An Idea to solve a limit involve integral

Question

Suppose $f(x)$ is a continuous function over $[0,1]$,and $0<a<b$ find the below limit $$\lim_{t\to 0^+}{\int_{at}^{bt}\frac{f(x)}{x}}=?$$ My work is below
$\int_{at}^{bt}\frac{f(x)}{x}=\int_{at}^{bt}f(x)d(\ln(x))$

Using the mean value theorem, and $f\in c^1[0,1]$ we have $$\exists c:a<c<b \ |f(ct)=\frac{\int_{at}^{bt}f(x)d(\ln(x))}{\ln(bt)-\ln(at)}$$ so $$\lim_{t\to 0^+}{\int_{at}^{bt}\frac{f(x)}{x}}=\lim_{t\to 0^+}{\int_{at}^{bt}f(x)d(\ln(x))}=\\ \lim_{t\to 0^+}f(ct)ln(\frac{bt}{at})=f(0)\ln(\frac ba) $$ My questions:
1:Is my working true?
2:Is there another Idea or method to solve the problem?
Thanks in advance

Hard to read the first displayed limit of an integral in your post. Now the integrand just looks like $\frac{(x)}{x}$ which simplified is just $1.$ Is there some function missing? Maybe $\ln(x)$ ? — coffeemath, Feb 19 '21 at 18:36
@CameronWilliams I don't immediately see the connection to Frullani's theorem. Could you elucidate the connection to this problem? — Isaac Browne, Feb 19 '21 at 19:07
@Khosrotash I believe this is the post CameronWilliams was referring to: https://math.stackexchange.com/q/61828/429987. In any case, you should be able to see my post on this question here still — Isaac Browne, Feb 19 '21 at 19:13
@CameronWilliams I see the connection now. This is an important step in the proof of Frullani's theorem, but I believe this question is different enough to be reopened. — Isaac Browne, Feb 19 '21 at 19:21
@IsaacBrowne Ah yeah I figured since the proof was contained in there, it was better just to direct it that way. — Cameron Williams, Feb 19 '21 at 19:54
Every time I see a limit question all of the answers always miss the easy $u$ substitution. Let $tu = x$ — Ninad Munshi, Feb 19 '21 at 20:25
can you not substitute $u=\frac xt$ then justify interchanging the limit and integral to get $$f(0)\int_a^b\frac 1u,du$$ — Henry Lee, Feb 20 '21 at 00:12

Isaac Browne · Answer 1 · 2021-02-19T19:09:01.777

A couple things. First the assumption only says $f$ is continuous, not that $f\in C^1[0,1]$, so mean value theorem doesn't work.

Second, I believe the bounds in the integrals should be $\ln(at)$ and $\ln(bt)$ when you've switched to $d(\ln(x))$. Though this could just be some convention I am unaware of.

A similar idea is to use inequalities to bound it i.e. $$\int_{at}^{bt} \frac{f(x)}{x} dx \leq \sup_{at \leq x \leq bt} f(x) \int_{at}^{bt}\frac{1}{x}dx = \ln(b/a)\sup_{at\leq x\leq bt}f(x) $$ and $$\int_{at}^{bt} \frac{f(x)}{x} dx \geq \inf_{at \leq x \leq bt} f(x) \int_{at}^{bt}\frac{1}{x}dx = \ln(b/a)\inf_{at\leq x\leq bt}f(x) $$ Since we are taking $t\to 0$ and $f$ is continuous, both the supremum and infimum will become $f(0)$ in the limit.

An Idea to solve a limit involve integral

1 Answers1