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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{f(ax)-f(bx)}{x}\mathrm{d}x$

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

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{f(ax)-f(bx)}{x}\mathrm{d}x=(f(\mathrm{\infty })-f(0))\ln \frac{a}{b}.$

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

Proof for continuously differentiable functions

A simple proof of the formula (under stronger assumptions than those stated above, namely $f\in {{𝒞}}^1(0,\mathrm{\infty })$) can be arrived at by using the Fundamental theorem of calculus to express the integrand as an integral of $f^{\prime }(xt)=\frac{\partial }{\partial t}\left(\frac{f(xt)}{x}\right)$:

$\begin{array}{rl}\frac{f(ax)-f(bx)}{x}& ={\left[\frac{f(xt)}{x}\right]}_{t=b}^{t=a}\\ & =\underset{b}{\overset{a}{\int }}{f}^{\prime }(xt)dt\\ \end{array}$

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

$\begin{array}{rl}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{f(ax)-f(bx)}{x}dx& =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{b}{\overset{a}{\int }}{f}^{\prime }(xt)dtdx\\ & =\underset{b}{\overset{a}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{f}^{\prime }(xt)dxdt\\ & =\underset{b}{\overset{a}{\int }}{\left[\frac{f(xt)}{t}\right]}_{x=0}^{x\to \mathrm{\infty }}dt\\ & =\underset{b}{\overset{a}{\int }}\frac{f(\mathrm{\infty })-f(0)}{t}dt\\ & =(f(\mathrm{\infty })-f(0))(\ln (a)-\ln (b))\\ & =(f(\mathrm{\infty })-f(0))\ln \left(\frac{a}{b}\right)\\ \end{array}$

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

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

Applications

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-x}-e^{-bx}}{x}\mathrm{d}x=(\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{e^n}-e^0)\ln (\frac{1}{b})=\ln (b)$

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.