User:TMM53/Frullani integral-2024-05-26

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 $a,b>0$:

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

Proof for continuously differentiable functions

Lemmas

This is a simple proof of the formula under stronger assumptions than the prior assumption $f\in {\mathcal {C}}^{1}(0,\infty )$. The first lemma arises from the Fundamental theorem of calculus.

${\begin{aligned}{\frac {f(ax)-f(bx)}{x}}&=\left[{\frac {f(xt)}{x}}\right]_{t=b}^{t=a}\,\\&=\int _{b}^{a}{\frac {1}{x}}{\frac {\partial f(xt)}{\partial t}}\,dt\\\end{aligned}}$

Lemma 1

${\frac {f(ax)-f(bx)}{x}}=\int _{b}^{a}{\frac {1}{x}}{\frac {\partial f(xt)}{\partial t}}\,dt$

The second lemma relates the partial derivatives involving variables ${\textstyle t}$ and ${\textstyle x}$ using the chain rule.

${\begin{aligned}u&=tx,\ f(u)=f(tx)\\{\frac {1}{x}}{\frac {f(tx)}{\partial t}}&={\frac {1}{x}}{\frac {\partial f(u)}{\partial u}}{\frac {\partial u}{\partial t}}\\&={\frac {1}{x}}{\frac {\partial f(u)}{\partial u}}\ x\\&={\frac {\partial f(u)}{\partial u}}\\{\frac {1}{t}}{\frac {f(tx)}{\partial x}}&={\frac {1}{t}}{\frac {\partial f(u)}{\partial u}}{\frac {\partial u}{\partial x}}\\&={\frac {1}{t}}{\frac {\partial f(u)}{\partial u}}\ t\\&={\frac {\partial f(u)}{\partial u}}\end{aligned}}$

Lemma 2

${\frac {1}{x}}{\frac {f(tx)}{\partial t}}={\frac {1}{t}}{\frac {f(tx)}{\partial x}}$

The third lemma arises from the fundamental theorem of calculus.

${\begin{aligned}\int _{0}^{\infty }{\frac {\partial f(xt)}{\partial x}}\ dx&=\left[{\underset {}{f(xt)}}\right]_{x=0}^{x\to \infty }\\&=f(\infty )-f(0)\\\end{aligned}}$

Lemma 3

$\int _{0}^{\infty }{\frac {\partial f(xt)}{\partial x}}\ dx=f(\infty )-f(0)$

Proof

Begin with the integral.

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

Substitute using lemma 1.

$\int _{0}^{\infty }\int _{b}^{a}{\frac {1}{x}}{\frac {\partial f(xt)}{\partial t}}\ dt\ dx$

Substitute using lemma 2.

$\int _{0}^{\infty }\int _{b}^{a}{\frac {1}{t}}{\frac {f(tx)}{\partial x}}\ dt\ dx$

Use Tonelli’s theorem to interchange the two integrals.

$\int _{b}^{a}\int _{0}^{\infty }{\frac {1}{t}}{\frac {f(tx)}{\partial x}}dx\ dt$

Place the integral in parentheses.

$\int _{b}^{a}{\frac {1}{t}}\left(\int _{0}^{\infty }{\frac {f(tx)}{\partial x}}dx\right)\ dt$

Substitute using lemma 3.

$\int _{b}^{a}{\frac {1}{t}}(f(\infty )-f(0))\ dt$

Place one factor outside the integral.

$(f(\infty )-f(0))\int _{b}^{a}{\frac {1}{t}}\ dt$

Apply the logarithm integration formula.

$(f(\infty )-f(0))(\operatorname {ln} (a)-\operatorname {ln} (b))$

Rewrite the logarithm expression.

$(f(\infty )-f(0))\operatorname {ln} \left({\frac {a}{b}}\right)$

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.