Let $f:[0,\infty) \rightarrow \mathbb{R}$ be a continuous function and let $g(x)=\frac{1}{x}\int_1^x f(t)dt$; $x>0$. Assume that $\lim_{x\rightarrow \infty} g(x)=B$ exists. Let $0 < a < b$ be two fixed numbers. Show $$\lim_{T\rightarrow \infty}\int_{Ta}^{Tb}\frac{f(x)}{x}dx=B\ln\left(\frac{b}{a}\right)$$
Here's my partial solution: $g(x)=\frac{1}{x}\left(F(x)-F(1)\right) \Rightarrow f(x)=g(x)+xg'(x)$, Therefore: \begin{align*} \lim_{T\rightarrow \infty}\int_{Ta}^{Tb}\frac{f(x)}{x}dx&=\lim_{T\rightarrow \infty}\int_{Ta}^{Tb}\frac{g(x)}{x}dx+\int_{Ta}^{Tb}g'(x)dx\\ &=\lim_{T\rightarrow\infty}\ln(Tb)g(Tb)-\ln(Ta)g(Ta)-\int_{Ta}^{Tb}g'(x)\ln(x)+\int_{Ta}^{Tb}g'(x)dx\\ &=\lim_{T\rightarrow \infty}B\ln\left(\frac{Tb}{Ta}\right)-\int_{Ta}^{Tb}g'(x)\ln(x)+\int_{Ta}^{Tb}g'(x)dx\\ &=B\ln\left(\frac{b}{a}\right)+\dots \end{align*}
I can see how $\lim_{T\rightarrow\infty}\int_{Ta}^{Tb}g'(x)dx=\lim_{T\rightarrow \infty}g(Tb)-g(Ta)=0$ but I can't see to make the $-\int_{Ta}^{Tb}g'(x)\ln(x)$ go to zero. Any help on that?
We have to show that $$\lim_{T\rightarrow \infty}\int_{Ta}^{Tb}\frac{f(x)}{x}dx=\lim_{T\rightarrow \infty}\int_{Ta}^{Tb}\frac{g(x)}{x}dx+\lim_{T\rightarrow \infty}\int_{Ta}^{Tb}g'(x)dx=B\ln\left(\frac{b}{a}\right).$$ You have already noted $$\lim_{T\rightarrow\infty}\int_{Ta}^{Tb}g'(x)dx=\lim_{T\rightarrow \infty}g(Tb)-g(Ta)=0.$$ As regards the other integral $$\begin{align*}\lim_{T\rightarrow \infty}\int_{Ta}^{Tb}\frac{g(x)}{x}dx &= \lim_{T\rightarrow \infty}\left(\int_{0}^{Tb}\frac{g(x)}{x}dx-\int_{0}^{Ta}\frac{g(x)}{x}dx\right)\\ &=\lim_{T\rightarrow \infty}\left(\int_{0}^{T}\frac{g(bt)}{bt}d(bt)-\int_{0}^{T}\frac{g(ax)}{ax}d(ax)\right)\\ &= \int_{0}^{\infty}\frac{g(bt)-g(at)}{t}dt\\ &=(B-g(0))\ln\left(\frac{b}{a}\right)=B\ln\left(\frac{b}{a}\right) \end{align*}$$ where at the last step we used the Frullani integral applied to the function $g$ (here we assume that $g(x)=\frac{1}{x}\int_1^x f(t)dt$ for $x\geq 1$ and $g$ is continuously extended to $0$ in $[0,1)$).