Play with functions 5

Let's define a function $g(x)$ . $\int_{1}^{x}f(t)dt = g(x)$ So the given equation becomes:- $g(xy)= yg(x) + x g(y)$ Put $y=2$ . $g(2x) = 2g(x) + x g(2)$ Differentiate the above equation. $2g'(2x) = 2g'(x) + g(2) \ldots (1)$ From the definition of $g(x)$ and Newton-Leibniz rule, we have:- $g'(x) = f(x)$ From (1):- $2f(2x) = 2 f(x)+ g(2)$ $f(2x) = f(x)+ \dfrac{g(2)}{2} \ldots (2)$ Put $x = 2^{0} , 2^{1}, 2^{2} \ldots 2^{n-1}$ in equation $(2)$ and add all the equations. We will get:- $f(2^{n}) = f(1) + \dfrac{n g(2)}{2}$ . Make the following substitution:- $2^{n} = x$ $n\ln2 = \ln x$ $n= \dfrac{\ln x}{\ln2}$ Hence our equation becomes:- $f(x) = f(1) + \dfrac{ g(2) \ln x}{2\ln2} \ldots (3)$ From the definition of $g(x)$ , we can write:- $\int_{1}^{2} f(t)dt = g(2)$ $\int_{1}^{2} \Bigg( f(1) + \dfrac{ g(2) \ln t}{2\ln2} \Bigg) dt = g(2)$ Solving above integral and then solving for $g(2)$ we have:- $g(2) = 2\ln2 f(1)$ From equation $(3)$ :- $f(x) = f(1)+\ln xf(1)$ $f(x) = f(1) ( 1+ \ln x)$ Put $x= e$ . $\boxed{f(e) = 2f(1)}$