Assume that $\int_{a}^{ab} f(x) dx$ is independent of $a$. Prove $f(x)=\frac{c}{x}$ Assume $f$ is integrable on $[0,\infty)$ and assume that for $a,b>0$, the value of  $\int_{a}^{ab} f(x) dx$ is independent of $a$. 
Prove that $f(x)=\frac{c}{x}$, where $c$ is a constant.
I have tried several things, like showing $g(x)=xf(x)$ must have derivative zero, but I am unsure how to use the integral assumption.
Thanks!

Consider the function $g(x) =\int_{1}^{x}f(t)\,dt$. By the given assumption we have $g(ab) =g(a) +g(b) $ for all positive $a, b$. Note that $g$ is continuous and it can be proved with some effort that $g$ is differentiable with derivative $g'(x) =g'(1)/x$ for $x>0$. Thus we have $f(x) =g'(1)/x$ at least at points of continuity of $f$ via FTC. 
