Frullani integral

Short description: Type of improper integral with general solution

In mathematics, Frullani integrals are a specific type of improper integral named after the Italian mathematician Giuliano Frullani. The integrals are of the form

[math]\displaystyle{ \int _{0}^{\infty}{\frac {f(ax)-f(bx)}{x}}\,{\rm {d}}x }[/math]

where [math]\displaystyle{ f }[/math] is a function defined for all non-negative real numbers that has a limit at [math]\displaystyle{ \infty }[/math], which we denote by [math]\displaystyle{ f(\infty) }[/math].

The following formula for their general solution holds under certain conditions:

[math]\displaystyle{ \int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\,{\rm {d}}x=\Big(f(\infty)-f(0)\Big)\ln {\frac {a}{b}}. }[/math]

Proof

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 [math]\displaystyle{ f'(xt) = \frac{\partial }{\partial t} \left(\frac{f(xt)}{x}\right) }[/math]:

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

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

[math]\displaystyle{ \begin{align} \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{align} }[/math]

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

Applications

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

[math]\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) }[/math]

The formula can also be generalized in several different ways.

References

• 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.