I was wondering when it held that

$$\int\limits_0^x fg=\left(\int\limits_0^xf\right)\left(\int\limits_0^xg\right)$$


$$P:= x \mapsto \int\limits_0^x fg$$ $$F:= x \mapsto \int\limits_0^x f$$ $$G:= x \mapsto \int\limits_0^x g$$

The equality becomes $$P=FG$$

Obviously $f\equiv 0$ or $g\equiv 0$ are solutions.

But are there other solutions?

If you search for solutions of the form

$$f(x)=\sum_{n=0}^{+\infty} a_nx^n$$ $$g(x)=\sum_{n=0}^{+\infty} b_nx^n$$

You get

$$fg(x)=\sum_{n=0}^{+\infty} c_nx^n$$ $$P(x)=\sum_{n=0}^{+\infty}C_nx^n$$ $$F=\sum_{n=0}^{+\infty} A_n x^n$$ $$G=\sum_{n=0}^{+\infty} B_n x^n$$ $$FG(x)=\sum_{n=0}^{+\infty} D_n x^n$$


$$\forall n \in \Bbb N, c_n=\sum_{k=0}^na_kb_{n-k}$$ $$\forall n \in \Bbb N^*, C_{n} = \cfrac{c_{n-1}}{n} = \cfrac{1}{n}\left(\sum_{k=0}^{n-1}a_kb_{n-k-1}\right)\text{ and } C_0 = 0$$ $$\forall n \in \Bbb N^*, A_{n} = \cfrac{a_{n-1}}{n}\text{ and } A_0 = 0$$ $$\forall n \in \Bbb N^*, B_{n} = \cfrac{b_{n-1}}{n}\text{ and } B_0 = 0$$ $$\forall n \in \Bbb N, D_{n} = \sum_{k=0}^nA_kB_{n-k}=\sum_{k=1}^{n-1}A_kB_{n-k}=\sum_{k=1}^{n-1}\cfrac{a_{k-1}}{k}\cfrac{b_{n-k-1}}{n-k}$$

And the equality becomes

$$\forall n \in \Bbb N, C_n=D_n$$

We always have that $$C_0=0=D_0$$

But we also have $$0=D_1=C_1=a_0b_0$$. Since the problem is symmetric in $f$ and $g$ and hence in $a_n$ and $b_n$, we can assume WLOG that $a_0=0$.

Which gives $$C_{n} = \cfrac{1}{n}\left(\sum_{k=1}^{n-1}a_kb_{n-k-1}\right)$$

So then the equality becomes

$$\forall n\in \Bbb N, \cfrac{1}{n}\left(\sum_{k=1}^{n-1}a_kb_{n-k-1}\right)=\sum_{k=1}^{n-1}\cfrac{a_{k-1}}{k}\cfrac{b_{n-k-1}}{n-k}$$

And then I'm not sure what I should do...

And I'd also like to know if there are solutions that aren't of this form.

Thank you in advance for your answers.

  • 4
    $\begingroup$ It's (more or less) the same as looking for solution of $(FG)'=F'G'$. $\endgroup$
    – Berci
    May 4 '13 at 11:47
  • $\begingroup$ Should there not be a '$dx$' after the integrals? Is it shorthand or more general notation? $\endgroup$
    – Meow
    May 4 '13 at 12:25
  • 1
    $\begingroup$ Well $\int\limits_a^bf = \int\limits_a^bf(t)dt$. It's just a shorthand. $\endgroup$
    – xavierm02
    May 4 '13 at 12:26

Differentiating we get $fg=Fg+Gf$, or $ Fg=f(g-G)=F'(g-G)$, or $\frac{F'}{F}=\frac{g}{g-G}$, so $$F=\pm C\exp \int \frac{g}{g-G}.$$ Now insert possible $G$ to see what happens really. As the comment below pointed, $F(0)=0$, so $C=0$, there is no non-trivial solution...

  • 2
    $\begingroup$ Can $F(0)=0$ for a solution of this form? [$F(0)=0$ from the definition of $F$] It would seem, because of the exp, that it would require $C=0$, and thus imply $F$ must be the zero function. $\endgroup$
    – coffeemath
    May 4 '13 at 12:04
  • $\begingroup$ I think you meant "no non-trivial solution" instead of "no trivial solution". $\endgroup$
    – xavierm02
    May 4 '13 at 12:45
  • 1
    $\begingroup$ I feel I should point out that $$f(x)=\begin{cases}1&x\in\mathbb{Q}\\0&\text{otherwise}\end{cases} \qquad g(x)=\begin{cases}1&x\pi\in\mathbb{Q}\\0&\text{otherwise}\end{cases}$$ is also a solution. $\endgroup$
    – Glen O
    May 4 '13 at 14:15
  • $\begingroup$ Indeed, any pair of functions for which one is non-zero only on a countable subset of the reals will form a solution, including $f(x)$ as given in my previous comment and $g(x)=1$. $\endgroup$
    – Glen O
    May 4 '13 at 14:35

For analytic functions, as Ma Ming has pointed out, only the trivial solution is possible. However, there are other functions that work to solve the problem, such as ones based around requiring that, at any point, one of the two integrals on the left must be zero. As such, this pair of continuous functions will satisfy the integral relation:

$$ f(x)=\begin{cases}\sin(x)&0<x<2\pi\\0&\text{otherwise}\end{cases}\\ g(x)=\begin{cases}x-2\pi&x>2\pi\\0&\text{otherwise}\end{cases} $$ This works because $f(x)g(x)=0$ and, for $x\leq2$, $\int_0^x g(x)dx = 0$ and $\int_0^x f(x)g(x)dx = 0$, while for $x>2$, $\int_0^x f(x)dx = 0$ and $\int_0^x f(x)g(x)dx = 0$.

EDIT: Indeed, one can construct everywhere-smooth functions that satisfy the integral relation. Consider the function: $$ k(x)=\begin{cases}e^{-\frac{1}{x(1-x)}}&0<x<1\\0&\text{otherwise}\end{cases} $$ Now, $k(x)$ is everywhere-smooth, infinitely-differentiable at every point. We can use this function to construct some interesting functions satisfying the integral relation. The first thing to note is that $$ \int_0^2 k(x-1)-k(x)dx=0 $$ Now, if we let $\bar x = x-4\lfloor x/4\rfloor$, and take $$ f(x)=\left[k(\bar x-1)-k(\bar x)\right]\times A\\ g(x)=\left[k(\bar x-3)-k(\bar x-2)\right]\times B $$ where $A$ and $B$ are constants, then we have two smooth functions, neither of which is zero on an interval larger than 2, that satisfy the integral relation. More complicated smooth functions can also be created on this basis.

  • $\begingroup$ What are $A$ and $B$? And I don't get how your last two functions can be smooth since you use ceil and since they are in $k$, I don't see how both ceil you somehow concel each other. $\endgroup$
    – xavierm02
    May 5 '13 at 8:33
  • 1
    $\begingroup$ $A$ and $B$ are constants. Sorry, I see that they can be read as functions, and that's not the intent - I'll move them to the end, for clarification. The floor function (not ceiling function) has its non-smoothness only where $k(x)$ is zero. Basically, if you look at $x-4\lfloor x/4\rfloor$, it's smooth between 0 and 4, which is where $k(x)$ is non-zero. It might also help if I alter the bracketing to make it a bit more readable. Changes will be made in a moment. $\endgroup$
    – Glen O
    May 5 '13 at 12:45
  • $\begingroup$ Ok I think I got it. Thanks :) $\endgroup$
    – xavierm02
    May 5 '13 at 12:59

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