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After viewing some topics on MSE concerning Frullani's integral I ask for a generalized version . I think it's too hard to get the result so I ask just for the convergence :

Let $f(x)$ be a continuous and differentiable function then determine when : $$\int_{0}^{\infty}\frac{f(ag(x))-f(bg(x))}{x}dx$$ Converges where $a>b>0$

I think it converges if $$\int_{0}^{\infty}\frac{f(g(x))}{x}dx$$ Converges but I cannot prove it since I have not the tool for .

Any help is appreciated .

Thanks a lot for all your contributions (and you) .

Mark Viola
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Barackouda
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  • What do you mean precisely by the phrase "then determine when." For example, it converges when $a=b$. It converges when $f(x)=x$. Are you asking for necessary conditions or sufficient conditions? What are you really asking here? – Mark Viola Jun 22 '20 at 16:51
  • @MarkViola I ask for necessary and sufficient condition such that the first integral converges . Is it clear ?Thanks . – Barackouda Jun 22 '20 at 17:04
  • Do you believe that Frullani's Theorem has both necessary and sufficient conditions? If so, what do you believe that those conditions are? For example, $\int_0^\infty \frac{\cos(x)}{x},dx$ fails to exist. And $\lim_{x\to\infty}\cos(x)$ fails to exist. But, $$\int_0^\infty \frac{\cos(ax)-\cos(bx)}{x},dx=\log(b/a)$$ – Mark Viola Jun 22 '20 at 17:06
  • And $\int_0^\infty \frac{\sin(x)}{x},dx$ exists, but $\int_0^\infty \frac{\sin(ax)-\sin(bx)}{x},dx=0\ne \log(b/a)$ unless $a=b$. So, from the previous comment, the convergence of $\int_0^\infty \frac{f(x)}{x},dx$ was not necessary and here we see that it is not sufficient. – Mark Viola Jun 22 '20 at 17:15
  • @MarkViola Mouarf You're right can you find a necessary condition ?Thanks for the down-vote (lol). – Barackouda Jun 22 '20 at 17:16
  • Who or what is "Mouarf?" – Mark Viola Jun 22 '20 at 17:17
  • @MarkViola It's an onomatopoeia to say you're right – Barackouda Jun 22 '20 at 17:19
  • So, we have determined that in the "standard" Frullani integral the condition on convergence of $\int_0^\infty \frac{f(x)}{x},dx$ is neither necessary or sufficient. That is to say, we need more conditions for the theorem to be valid. – Mark Viola Jun 22 '20 at 17:23
  • And I had intended to up vote, but inadvertently hit the down vote. I had to edit your question to set things right. ;-) – Mark Viola Jun 22 '20 at 17:25

1 Answers1

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We first cite the following theorem:[1]

Theorem. (Agnew, 1951) Let $f : (0, \infty) \to \mathbb{R}$ be locally integrable. Then the followings are equivalent:

(1) $\int_{0}^{\infty} \frac{f(at) - f(bt)}{t} \, \mathrm{d}t$ converges as an improper integral for each $\lambda = \log(a/b)$ in a set of positive measure.

(2) Both $\lim_{x \to \infty} \frac{1}{x}\int_{1}^{x} f(t) \, \mathrm{d}t$ and $\lim_{x \to 0^+} x \int_{x}^{1} \frac{f(t)}{t^2} \, \mathrm{d}t$ converge.

(3) Both $\lim_{x \to \infty} \frac{1}{x}\int_{0}^{x} f(t) \, \mathrm{d}t$ and $\lim_{x \to 0^+} \frac{1}{x}\int_{0}^{x} f(t) \, \mathrm{d}t$ converge.

Moreover, in such case, we have

\begin{align*} C_{\infty} &: = \lim_{x \to \infty} \frac{1}{x}\int_{0}^{x} f(t) \, \mathrm{d}t = \lim_{x \to \infty} \frac{1}{x}\int_{1}^{x} f(t) \, \mathrm{d}t, \\ C_{0} &: = \lim_{x \to 0^+} \frac{1}{x}\int_{0}^{x} f(t) \, \mathrm{d}t = \lim_{x \to 0^+} x \int_{x}^{1} \frac{f(t)}{t^2} \, \mathrm{d}t \end{align*} ad well as $$\int_{0}^{\infty} \frac{f(at) - f(bt)}{t} \, \mathrm{d}t = \lambda (C_{\infty} - C_0)$$ for each pair of positive numbers $a$ and $b$.

So the convergence of the standard Frullani's integral is intimately related to the Cesàro means near $0$ and $\infty$. The proof essentially hinges on the property of the Haar measure $x^{-1} \, \mathrm{d}x$ on the multiplicative group $(0, \infty)$. That being said, it seems unlikely to me that there exists a useful necessary condition for the existence of OP's generalized Frullani's integral.


Still, we may try to produce some sufficient conditions: To make the argument simple, we impose the following assumptions:

  • $f \in C([0,\infty])$, meaning that that $f$ is continuous on $[0, \infty)$ and $f(\infty):=\lim_{x\to\infty}f(x)$ converges.

  • $g : [0, \infty) \to [0, \infty)$ is a strictly increasing, continuous bijection with the inverse $h = g^{-1}$.

Then for any $a, b > 0$ and $0 < r < R$,

\begin{align*} &\int_{r}^{R} \frac{f(ag(x)) - f(bg(x))}{x} \, \mathrm{d}x \\ &= \int_{r}^{R} \frac{f(ag(x))}{x} \, \mathrm{d}x - \int_{r}^{R} \frac{f(bg(x))}{x} \, \mathrm{d}x \\ &= \int_{ag(r)}^{ag(R)} f(u) \, \mathrm{d}\log h(u/a) - \int_{bg(r)}^{bg(R)} f(u) \, \mathrm{d}\log h(u/b) \\ &= \int_{ag(r)}^{bg(r)} f(u) \, \mathrm{d}\log h(u/a) - \int_{ag(R)}^{bg(R)} f(u) \, \mathrm{d}\log h(u/b) \\ &\quad + \int_{bg(r)}^{ag(R)} f(u) \, \mathrm{d}\log \left( \frac{h(u/a)}{h(u/b)} \right). \end{align*}

Using the continuity of $f$, we may provide a sufficient condition for which the above expression converges as $r \to 0^+$ and $R \to \infty$ for arbitrary $a, b > 0$:

Condition. For any $c > 0$, each of the following converges:

$$ \lim_{r \to 0^+} \frac{h(cr)}{h(r)}, \qquad \lim_{R \to 0^+} \frac{h(cR)}{h(R)}, \qquad \int_{0}^{\infty} \left| \mathrm{d}\log\left(\frac{h(cu)}{h(u)} \right) \right| $$

Indeed, if this condition holds, then Karamata's characterization theorem for regularly varying functions tells that there exist $p, q \geq 0$ satisfying

$$ \lim_{r \to 0^+} \frac{h(cr)}{h(r)} = c^{p} \qquad \text{and} \qquad \lim_{R \to 0^+} \frac{h(cR)}{h(R)} = c^{q} $$

for all $c > 0$. Then it is not hard to prove that

\begin{align*} \int_{0}^{\infty} \frac{f(ag(x)) - f(bg(x))}{x} \, \mathrm{d}x &= (q f(\infty) - p f(0)) \log ( a/b ) + \int_{0}^{\infty} f(u) \, \mathrm{d}\log \left( \frac{h(u/a)}{h(u/b)} \right). \end{align*}

  • Example 1. If $g(x) = x^{d}$ for some $d > 0$, then $p = q = \frac{1}{d}$ and $\mathrm{d}\log \left( \frac{h(u/a)}{h(u/b)} \right) = 0$. So we get \begin{align*} \int_{0}^{\infty} \frac{f(ax^{d}) - f(bx^{d})}{x} \, \mathrm{d}x &= \frac{(f(\infty) - f(0)) \log ( a/b )}{d}. \end{align*} This can also be proved directly from the standard Frullani's integral applied to $x \mapsto f(x^d)$.

  • Example 2. If $g(x) = \frac{x+\sqrt{x^2+4x}}{2}$, then $h(u) = \frac{u^2}{u+1}$, and so, $p = 2$ and $q = 1$. So we get \begin{align*} \int_{0}^{\infty} \frac{f(ag(x)) - f(bg(x))}{x} \, \mathrm{d}x &= (f(\infty) - 2f(0)) \log ( a/b ) + \int_{0}^{\infty} \frac{(a-b)f(u)}{(u+a)(u+b)} \, \mathrm{d}u. \end{align*} As a sanity check, plugging $f \equiv 1$ shows that both sides are zero.


References.

  1. Agnew, Ralph Palmer. "Mean Values and Frullani Integrals." Proceedings of the American Mathematical Society 2, no. 2 (1951): 237-41. Accessed June 22, 2020. doi:10.2307/2032493.
Sangchul Lee
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  • It exceeds my hope and my level in mathematics for this nice and complete answer . You speak about the Cesàro means Looks it like a discrete analogue ? Thanks to learn me something good !(+1) – Barackouda Jun 23 '20 at 08:13