Let $f (0,\infty)\rightarrow \mathbb{R}$ be a continous function such that for all $x \in (0,\infty)$ $f(x)=f(3x)$ Define $g(x)= \int_{x}^{3x} \frac{f(t)}{t}dt$ for $x \in (0,\infty)$ is a constant function
Use leibniz rule for differentiation under integral sign
using leibniz rule for differentiation under integral sign we get
$g'(x)=f(3x)-f(x)$
$\Rightarrow g'(x)=0$ [ Because f(3x)=f(x)]
Since the derivative of $g(x)$ is $0$ for all $x$, Hence $g(x)$ is a constant function
Let $f (0,\infty)\rightarrow \mathbb{R}$ be a continous function such that for all $x \in (0,\infty)$ $f(x)=f(3x)$ Define $g(x)= \int_{x}^{3x} \frac{f(t)}{t}dt$ for $x \in (0,\infty)$ is a constant function
Use leibniz rule for differentiation under integral sign
using leibniz rule for differentiation under integral sign we get
$g'(x)=f(3x)-f(x)$
$\Rightarrow g'(x)=0$ [ Because f(3x)=f(x)]
Since the derivative of $g(x)$ is $0$ for all $x$, Hence $g(x)$ is a constant function
