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?

  • $\begingroup$ 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. $\endgroup$ – 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 $[0,1)$).

  • $\begingroup$ @FatFeynman I revised my answer. $\endgroup$ – Robert Z Feb 14 '19 at 18:33
  • $\begingroup$ I will revise this later and perhaps prove Frullani integral to be able to apply the result and see where that gets me in regards to the rest of the problem thank you! $\endgroup$ – FatFeynman Feb 14 '19 at 20:53

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