A simple proof of the formula can be arrived at by using the Fundamental theorem of calculus to express the integrand as an integral of ${\displaystyle f'(xt)={\frac {\partial }{\partial t}}\left({\frac {f(xt)}{x}}\right)}$ :

${\displaystyle {\begin{aligned}{\frac {f(ax)-f(bx)}{x}}&=\left[{\frac {f(xt)}{x}}\right]_{t=b}^{t=a}\,\\&=\int _{b}^{a}f'(xt)\,dt\\\end{aligned}}}$ 

and then use Tonelli’s theorem to interchange the two integrals:

${\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\,dx&=\int _{0}^{\infty }\int _{b}^{a}f'(xt)\,dt\,dx\\&=\int _{b}^{a}\int _{0}^{\infty }f'(xt)\,dx\,dt\\&=\int _{b}^{a}\left[{\frac {f(xt)}{t}}\right]_{x=0}^{x\to \infty }\,dt\\&=\int _{b}^{a}{\frac {f(\infty )-f(0)}{t}}\,dt\\&={\Big (}f(\infty )-f(0){\Big )}{\Big (}\ln(a)-\ln(b){\Big )}\\&={\Big (}f(\infty )-f(0){\Big )}\ln {\Big (}{\frac {a}{b}}{\Big )}\\\end{aligned}}}$ 

Note that the integral in the second line above has been taken over the interval ${\displaystyle [b,a]}$ , not ${\displaystyle [a,b]}$ .

The formula can be used to derive an integral representation for the natural logarithm ${\displaystyle \ln(x)}$  by letting ${\displaystyle f(x)=e^{-x}}$  and ${\displaystyle a=1}$ :

${\displaystyle {\int _{0}^{\infty }{\frac {e^{-x}-e^{-bx}}{x}}\,{\rm {d}}x={\Big (}\lim _{n\to \infty }{\frac {1}{e^{n}}}-e^{0}{\Big )}\ln {\Big (}{\frac {1}{b}}}{\Big )}=\ln(b)}$ 

The formula can also be generalized in several different ways.^[1]

• G. Boros, Victor Hugo Moll, Irresistible Integrals (2004), pp. 98
• Juan Arias-de-Reyna, On the Theorem of Frullani (PDF; 884 kB), Proc. A.M.S. 109 (1990), 165-175.
• ProofWiki, proof of Frullani's integral.