A function/distribution which satisfies an integral equation. (sounds bizzare)

Question

I think I need a function, $f(x)$ such that $\large{\int_{x_1}^{x_2}f(x)\,d{x} = \frac{1}{(x_2-x_1)}}$ $\forall x_2>x_1>0$. Wonder such a function been used or studied by someone, or is it just more than insane to ask for such a function? I guess it has to be some distribution, and if so, I'd like to know more about it.

Answer 1

Note that for your function you can easely find $x_1 <x_2 < x_3$ so that

$$\int_{x_1}^{x_2} f(x) \, dx + \int_{x_2}^{x_3} f(x) \, dx \neq \int_{x_1}^{x_3} f(x) \, dx$$

so no such function can exists.

Actually you can prove this way that there exists no measure $m$ so that

$$ m([x_1,x_2])=\frac{1}{x_2-x_1} \,.$$

so in particular there is no absolutely continuous measure to satisfy that relation.

Answer 2

No such $f(x)$ can exist. Such a function $f(x)$ would have antiderivative $F(x)$ that satisfies $$F(x_2)-F(x_1)=\frac{1}{x_2-x_1}$$ for all $x_2>x_1>0$.

Take the limit as $x_2\rightarrow \infty$ of both sides; the right side is 0 (hence the limit exists), while the left side is $$\left(\lim_{x\rightarrow \infty} F(x)\right)-F(x_1)$$

Hence $F(x_1)=\lim_{x\rightarrow \infty} F(x)$ for all $x_1$. Hence $F(x)$ is constant, but no constant function satisfies the relationship desired.