dfrac{d}{dx}\left[F(x)G(x)\right]=G(x)f(x)+F(x)g(x) \ne f(x)g(x) $$

Emilio Novati
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    What the OP is asking is how to prove that $Gf+Fg$ will be generally different from $fg$. Certainly there are some lucky cases where they are equal. – fonini May 21 '15 at 22:04