TITLE: If $f(17) = 17$, calculate $A(x) = \int_{1}^{x} f(t)dt$ for $x>0$.
QUESTION [5 upvotes]: Let $f: (0, \infty) \to \mathbb{R} $ be a continuous function. For $x>0$, $y>0$ and any $x'$ in $(0, \infty)$ we have that 
$$\int_{x}^{xy} f(u) \ du = \int_{x'}^{x'y} f(u) \ du.$$ 
If $f(17) = 17$, calculate $A(x) = \int_{1}^{x} f(t)dt$ for $x>0$.
I got this:
$$\begin{align}
A(xy) &= \int_{1}^{xy} f(t)dt \\
&=\int_{1}^{x} f(t)dt + \int_{x}^{xy} f(t)dt \\
&= \int_{1}^{x} f(t)dt + \int_{1}^{y} f(t)dt \\
&= A(x) + A(y)
\end{align}$$
and $$\frac{dA(xy)}{dy} = \frac{dA(xy)}{dy} y = A'(x).$$
With this we have that $A'
(17y)= \frac{f(17)}{y}$. 
Here I got stuck. What else I can do?

REPLY [3 votes]: We can differentiate the original identity with respect to $y$ to obtain:
$$xf(xy) = \frac{d}{dy}\int_{x}^{xy}f(t)\,dt = \frac{d}{dy}\int_{x'}^{x'y}f(t)\,dt = x'f(x'y)$$
for all $x,x',y > 0$. Plugging in $x' = 17$ and $y = 1$ gives
$$xf(x) = 17f(17) = 17^2 = 289$$
or $$f(x) = \frac{289}{x}$$
for all $x > 0$.
Now for all $x > 0$ we have
$$A(x) = \int_1^x f(t)\,dt = \int_1^x\frac{289}{t}\, = 289\ln x$$