# 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!

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

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

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

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

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 $$c=f(1)$$.

Hint

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

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

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?

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.