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I know that there are a lot of posts that derive Frulani's integral but I would really like to know how to do it for myself. Here is the problem:

Lef $f:[0,\infty)\to\mathbb{R}$ a $C^{1}$ function such that $\displaystyle\int_{1}^{\infty} \frac{f(u)}{u}\,d u$ converges. Prove, using the Fubuni's theorem that if $0<a<b$ then $$\int_{0}^{\infty}\dfrac{f(ax)-f(bx)}{x}\, dx=f(0)\log\left(\frac{b}{a} \right)$$

Proof: By definition, $I=\displaystyle\int_{0}^{\infty}\dfrac{f(ax)-f(bx)}{x}\, dx=\lim\limits_{t\to\infty}\displaystyle\int_{0}^{t}\dfrac{f(ax)-f(bx)}{x}\, dx $. By the fact that $\dfrac{d\left(\frac{f(xy)}{x} \right)}{dy}=f'(xy)$, then follows that $\displaystyle\int_{b}^{a} f'(xy)\, dy=\dfrac{f(ax)-f(bx)}{x}$ and therefore we can write $I=\lim\limits_{t\to\infty} \displaystyle\int_{0}^{t}\left( \displaystyle\int_{b}^{a} f'(xy)\, dy\right)\, dx$. By Fubini's theorem, follows that $I=\lim\limits_{t\to\infty} \displaystyle\int_{b}^{a}\left( \displaystyle\int_{0}^{t} f'(xy)\, dx\right)\, dy=\lim\limits_{t\to\infty} \displaystyle\int_{b}^{a}\left( \dfrac{f(xy)}{y} \ \Big|_{0}^{t}\right)\, dy=\lim\limits_{t\to\infty} \displaystyle\int_{b}^{a}\left( \dfrac{f(ty)-f(0)}{y} \right)\, dy$. Then $I=\lim\limits_{t\to\infty}\left( \displaystyle\int_{b}^{a} \dfrac{f(ty)}{y}\, dy-\displaystyle\int_{b}^{a} \dfrac{f(0)}{y}\, dy \right)=\lim\limits_{t\to\infty}\left( \displaystyle\int_{b}^{a} \dfrac{f(ty)}{y}\, dy+f(0)\log\left(\dfrac{b}{a} \right) \right)$. Here is where I'm stuck. I think that the limit $\lim\limits_{t\to\infty}\left( \displaystyle\int_{b}^{a} \dfrac{f(ty)}{y}\, dy\right)$ must converge to $0$ to obtain the desired result, but I'm unable to prove that. Any hints? Thanks. Clearly I'm still not using the fact that $\displaystyle\int_{1}^{\infty} \frac{f(u)}{u}\,d u$ converges.

Gary
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Carlos Jiménez
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1 Answers1

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Too long for a comment

We can use the Mean value theorem. Let $f(x)$ is a continuous function ($C^0$), and $b>a$

Then $$ \int_\delta^\Delta\frac{f(ax)-f(bx)}{x}dx= \int_{\delta a}^{\Delta a}\frac{f(t)}{t}dt-\int_{\delta b}^{\Delta b}\frac{f(t)}{t}dt$$ $$=\int_{\delta a}^{\delta b}\frac{f(t)}{t}dt+\int_{\delta b}^{\Delta a}\frac{f(t)}{t}dt-\int_{\delta b}^{\Delta a}\frac{f(t)}{t}dt-\int_{\Delta a}^{\Delta b}\frac{f(t)}{t}dt=\int_{\delta a}^{\delta b}\frac{f(t)}{t}dt -\int_{\Delta a}^{\Delta b}\frac{f(t)}{t}dt$$ According to the Mean value theorem for integrals, there exist $c_1\in[\delta a;\delta b] , c_2\in[\Delta a;\Delta b]$ such that $$=f(c_1)\int_{\delta a}^{\delta b}\frac{dt}{t} -f(c_2)\int_{\Delta a}^{\Delta b}\frac{dt}{t}=\Big(f(c_1)-f(c_2)\Big)\ln\frac{b}{a}$$ and $$\underset{x\in[\delta a;\delta b]}{\min f(x)}<f(c_1)<\underset{x\in[\delta a;\delta b]}{\max f(x)}\quad\text{and}\quad \underset{x\in[\Delta a;\Delta b]}{\min f(x)}<f(c_2)<\underset{x\in[\Delta a;\Delta b]}{\max f(x)}$$ Now let's lead $\delta\to0$ and $\Delta\to\infty$

John Bentin
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Svyatoslav
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    This answer does not use the condition specified in the question, which concerned the OP, namely the convergence of $$\int_1^\infty\frac {f(u)}u\mathrm du.$$ Instead, the answer implicitly uses a different condition: that $f(u)\to0$ as $u\to\infty$. Neither condition implies the other. The original condition can be used by modifying your proof: Rather than applying the MVT to the subtracted integral, just keep the integral in its original form, and let $\Delta\to\infty$. – John Bentin Feb 28 '24 at 11:27
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    @John Bentin The answer uses the condition that $\lim_{x\to\infty}f(x)$ exists (not necessary equal 0). I wrote in the comment (please, see above) that we can prove the Frullani formula for continuous functions (not necessarily $C^1$). But you are right: the OP conditions are different. Thank you for your comment. – Svyatoslav Feb 28 '24 at 11:48
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    If $\lim_{x\to\infty}f(x)$ exists but is nonzero, then $f(c_2)$ will also not go to zero in the limit, and your result will be contrary to the required result. What I wrote was correct, but I won't argue about it any more. – John Bentin Feb 28 '24 at 17:27