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!
 A: Most probably you mean that $f$ is continuous.
Let $F$ be an antiderivative of $f$.
Hence, for any $t>0$ we have
$$\int_a^{at}f(x)\;dx = F(at)-F(a)$$
Since the integral is independent of $a$ you have
$$\partial_a(F(at)-F(a))=tf(at) - f(a) = 0 \Rightarrow f(at)=\frac{f(a)}{t}$$
Now, seting $x=at$ you get
$$f(x) = \frac{af(a)}{x} =\frac{c}{x}$$
A: Hint. Let $F$ be the antiderivative of $f$, then the condition is $F(ab)-F(a)=g(b)$ for some $g$. Differentiating with respect to $a$, we get $bf(ab)-f(a)=0$, hence $f(ab)/f(a)=1/b$.
A: 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 $c=f(1)$. 
A: Hint
$$F(b):=\int_{a}^{ab} f(x) dx=\int_1^b af(au) du$$
is independent on $a$. Find $F'(b)$.
A: 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?
A: 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. 
