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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!

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    $\begingroup$ But $f(x) = \frac{c}{x}$ is not integrable on $[0,\infty)$ $\endgroup$
    – GA316
    Commented Feb 10, 2020 at 5:38

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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

$$f(x) = \frac{af(a)}{x} =\frac{c}{x}$$

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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 $f(ab)/f(a)=1/b$.

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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 $c=f(1)$.

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Hint

$$F(b):=\int_{a}^{ab} f(x) dx=\int_1^b af(au) du$$

is independent on $a$. Find $F'(b)$.

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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?

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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.

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