# Improper Integrals in Analysis

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?

• The line before the last one looks suspicious: you have there $\;B\log\frac{Tb}{Ta}\;$ , so it seems to be you took the limit of $\;g\;$ without taking the limit of the logarithm . This can't be done in general or else demands proof. – DonAntonio Feb 14 '19 at 18:09

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 $$[0,1)$$).