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

Suppose $f,g$ are real valued on $\mathbb{R}$ (and no further restrictions apart from the obvious requirement that the integrals exist), then when does $\displaystyle\int f(x)g(x)\,dx = \int f(x) \, dx \int g(x) \, dx$? (all integrals indefinite)

This question was posed on another site, and I was wondering whether there are any large classes of functions $f,g$ that work.

Aside from the trivial solution, things like $f(x) = e^{nx}, g(x) = e^{\frac{n}{n-1}x}$ $(n\not= 1)$ work by inspection.

I've tried re-casting the problem as:

$$\int F^2 - \left(\int F\right)^2 = \int G^2 - \left(\int G\right)^2$$

With $F=f+g, G=f-g$, to introduce some symmetry. This gives rise to similarities with the variance formulae but nothing more than that.

Using power series $F(x) = \displaystyle\sum_{n=0}^{\infty} a_nx^n, G(x) = \sum_{n=0}^{\infty} b_nx^n$, and the cauchy product, this equation boils down to:

$$\sum_{m=0}^n \frac{(n-m)(a_ma_{n-m} - b_mb_{n-m})}{(n+1)(n-m+1)}= 0\qquad \forall n\geq 0$$

But again, progress is limited.

• From your recasting, I would rather try to solve $\int F^2 - (\int F)^2 = H$ for $F$
– user14972
May 21, 2014 at 16:45

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.
• Other details to deal with include points where $f(x)=0$ or $g(x)=0$, or where $G'(x) = G(x)$.
• Assuming $g$ is analytic and is not a constant times the exponential function, such points will be isolated singularities, and the equation will be true away from those singularities. May 22, 2014 at 1:37
• Hmmm... poles of $g$ and $g/(g - G)$ may become branch points of their integrals, which do complicate the matter: you'll have to be careful about the branches of $G$ and $F$. May 22, 2014 at 1:55
For example, $f_n(x) = e^{1 2 \pi n x}$ on $[0,1]$. Then $\int f_n = 0$ when $n \ne 0$ and $\int f_n f_m = 0$ when $n \ne -m$. Therefore $\int (f_n f_m) = 0 = \int f_n \int f_m$ when $n \ne -m$.