Integration problem involving a functional equation: $ \int _ x ^ { k x } f ( t ) \, \mathrm d t = c $ implies $ k f ( k x ) = f ( x ) $ 
Let $ k > 0 $ be fixed. Determine all continuous functions $ f : \mathbb R ^ + \to \mathbb R $ such that $ \int _ x ^ { k x } f ( t ) \, \mathrm d t $ is constant for any $ x $.

By the fundamental theorem of calculus, it boils down to solving the functional equation $ k f ( k x ) = f ( x ) $. I am not being able to proceed any further. Is there any other way of solving this problem without using the fundamental theorem?
 A: $$y=\int_x^{kx} f(t)dt $$
$$\frac{dy}{dx}=kf(kx)-f(x)=0 $$
$$f(kx)=\frac{1}{k}\cdot f(x) $$
An example of such function is $\frac{1}{x}$
A: Consider a constant $ k > 0 $ and a function $ f : \mathbb R ^ + \to \mathbb R $ satisfying
$$ k f ( k x ) = f ( x ) \tag 0 \label 0 $$
for all $ x \in \mathbb R ^ + $. Define $ g : \mathbb R ^ + \to \mathbb R $ with $ g ( x ) = x f ( x ) $ for all $ x \in \mathbb R ^ + $. Then, multiplying both sides of \eqref{0} by $ x $, you get
$$ g ( k x ) = g ( x ) \tag 1 \label 1 $$
for all $ x \in \mathbb R ^ + $. Again, define $ h : \mathbb R \to \mathbb R $ with $ h ( x ) = g ( k ^ x ) $ for all $ x \in \mathbb R $. Substituting $ k ^ x $ for $ x $ in \eqref{1}, you have
$$ h ( x + 1 ) = h ( x ) \tag 2 \label 2 $$
for all $ x \in \mathbb R $. Therefore,
$$ f ( x ) = \frac { h ( \log _ k x ) } x \tag 3 \label 3 $$
for some periodic function $ h $ with period $ 1 $, which is continuous iff $ f $ is so. Conversely, for any periodic function $ h : \mathbb R \to \mathbb R $ with period $ 1 $, you can define $ g : \mathbb R ^ + \to \mathbb R $ with $ g ( x ) = h ( \log _ k x ) $ for all $ x \in \mathbb R $, and conlude \eqref{1} from \eqref{2}. Again, defining $ f : \mathbb R ^ + \to \mathbb R $ with $ f ( x ) = \frac { g ( x ) } x $ for all $ x \in \mathbb R ^ + $, you can get \eqref{0} from \eqref{1}. Also, $ f $ will be continuous iff $ h $ is so. Therefore, \eqref{2} and \eqref{3} characterize all the possible solutions of \eqref{0}.
Note that the class of all (continuous) solutions is as large as the class of all (continuous) periodic functions with period $ 1 $, and therefore there are uncountably many solutions to the problem. Even if you require the solution to be more regular, e.g. smooth or even analytic, there are uncountably many solutions (think of functions like $ h ( x ) = a \cos ( 2 n \pi x ) $ and $ h ( x ) = a \sin ( 2 n \pi x ) $ for any $ a \in \mathbb R $ and any $ n \in \mathbb Z ^ + $, and consider Fourier series).
