$f,g$ such that $\int fg = \int f \int g$

Solution 1:

Differentiate and divide by $f(x) g(x)$, and you get $$ \dfrac{1}{f(x)} \int f(x)\; dx + \dfrac{1}{g(x)} \int g(x)\; dx = 1 $$ If $F(x) = \int f(x)\; dx$ and $G(x) = \int g(x)\; dx$, this says $$ \dfrac{d}{dx} \ln F(x) = \dfrac{F'(x)}{F(x)} = \dfrac{G'(x)}{G'(x) - G(x)}$$ and thus $$F(x) = c \exp \left( \int \dfrac{G'(x)}{G'(x) - G(x)}\; dx \right)$$ or $$ f(x) = \dfrac{c\; g(x)}{g(x) - \int g(x)\; dx} \exp \left( \int \dfrac{g(x)}{g(x) - \int g(x)\; dx}\; dx \right)$$ Since this was obtained by differentiating the original equation, you may have to adjust the constants of integration to make this work.