# An Extended Frullani Integral

In the development of Methodology $$2$$ of This Answer, I found a possible new extension of Frullani's Integral (See Here).

Theorem: Let $$f$$ be Riemann integrable on $$[0,x]$$ for all $$x>0$$ and let $$a>0$$ and $$b>0$$. Furthermore, let $$F(x)=\int_0^x f(t)\,dt$$ denote an antiderivative of $$f(t)$$ and $$\bar F(x)=\frac1xF(x)$$ be the average value of $$f$$ on the interval $$[0,x]$$.

If the integral $$\int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx$$ exists and $$\lim_{x\to\infty}\bar F(x)=\bar F_\infty$$ and $$\lim_{x\to0^+}\bar F(x)= F'(0)$$ exist and are finite, then

$$\int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx=(F'(0)-\bar F_\infty)\log(b/a)\tag1$$

Proof:

Integrating by parts the integral on the left-hand side of $$(1)$$ with $$u=\frac1x$$ and $$v=\frac1aF(ax)-\frac1bF(bx)$$, we obtain

$$\int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx=\int_0^\infty \frac{\bar F(ax)-\bar F(bx)}{x}\,dx\tag2$$

Note that $$(2)$$ is a standard Frullani integral and the result in $$(1)$$ follows immediately.

Example Applications:

Alongside the integral in This Answer, namley $$\int_0^\infty \frac{|\sin(\sqrt{qx})|-|\sin(\sqrt{px})|}{x}\,dx=\frac2\pi \log(q/p)$$, we can use $$(1)$$ to evaluate the companion integral

$$\int_0^\infty \frac{|\cos(\sqrt{qx})|-|\cos(\sqrt{px})|}{x}\,dx=\left(1-\frac2\pi\right)\log(p/q)$$

QUESTIONS:

Is there a reference to the Theorem herein or is this a new result? If so, please advise?

Alongside the two examples I cited, are there other non-standard Frullani integrals to which this result would not only apply, but facilitate evaluation?

Can the conditions of the theorem be relaxed?

Can the result be generalized further? For example, in This Answer, I generalized Frullani for complex $$a$$ and $$b$$, but I required $$f$$ to be analytic.

• Hi, glad to see that you are still active in this trying time! As for the first question, your condition turns out to be essentially an equivalence condition for the Frullani's integral to exists. I described the relevant theorem and its reference in this answer. – Sangchul Lee Jul 10 at 18:37
• @SangchulLee Hi Sangchul. It's great to see you're here also. I've had a cursory look at your posted solution and it is strikingly similar. Herein, integration by parts seems to be a trivial way to prove the "extended theorem." It is of interest to note that the equivalence of existence does not require $f$ to be continuous, only Riemann integrable. Finally, the question I posed on applications of the extended theorem is still of interest to me. I gave two examples and would love to see others that are not merely contrived, but are motivated from something else. – Mark Viola Jul 10 at 18:47
• It is also surprising that the convergence of 1-Cesaro integral means near $0$ and $\infty$ is both necessary ans sufficient, because I initially expected that higher-order Cesaro means would work as well. I would also love to see more live examples of this general version. – Sangchul Lee Jul 10 at 19:12
• Thanks Sangchul for all of your input. As always, it is a true pleasure reading your posts. – Mark Viola Jul 10 at 19:32