Frullani integral

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

$\int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\,{\rm {d}}x$

where $f$ is a function defined for all non-negative real numbers that has a limit at $\infty$, which we denote by $f(\infty )$.

The following formula for their general solution holds if $f$ is continuous on $(0,\infty )$, has finite limit at $\infty$, and $a,b>0$:

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

If $f(\infty )$ does not exist, but $\int _{c}^{\infty }{\frac {f(x)}{x}}dx$ exists for some $c>0$, then $$\int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\,{\rm {d}}x=-f(0)\ln {\frac {a}{b}}.$$

Proof for continuously differentiable functions

A simple proof of the formula (under stronger assumptions than those stated above, namely $f\in {\mathcal {C}}^{1}(0,\infty )$) can be arrived at by using the Fundamental theorem of calculus to express the integrand as an integral of $f'(xt)={\frac {\partial }{\partial t}}\left({\frac {f(xt)}{x}}\right)$:

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

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

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

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

Ramanujan's generalization

Ramanujan, using his master theorem, gave the following generalization.

Let $f,g$ be functions continuous on $[0,\infty ]$.$$f(x)-f(\infty )=\sum _{k=0}^{\infty }{\frac {u(k)(-x)^{k}}{k!}}\quad {\text{ and }}\quad g(x)-g(\infty )=\sum _{k=0}^{\infty }{\frac {v(k)(-x)^{k}}{k!}}$$Let ${\textstyle u(x)}$ and ${\textstyle v(x)}$ be given as above, and assume that ${\textstyle f}$ and ${\textstyle g}$ are continuous functions on ${\textstyle [0,\infty )}$. Also assume that ${\textstyle f(0)=g(0)}$ and ${\textstyle f(\infty )=g(\infty )}$. Then, if ${\textstyle a,b>0}$,

$$\lim _{n\rightarrow 0}\int _{0}^{\infty }x^{n-1}\{f(ax)-g(bx)\}dx=\{f(0)-f(\infty )\}\left\{\log \left({\frac {b}{a}}\right)+{\frac {d}{ds}}\left(\log \left({\frac {v(s)}{u(s)}}\right)\right)_{s=0}\right\}$$

Applications

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

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

The formula can also be generalized in several different ways.

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.