TITLE: Assume that $\int_{a}^{ab} f(x) dx$ is independent of $a$. Prove $f(x)=\frac{c}{x}$
QUESTION [6 upvotes]: 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!

REPLY [3 votes]: Define $$g(t)=\int_t^{tb}f(x)\operatorname dx$$ 
Then $g(t)=c$, so $g'(t)=0$.  
By FTC, $$g'(t)=bf(tb)-f(t)$$
$$\implies f(tb)=\frac1{bf(t)}$$  Let $t=1$.  Then $f(b)={f(1)\over b}$.
$$\implies \boxed{f(x)=\frac cx}$$ with $c=f(1)$.