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!

I suppose the assumption is for all $b$. If it holds for just $b=1$ we cannot prove that $f$ has the desired form.
By Lebesgue's theorem on differentiation of indefinite integrals we can differentiate w.r.t. $a$ and get $bf(ab)-f(a)=0$ a.e.. If $g(x)=xf(x)$ we get $g(ab)=g(a)$ a.e... Can you finish?
