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 $$a,b > 0$$:


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

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{align}

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

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{align}$$ The third lemma arises from the fundamental theorem of calculus.
 * $$\begin{align}

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

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.
 * $$\int^a_b \frac{1}{t} \left( \int^{\infty}_0 \frac{f(tx)}{\partial x} dx \right) \ dt $$

Substitute using lemma 3.
 * $$\int^a_b \frac{1}{t} (f(\infty)-f(0)) \ dt $$

Place one factor outside the integral.
 * $$(f(\infty)-f(0)) \int^a_b \frac{1}{t} \ dt $$

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.